circuits and analog electronics 电路与模拟电子技术

70
Circuits and Analog Circuits and Analog Electronics Electronics 电电电电电电电电电 电电电电电电电电电 Prof. Li Chen, School of Information Science and Technology, Sun Yat-sen University 电电电电电电电电电电电电电 电电电电电 Email: [email protected] 10 10 电电电电电电 电电电电电电 2 2 2 2

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10 级计算机科学 2 + 2. Circuits and Analog Electronics 电路与模拟电子技术. Prof. Li Chen, School of Information Science and Technology, Sun Yat-sen University 中山大学信息科学与技术学院 陈立副教授 Email: [email protected]. Circuits and Analog Electronics. References : - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Circuits and Analog Electronics 电路与模拟电子技术

Circuits and Analog ElectronicsCircuits and Analog Electronics

电路与模拟电子技术电路与模拟电子技术

Prof Li Chen School of Information Science and Technology Sun Yat-sen University

中山大学信息科学与技术学院 陈立副教授

Email chenli55mailsysueducn

10 10 级计算机科学 级计算机科学 22 ++ 22

References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering

Circuit Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7

bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3

bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7

Circuits and Analog ElectronicsCircuits and Analog Electronics

Handouts available at

sistsysueducn~chenli

References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering Circuit

Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7

bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3

bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7

Circuits and Analog ElectronicsCircuits and Analog Electronics

Weeks Chapters References

1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5

3 4 Basis analysis methods to circuits Hayt Ch 3 4

5 Basis RL amp RC circuits Hayt Ch 6

6 7 8 Sinusoidal steady state analysis Hayt Ch 7

9 Midterm

10 Diodes and diodes circuits Boylestad Ch 1 2

11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6

14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11

17 Review

Teaching ScheduleTeaching Schedule

Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits

11 Basic Concepts and Electric Circuits

12 Basic Quantities

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

References Hayt Ch1 2 5 Gao Ch1

Circuits and Analog ElectronicsCircuits and Analog Electronics

Signal processing and transmissiontransmission

Amplifiers

11 Basic Concepts and Electric Circuits

Electrical powerElectrical power conversion and transmissiontransmission

Power Supplies

TransmissionTransmission Loads

Circuits KinescopeAntenna

Speakertransmitter

11 Basic Concepts and Electric Circuits Electrical power conversion and transmission

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 2: Circuits and Analog Electronics 电路与模拟电子技术

References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering

Circuit Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7

bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3

bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7

Circuits and Analog ElectronicsCircuits and Analog Electronics

Handouts available at

sistsysueducn~chenli

References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering Circuit

Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7

bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3

bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7

Circuits and Analog ElectronicsCircuits and Analog Electronics

Weeks Chapters References

1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5

3 4 Basis analysis methods to circuits Hayt Ch 3 4

5 Basis RL amp RC circuits Hayt Ch 6

6 7 8 Sinusoidal steady state analysis Hayt Ch 7

9 Midterm

10 Diodes and diodes circuits Boylestad Ch 1 2

11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6

14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11

17 Review

Teaching ScheduleTeaching Schedule

Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits

11 Basic Concepts and Electric Circuits

12 Basic Quantities

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

References Hayt Ch1 2 5 Gao Ch1

Circuits and Analog ElectronicsCircuits and Analog Electronics

Signal processing and transmissiontransmission

Amplifiers

11 Basic Concepts and Electric Circuits

Electrical powerElectrical power conversion and transmissiontransmission

Power Supplies

TransmissionTransmission Loads

Circuits KinescopeAntenna

Speakertransmitter

11 Basic Concepts and Electric Circuits Electrical power conversion and transmission

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 3: Circuits and Analog Electronics 电路与模拟电子技术

Handouts available at

sistsysueducn~chenli

References bull W H Hayt Jr J E Kemmerly and S M Durbin Engineering Circuit

Analysis McGraw-Hill 2005 ISBN 978-7-121-01667-7

bull R L Boylestad and L Nashelsky Electronic Devices and Circuit Theory Pearson Education 2007 ISBN 978-7-121-04396-3

bull 高玉良 电路与模拟电子技术 高教出版社 2004 ISBN 7-04-014536-7

Circuits and Analog ElectronicsCircuits and Analog Electronics

Weeks Chapters References

1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5

3 4 Basis analysis methods to circuits Hayt Ch 3 4

5 Basis RL amp RC circuits Hayt Ch 6

6 7 8 Sinusoidal steady state analysis Hayt Ch 7

9 Midterm

10 Diodes and diodes circuits Boylestad Ch 1 2

11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6

14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11

17 Review

Teaching ScheduleTeaching Schedule

Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits

11 Basic Concepts and Electric Circuits

12 Basic Quantities

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

References Hayt Ch1 2 5 Gao Ch1

Circuits and Analog ElectronicsCircuits and Analog Electronics

Signal processing and transmissiontransmission

Amplifiers

11 Basic Concepts and Electric Circuits

Electrical powerElectrical power conversion and transmissiontransmission

Power Supplies

TransmissionTransmission Loads

Circuits KinescopeAntenna

Speakertransmitter

11 Basic Concepts and Electric Circuits Electrical power conversion and transmission

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 4: Circuits and Analog Electronics 电路与模拟电子技术

Weeks Chapters References

1 2 Basis concepts and laws of electronics Hayt Ch 1 2 5

3 4 Basis analysis methods to circuits Hayt Ch 3 4

5 Basis RL amp RC circuits Hayt Ch 6

6 7 8 Sinusoidal steady state analysis Hayt Ch 7

9 Midterm

10 Diodes and diodes circuits Boylestad Ch 1 2

11 12 13 Basic BJT amplifier circuits Boylestad Ch 3-6

14 15 16 Operational amplifier and Op Amp circuits Boylestad Ch 11

17 Review

Teaching ScheduleTeaching Schedule

Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits

11 Basic Concepts and Electric Circuits

12 Basic Quantities

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

References Hayt Ch1 2 5 Gao Ch1

Circuits and Analog ElectronicsCircuits and Analog Electronics

Signal processing and transmissiontransmission

Amplifiers

11 Basic Concepts and Electric Circuits

Electrical powerElectrical power conversion and transmissiontransmission

Power Supplies

TransmissionTransmission Loads

Circuits KinescopeAntenna

Speakertransmitter

11 Basic Concepts and Electric Circuits Electrical power conversion and transmission

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 5: Circuits and Analog Electronics 电路与模拟电子技术

Ch1 Ch1 Basic Concepts and Laws of Electric Basic Concepts and Laws of Electric CircuitsCircuits

11 Basic Concepts and Electric Circuits

12 Basic Quantities

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

References Hayt Ch1 2 5 Gao Ch1

Circuits and Analog ElectronicsCircuits and Analog Electronics

Signal processing and transmissiontransmission

Amplifiers

11 Basic Concepts and Electric Circuits

Electrical powerElectrical power conversion and transmissiontransmission

Power Supplies

TransmissionTransmission Loads

Circuits KinescopeAntenna

Speakertransmitter

11 Basic Concepts and Electric Circuits Electrical power conversion and transmission

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 6: Circuits and Analog Electronics 电路与模拟电子技术

Signal processing and transmissiontransmission

Amplifiers

11 Basic Concepts and Electric Circuits

Electrical powerElectrical power conversion and transmissiontransmission

Power Supplies

TransmissionTransmission Loads

Circuits KinescopeAntenna

Speakertransmitter

11 Basic Concepts and Electric Circuits Electrical power conversion and transmission

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 7: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits Electrical power conversion and transmission

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 8: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

Question What is the current through the bulb

Concept of Abstraction

Solution

In order to calculate the current we can replace the bulb with a resistor

R is the only subject of interest which serves as an abstraction of the bulb

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 9: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

Lumped circuit abstraction

bull A resistor is a circuit element that transforms the electrical energy (eg electricity heat)

bull Commonly used devices that are modeled as resistors include incandescent heaters wires and etc

bull A circuit consists of sources resistors capacitors inductors and conductors

bull Elements are lumped

bull Conductors are perfect

Resistance R = VI 1 =1VA ohm

Conductance G = 1R = 1AV siemens (S)

1S = 1AV i(t) = G times v(t) Instantaneous current and voltage at time t

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 10: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

Understanding the AM radio requires knowledge of several concepts

bull Communicationssignal processing (frequency domain analysis)

bull Electromagnetics (antennas high-frequency circuits)

bull Power (batteries power supplies)

bull Solid state (miniaturization low-power electronics)

The AM Radio SystemThe AM Radio System

Transmitter Receiver

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 11: Circuits and Analog Electronics 电路与模拟电子技术

Example 1 The AM audio system

Example 2 The telephone system

11 Basic Concepts and Electric Circuits

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 12: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System A signal is a quantity that may vary with time

Voltage or current in a circuit

Sound (sinusoidal wave traveling through air)

Light or radio waves (electromagnetic energy traveling through free space)

The analysis and design of AM radios (and communication systems in general) is usually conducted in the frequency domain using Fourier analysis which allows us to represent signals as combinations of sinusoids (sines and cosines)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 13: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Frequency is the rate at which a signal oscillatesDuration of the signal T frequency of the signal f = 1T

High Frequency Low Frequency

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 14: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Visible light is the electromagnetic energy with frequency between 380THz (Terahertz) and 860THz Our visual system perceives the frequency of the electromagnetic energy as color is 460THz is 570THz and is 630THz An AM radio signal has a frequency of between 500kHz and 18MHz

FM radio and TV uses different frequencies

Mathematical analysis of signals in terms of frequency

Most commonly encountered signals can be represented as a Fourier series or a Fourier transform A Fourier series is a weighted sum of cosines and sines

red green blue

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 15: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemFourier Series A Fourier series decomposes a periodic function (or signal) into the sum of a set of sines and cosines Given function f(t) with angular frequency ω and period T its Fourier series can be written as

f(t) = A0 + A1msin(ωt + ψ1) + A2msin(2ωt + ψ2) +

=

10

1 10

10

cossin

sincoscossin

)sin(

kkmkm

k kkkmkkm

kkkm

tkCtkBA

tkAtkAA

tkAA

0 0

0

0

1

2sin

2cos

T

T

km

T

km

A f t dtT

B f t k tdtT

B f t k tdtT

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 16: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

21

01)(

t

ttfExample Given function during a period

2 3 t

1

)12sin(12

14]5sin

5

13sin

3

1[sin

4)(

l

tll

ttttf

For the example 2 2

0 0 0

1 1 11 1 0

2 2 2A f t d t d t d t

2 2

0 0

00

1 1cos 1 cos 1 cos

2 2 cos sin 0

kmC f t k td t k td t k td t

k td t k tk

2 2

0 0

00 40

1 1sin 1 sin 1 sin

2 2 2 sin cos 1 cos

km

k

B f t k td t k td t k td t

k td t k t kk k

k is even

k is odd

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 17: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Example-Fourier SeriesExample-Fourier Series

基波

3次谐波

基波+3 次谐波

bull Signals can be represented in terms of their frequency components

bull The AM transmitter and receiver are analyzed in terms of their effects on the frequency components signals

1st series + 3rd series

1st series (k = 1)

3rd series (k = 3)

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 18: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

The modulator converts the frequency of the input signal from the audio range (0-5kHz) to the carrier frequency of the station (ie 605kHz-615kHz)

freq5kHz

Frequency domain representation of input

Frequency domain representation of output

freq610kHz

ModulatorModulator

Signal

SourceModulator

Power

Amplifier

Antenna

Transmitter Block DiagramTransmitter Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 19: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Input Signal

Output Signal

Modulator Time DomainModulator Time Domain

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 20: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull A typical AM station broadcasts several kWndash Up to 50kW-Class I or Class II stationsndash Up to 5kW-Class III stationndash Up to 1kW-Class IV station

bull Typical modulator circuit can provide at most a few mWbull Power amplifier takes modulator output and increases its magnitude

Power AmplifierPower Amplifier

The antenna converts a current or a voltage signal to an electromagnetic signal which is radiated through the space

AntennaAntenna

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 21: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

Audio

Amplifier

Antenna

Speaker

Receiver Block DiagramReceiver Block Diagram

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 22: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull The antenna captures electromagnetic energy and converts it to a small voltage or current

bull In the frequency domain the antenna output is

0 frequency

Undesired SignalsDesired Signal

Carrier Frequencyof desired station

AntennaAntenna

interferences interferences

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 23: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull RF Amplifier amplifies small signals from the antenna to voltage levels appropriate for transistor circuits

bull RF Amplifier also performs as a Bandpass filter for the signal

ndash Bandpass filter attenuates the other components outside the frequency range that contains the desired station

RF (Radio Frequency) AmplifierRF (Radio Frequency) Amplifier

0 frequency

Undesired Signals

Desired Signal

Carrier Frequency of desired station

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 24: Circuits and Analog Electronics 电路与模拟电子技术

The AM Radio SystemThe AM Radio System

0 frequency

Undesired Signals

Desired Signal

455 kHz

IF (Intermediate Frequency) MixerIF (Intermediate Frequency) Mixerbull The IF Mixer shifts its input in the frequency domain from the carrier

frequency to an intermediate frequency of 455kHz

bull The IF amplifier bandpass filters the output of the IF mixer eliminating all of the undesired signals

IF AmplifierIF Amplifier

0 frequency

Desired Signal

455 kHz

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 25: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

bull Computes the envelope of its input signal

Envelope DetectorEnvelope Detector

Output Signal

Input Signal

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
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Page 26: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemAudio AmplifierAudio Amplifier

bull Amplifies signal from envelope detector

bull Provides power to drive the speaker

Hierarchical System ModelsHierarchical System Modelsbull Modelling at different levels of abstraction

bull Higher levels of the model describe overall function of the system

bull Lower levels of the model describe necessary details to implement the system

bull In the AM receiver the input is the antenna voltage and the output is the sound energy produced by the speaker

bull In EE a system is an electrical andor mechanical device a process or a mathematical model that relates one or more inputs to one or more outputs

SystemInputs Outputs

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 27: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio SystemTop Level ModelTop Level Model

AM ReceiverInput Signal Sound

Second Level ModelSecond Level Model

RFAmplifier

IFMixer

IFAmplifier

EnvelopeDetector

AudioAmplifier

Antenna

Speaker

Power Supply

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 28: Circuits and Analog Electronics 电路与模拟电子技术

11 Basic Concepts and Electric Circuits

The AM Radio SystemThe AM Radio System

Half-waveRectifier

Low-passFilter

Low Level Model Envelope DetectorLow Level Model Envelope Detector

Circuit Level Model Envelope DetectorCircuit Level Model Envelope Detector

+

-R C

+

-VoutVin

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 29: Circuits and Analog Electronics 电路与模拟电子技术

12 Basic Quantities

UnitsUnitsbull Standard SI Prefixes

ndash 10-12 pico (p)

ndash 10-9 nano (n)

ndash 10-6 micro ()

ndash 10-3 milli (m)

ndash 103 kilo (k)

ndash 106 mega (M)

ndash 109 giga (G)

ndash 1012 tera (T)

bull Electric charge (q)

ndash in Coulombs (C)

bull Current (I)

ndash in Amperes (A)

bull Voltage (V)

ndash in Volts (V)

bull Energy (W)

ndash in Joules (J)

bull Power (P)

ndash in Watts (W)

I t q

VI

R

IR V

W qV Pt V I t

P VI

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 30: Circuits and Analog Electronics 电路与模拟电子技术

CurrentCurrent

bull Time rate of change of charge t

qI Constant current tIq

dttdqti )()( Time varying current

t

dxxitq )()(

Unit mAA 3101 AmA 3101 (1 A = 1 Cs)

12 Basic Quantities

bull Notation Current flow represents the flow of positive chargebull Alternating versus direct current (AC vs DC)

i(t) i(t)

t t

DCACTime ndash varying current Steady current

bull A mount of electric charges flowing through the surface per unit time

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 31: Circuits and Analog Electronics 电路与模拟电子技术

CurrentCurrent

Positive versus negative currentPositive versus negative current

2 A -2 A

P11 In the wire electrons moving left to right to create a current of 1 mA Determine I1 and I2

Ans Ans II11 = -1 mA = -1 mA II2 2 = +1 = +1

mAmA

12 Basic Quantities

Current is always associated with arrows (directions)

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Negative charge of -2Cs moving

Positive charge of 2Cs moving or

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 32: Circuits and Analog Electronics 电路与模拟电子技术

Voltage(Potential)Voltage(Potential)

baab VVV

b

a

b

aab ldE

q

ldF

q

WV

VoltageVoltage Units 1 V = 1 JC

Positive versus negative voltagePositive versus negative voltage

+

ndash

ndash

+

2 V -2 V

12 Basic Quantities

bull Energy per unit chargebull It is an electrical force drives an electric current

+- of voltage (V) tell the actual polarity of a certain point DN

Two ldquoDo Not (DN)rdquo

+- of current (I) tell the actual direction of particlersquos movement DN

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 33: Circuits and Analog Electronics 电路与模拟电子技术

Voltage (Potential)Voltage (Potential)

a

b

VVab 5 a b which pointrsquos potential is higher

b

a

V6aV V4bV Vab =

a b +Q from point b to point a get energy Point a is

Positive or Positive or negativenegative

12 Basic Quantities

Example

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
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Page 34: Circuits and Analog Electronics 电路与模拟电子技术

Voltage (Potential)Voltage (Potential)

ab

cacute

c d

dacute

2211

21

221121222

2

21112

1111

111

1b1bb

0

)(

)(

0

rRrR

EEI

rRrRIEEIrEVIrVV

EVV

RrRIEIRVV

rRIEIrVV

IREVEV

IRVIRVVVV

V

dda

dd

cd

cc

bc

aab

a

12 Basic Quantities

Example

I

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 35: Circuits and Analog Electronics 电路与模拟电子技术

Voltage (Potential)Voltage (Potential)

K Open

K Close

Va=)V(521

)V(18

a

a

V

V

12 Basic Quantities

Example

I

I

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
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Page 36: Circuits and Analog Electronics 电路与模拟电子技术

I

11 2

a

Ev E R

R R

12 Basic Quantities

ExampleExample

I

1 21 1

1 2a

E Ev E R

R R

1 2 3 1 2 3 2 1 3 3 1 2

1 2 3 1 2 3 2 3 1 2 1 3

a a a aa

v E v E v E v E R R R E R R R E R R Rv

R R R R R R R R R R R R R R R R

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 37: Circuits and Analog Electronics 电路与模拟电子技术

PowerPower

bull One joules of energy is expanded per second

bull Rate of change of energy

P = Wt )()()()()( titVdt

dqtVdttdwtp abab

bull Used to determine the electrical power is being absorbed or supplied

ndash if P is positive (+) power is absorbed

ndash if P is negative (ndash) power is supplied

+

ndash

v(t)

i(t)p(t) = v(t) i(t)

v(t) is defined as the voltage with positive reference at the same terminal that the current i(t) is entering

12 Basic Quantities

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 38: Circuits and Analog Electronics 电路与模拟电子技术

PowerPower

Example

12 Basic Quantities

2A+

ndash

-5V 5 2 10WP Power is supplied delivered power to external element

+

ndash

5V

2A

5 2 10WP Power is absorbed Power delivered to

Note +

ndash

+5V

+

ndash

-5V

2A

-2A

Power absorbed

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 39: Circuits and Analog Electronics 电路与模拟电子技术

PowerPower

bull Power absorbed by a resistor

)()()( titvtp )(2 tiR

Rtv )(2)(2 tvG

Gti )(2

12 Basic Quantities

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 40: Circuits and Analog Electronics 电路与模拟电子技术

PowerPower

1

2

3 4

5

I1 I2 I3+

-

-

-

-

-

+

+

+

+-

+

+

-

+-

P15 Find the power absorbed by each element in the circuit

12 Basic Quantities

A21 I A12 IA13 I

V35 V

V41 V

V82 V V43 V

V74 V

3

16

7

4

8

535

212

734

323

111

WVIP

WVIP

WVIP

WVIP

WVIP

Supply energy element 1 3 4 Absorb energy element 2 5

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 41: Circuits and Analog Electronics 电路与模拟电子技术

Open CircuitOpen Circuit R=

I=0 V=E P=0E

R0

Short CircuitShort Circuit R=0

E

R0

R = 0 0R

EI 00 IREV

02RIPE

12 Basic Quantities

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 42: Circuits and Analog Electronics 电路与模拟电子技术

RR

EI

o

0IREIRV

02RIEIVI

Loaded CircuitLoaded Circuit

E

R0 R

I

0PPP E

12 Basic Quantities

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 43: Circuits and Analog Electronics 电路与模拟电子技术

13 Circuit ElementsCircuit Elements

Key Words Resistors Capacitors Inductors Resistors Capacitors Inductors voltage source current source

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 44: Circuits and Analog Electronics 电路与模拟电子技术

bull Passive elements (cannot generate energy)

ndash eg resistors capacitors inductors etc

bull Active elements (capable of generating energy)

ndash batteries generators etc

bull Important active elements

ndash Independent voltage source

ndash Independent current source

ndash Dependent voltage source

bull voltage dependent and current dependent

ndash Dependent current source

bull voltage dependent and current dependent

13 Circuit ElementsCircuit Elements

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 45: Circuits and Analog Electronics 电路与模拟电子技术

ResistorsResistors

Dissipation ElementsElements

S

lR v=iR P=vi=Ri2=v2R gt0

v-i relationship

v

i

13 Circuit ElementsCircuit Elements

Resistors connected in series

ndash Equivalent Resistance is found by Req= R1 + R2 + R3 + hellip

R1 R2 R3

Resistors connected in parallel 1Req=1R1 + 1R2 + 1R3 + hellip

R1 R2 R3

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 46: Circuits and Analog Electronics 电路与模拟电子技术

Capacitors

bull Capacitance occurs when two conductors (plates) are separated by a dielectric (insulator)

bull Charge on the two conductors creates an electric field that stores energy

bull The voltage difference between the two conductors is proportional to the charge q = C v

bull The proportionality constant C is called capacitance

bull Units of Farads (F) - CV

bull 1F= one coulomb of charge of each conductor causes a voltage of one volt across the device

1F=106F 1F=106PF

13 Circuit ElementsCircuit Elements

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 47: Circuits and Analog Electronics 电路与模拟电子技术

Capacitors

store energy in an electric field

v-i relationship

dt

dqti =)(

dt

dvC

t

dxxiC

tv )(1

)(

i(t)+

-

v(t)

Therestofthe

circuit

dt

dvcvivp 2

2

1cvcvdvpdtwEnergy stored

13 Circuit ElementsCircuit Elements

Capacitors connected in seriesndash Equivalent capacitance is found

by 1Ceq=1C1 + 1C2 + 1C3 + hellip

series

parallel

Capacitors connected in parallel Ceq= C1 + C2 + C3 + hellip

vC(t+) = vC(t-)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
  • Slide 2
  • Slide 3
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Page 48: Circuits and Analog Electronics 电路与模拟电子技术

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

v(t)5V

1s 2s(1)

00

0

1

0

2

1

1

0

1

0

1

0 0 0

11 1 0 5 1 0 5

021

2 1 5 5 2 1 5 002

0 1s

11 0 5 1 5

021s 2s

11 5 10 5 2 0

02

t

tv t i t dt v t

Ct v

v dt

v dt

t

v t dt t v

t

v t dt t v

For (1)

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 49: Circuits and Analog Electronics 电路与模拟电子技术

Capacitors

t

i(t)1A

-1A 1s

2s

i(t)

+

-

v(t)02F

P17

13 Circuit ElementsCircuit Elements

t

w (t)

25J

1s 2s(2)

0 0

0

2 20

20

1

2

1 If 0

2Now 0 0 1 5 2 0

1 01 25 25

2 01 0 0

t t

t t

t

t

dvw t Pdt C v dt

dt

C vdv C v t v t

v t w t C v t

v v v

w

w

For (2)

For (1) (2)

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 50: Circuits and Analog Electronics 电路与模拟电子技术

dt

tdiLtv

)()(

t

dxxvL

ti )(1

)(

Inductors

store energy in a magnetic field that is created by electric passing through it

v-i relationship i(t) +

-

v(t)L

Inductors connected in series Leq= L1 + L2 + L3 + hellip

Inductors connected in parallel 1Leq=1L1 + 1L2 + 1L3 + hellip

13 Circuit ElementsCircuit Elements

dt

diLiivP )(

2

1)( 2 tLitwL Energy stored

022

000 2)( titi

LidiLdt

dt

diiLPdttw

ti

tv

t

t

t

t

iL(t+) = iL(t-)

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
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Page 51: Circuits and Analog Electronics 电路与模拟电子技术

Independent voltage source

+VS

RS = 0

v

i

VS

Ideal

sS

sS

IRVV

IRV

practical

13 Circuit ElementsCircuit Elements

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 52: Circuits and Analog Electronics 电路与模拟电子技术

Independent current source

I

v

iIS

RS infin=

Ideal

SS

SS

RVII

RVI

practical

13 Circuit ElementsCircuit Elements

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
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Page 53: Circuits and Analog Electronics 电路与模拟电子技术

n

kSkS VV

1

Voltage source connected in series

n

kSkS RR

1

Voltage source connected in parallel

n

kSkS II

1

SnSSS

SnSSS

RRRR

RRRR

1111

21

21

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
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Page 54: Circuits and Analog Electronics 电路与模拟电子技术

Voltage controlled (dependent) voltage source (VCVS)

+_

_

+

Sv Svv

Current controlled (dependent) voltage source (CCVS)

+_ Sriv Si

Q What are the units for and r

13 Circuit ElementsCircuit Elements

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
  • Slide 2
  • Slide 3
  • Slide 4
  • Slide 5
  • Slide 6
  • Slide 7
  • Slide 8
  • Slide 9
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Page 55: Circuits and Analog Electronics 电路与模拟电子技术

Voltage controlled (dependent) current source (VCCS)

Current controlled (dependent) current source (CCCS)

_

+

SvSgvi

Si Sii

Q What are the units for and g

13 Circuit ElementsCircuit Elements

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 56: Circuits and Analog Electronics 电路与模拟电子技术

Independent source

dependent source

Can provide power to the circuit

Excitation to circuit

Output is not controlled by external

Can provide power to the circuit No excitation to circuit

Output is controlled by external

13 Circuit ElementsCircuit Elements

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 57: Circuits and Analog Electronics 电路与模拟电子技术

bull So far we have talked about two kinds of circuit elements

ndash Sources (independent and dependent)

bull active can provide power to the circuit

ndash Resistors

bull passive can only dissipate power

Review

The energy supplied by the active elements is equivalent to the energy absorbed by the passive elements

13 Circuit ElementsCircuit Elements

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 58: Circuits and Analog Electronics 电路与模拟电子技术

14 Kirchhoffs Current and Voltage Laws

Key Words Nodes Branches Loops KCL KVL

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

  • Slide 1
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Page 59: Circuits and Analog Electronics 电路与模拟电子技术

Nodes Branches Loops mesh

Node point where two or more elements are joined (eg big node 1)

Loop A closed path that never goes twice over a node (eg the blue line)

Branch Component connected between two nodes (eg component R4)

The red path is NOT a loop

Mesh A loop that does not contain any other loops in it

14 Kirchhoffs Current and Voltage Laws

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 60: Circuits and Analog Electronics 电路与模拟电子技术

Nodes Branches Loops mesh

bull A circuit containing three nodes and five branches

bull Node 1 is redrawn to look like two nodes it is still one nodes

P18

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 61: Circuits and Analog Electronics 电路与模拟电子技术

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

KCL MathematicallyKCL Mathematicallyi1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

n

jjI

1

0

14 Kirchhoffs Current and Voltage Laws

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 62: Circuits and Analog Electronics 电路与模拟电子技术

bull sum of all currents entering a node is zero

bull sum of currents entering node is equal to sum of currents leaving node

KCL

P19

DCBA iiii

14 Kirchhoffs Current and Voltage Laws

In

Out

0A B C O

I

I

i i i i

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 63: Circuits and Analog Electronics 电路与模拟电子技术

KCL

+

-120V

50 1W Bulbs

Is

P110

bull Find currents through each light bulb

IB = 1W120V = 83mA

bull Apply KCL to the top node

IS - 50IB = 0

bull Solve for IS IS = 50 IB = 417mA

KCL-Christmas LightsKCL-Christmas Lights

14 Kirchhoffs Current and Voltage Laws

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 64: Circuits and Analog Electronics 电路与模拟电子技术

KCL

P111 We can make supernodes by aggregting node

0

0

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii3 amp 2 Adding

14 Kirchhoffs Current and Voltage Laws

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 65: Circuits and Analog Electronics 电路与模拟电子技术

KCL

Current dividerCurrent divider

N VG1

G2

I+

-

I1I2

IGG

GG

G

IVGI

21

1111

IGG

GVGI

21

222

I

G

GI

n

kk

kk

1

121

21

111

11

RRR

RRI

RRI

R

VI

I

RR

RI

21

12

14 Kirchhoffs Current and Voltage Laws

In case of parallel 1 21 2

1 1 1 V=

I IG G G

R R R R G

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 66: Circuits and Analog Electronics 电路与模拟电子技术

sum of voltages around any loop in a circuit is zero

KVL

bull A voltage encountered + to - is positivebull A voltage encountered - to + is negative

KVL Mathematically 0)(1

n

jj tv 0

1

n

jjV

14 Kirchhoffs Current and Voltage Laws

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 67: Circuits and Analog Electronics 电路与模拟电子技术

KVL is a conservation of energy principle

KVL

A positive charge gains electrical energy as it moves to a point with higher voltage and releases electrical energy if it moves to a point with lower voltage

AV

BBV)( AB VVqW

q

abV

a bq

abqVW LOSES

cdV

c dq

cdqVW GAINS

AV

BBV

q

CV

ABV

BC

V

CAV

If the charge comes back to the same Initial point the net energy gain Must be zero

0)( CABCAB VVVq

14 Kirchhoffs Current and Voltage Laws

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 68: Circuits and Analog Electronics 电路与模拟电子技术

KVL

P113 Determine the voltages Vae and Vec

14 Kirchhoffs Current and Voltage Laws

10 24 0aeV

16 12 4 6 0aeV

4 + 6 + Vec = 0

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 69: Circuits and Analog Electronics 电路与模拟电子技术

KVL

Voltage dividerVoltage divider

R1

R2

-

V1

+

+

-

V2

+

-

V

21

111 RR

RVIRV

21

222 RR

RVIRV

Important voltage Divider equations

NV

R

RV n

kk

kk

1

14 Kirchhoffs Current and Voltage Laws

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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Page 70: Circuits and Analog Electronics 电路与模拟电子技术

KVLVoltage dividerVoltage divider

kR 151

Volume control

P114 Example Vs = 9V R1 = 90kΩ R2 = 30kΩ

14 Kirchhoffs Current and Voltage Laws

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