emlab 1 회로 이론 (2014). emlab 2 review of circuit theory i 1.linear system 2.kirchhoff’s law...

EMLAB

1

회로 이론 (2014)

EMLAB

2

Review of circuit theory I

1. Linear system

2. Kirchhoff’s law

3. Nodal & loop analysis

4. Superposition

5. Thevenin’s and Norton’s theorem

6. Resistor, Inductor, Capacitor

7. Operational amplifier

8. First and second order transient circuit

EMLAB

3

Linear system

(t)1(t)1iLinear systemL

(t)(t) 21 BA (t)21 BiAi Linear systemL

(t)2(t)2iLinear systemL

A system satisfying the above statements is called as a linear system. Resis-tors, Capacitors, Inductors are all linear systems. An independent source is not a linear system.

All the circuits in the circuit theory class are linear sys-tems!

EMLAB

4

R1

1k

Resistor

Examples of Linear system

-)(tR

)(1 ti

-)(tC

)(2 ti

11 )()( RtitR

t

02 )(

1)( dtti

CtC

C1

1n

Capacitor

input

output

R1

C1

input

EMLAB

5

Inductor

-(t)L

)(2 ti

dt

diLtL

2)(

EMLAB

6

Kirchhoff’s Voltage law

-(t)1 -(t)2

-

(t)s

0)( n

n t 0C

drE

Sum of voltage drops along a closed loop should be equal to zero!

0)(-)()( 21 ttt s

R1 C1

EMLAB

7

Kirchhoff’s Current law

0)( n

n tI 0S

daJ

(t)0

)(2 ti)(1 ti

)(3 ti3

303

2

202

1

101

321

-)(,

-)(,

-)(

0)()()()(

Rti

Rti

Rti

titititIn

n

Ri ba

ab

RCurrent definition

Sum of outgoing(incoming) currents from any node should be equal to zero!

• To define a current, a direction can be chosen arbitrarily.

• The value of a current can be obtained from a voltage drop along the direction of current divided by a resistance met.

R2

)(1 t )(2 t

)(3 t

R1

R3

0---

3

30

2

20

1

10 RRR

EMLAB

8

Nodal analysis

• Unknowns : node voltages

• Kirchhoff’s current law is utilized to form matrix equations.

• For each node, the sum of out-going currents become zero.

1

2

0612

0-

6

0- 11 mkk

0

4

0-

4

0-6 22

kkm

EMLAB

9Loop analysis

• Unknowns : loop currents

• Matrix equations are formed by Kirchhoff’s voltage law.

• For each loop, the sum of voltage drops are equal to zero.

1i 2i 3i

0)(126 211 iikik

mi 62

04)(4 323 ikiik

EMLAB

10Superposition

• Superposition is utilized to simplify the original linear circuits.

• If a voltage source is eliminated, it is replaced by a short circuit connected to the original terminals.

• If a current source is eliminated, it is replaced by an open circuit.

Circuit Circuit Circuit

L

1L

2L+=

21 LLL

Circuit with currentsource set to zero(OPEN) Circuit with voltage source

set to zero (SHORT CIRCUITED)

EMLAB

11Example

VVkkk

k412

666

601

01

02

Vkm

kk

k 243612

266

121

61

61

02

V20244020102

EMLAB

12Thevenin and Norton equivalent

Open circuited voltagemeasured by voltmeter

Short circuited current measured by ammeter

Resistance obtained with voltage source shorted and current source open

AV

EMLAB

13Example

VVkk

kVTh 612

22

2

kkkRTh 12||2

kkkRTh 12||2

mAk

VIN 6

2

12

EMLAB

14Resistor – Input output relationship

2 4 6 80 10

5

10

15

0

20

time, sec

Vou

t, V

2 4 6 80 10

2

4

6

0

8

time, sec

I_P

robe

1.i,

A

VoutRR1R=2 Ohm

ItPWLSRC1I_Tran=pwl(time, 0,0, 2,2, 4,1,6,2.5,8,2 , 10,8 )

I_ProbeI_Probe1

i(t)

i(t)

v(t)v(t)

)()( tiRt

EMLAB

15

t

2 4 6 80 10

2

4

6

0

8

time, sec

I_P

robe

1.i,

A

2 4 6 80 10

5

10

15

20

0

25

time, sec

Vou

t, V

VoutCC1C=1F

ItPWLSRC1I_Tran=pwl(time, 0,0, 2,2, 4,1,6,2.5,8,2 , 10,8 )

I_ProbeI_Probe1

i(t)

v(t)

t diC

t 0 )(1

)(

i(t)

v(t)

Capacitor – Input output relationship

EMLAB

16

dt

diLt )(

2 4 6 80 10

2

4

6

0

8

time, sec

I_P

robe

1.i,

A

2 4 6 80 10

0

1

2

3

-1

4

time, sec

Vou

t, V

VoutLL1

R=L=1.H

ItPWLSRC1I_Tran=pwl(time, 0,0, 2,2, 4,1,6,2.5,8,2 , 10,8 )

I_ProbeI_Probe1

i(t)

v(t)

i(t)

v(t)

Inductor – Input output relationship

EMLAB

17First order transient circuit

The voltage drop across a capacitor cannot change instantaneously.

The current through an inductor cannot change instantaneously.

tRC

C

ststC

SCC

C

C

AevRC

s

RCsAeAev

Vtvdt

dvRCv

iRv

1

,1

0)1(,

)0(,0

0

]1[)(

)(,0

0)0(),0(

tL

RS

St

L

R

S

eR

Vti

R

VAeti

dt

diLiR

titVdt

diLiR

EMLAB

18

Second order transient circuit

02 2002

2

idt

di

dt

id 0

)0(0

)0()0()(

1

2

2

0

LC

i

dt

di

L

R

dt

id

t

tdtti

CiR

dt

diL s

t

C

Normalized form

L

R

LC 0

20 2

,1

02)( 200

2 sseti st

12002,1 s

LR

C

outst

out dttiC 0

)(1

L

R

LC 0

20 2

,1

EMLAB

19

tsts eKeKti 21

21)(

0)0( 21 KKi

sKsKsdt

diL 2211)0(

tAtAeti ddt sincos)( 21

0

0)0( 1 Ai

sd AAdt

diL 210)0(

tetBBti 0

21)(

1)0( Bi

sBBdt

diL 210)0(

Solution

Critically damped : ζ = 1

Under-damped : ζ <1Over-damped : ζ > 1

2

0 1 d

EMLAB

20

t

t

ttt

eBtBti

BtBtie

tietiee

titdt

dit

dt

id

0

0

00

0

)()(

)(

0])([0])([1

0)()(2)(

21

21

2002

2

Critically damped: ζ=1 인 경우

EMLAB

5 10 15 20 250 30

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0.0

1.6

time, sec

Vou

t, V

21

25.1

25.0

1

Transient response

Critically damped

Under-damped

Over-damped

EMLAB

22

LR

C

outs

Ringing in digital logics

in s

EMLAB

23Contents : Circuit theory 2

1. AC steady-state analysis : 60Hz sinusoidal input signal

• Power factor

2. Magnetically coupled networks : transformer

3. Poly-phase circuits : power distribution

• Single phase two wire

• Three phase 4 wire power distribution

4. Arbitrary input signal

• Fourier series and Fourier transform

• Laplace transform

5. Two-port network : black box

EMLAB

24

Chapter 8

AC steady-state analysis

EMLAB

25

)0(cos

)0(0)()(2)( 2

002

2

ttV

ttxt

dt

dxt

dt

xd

ss

Sinusoidal input signal

입력 전압의 형태

LR

C

outs

0t

EMLAB

26Sinusoids

tXtx M sin)(

(radians)argument

(rads/sec)frequency angular

valuemaximumor amplitude

t

XM

Dimensionless plot As function of time

tTtxtxT ),()( Period 2

)(cycle/sec Hertzin frequency 2

1

Tf

f 2

0t

)(sin)( 0ttXtx Mlag

“Lag by t0”

)(sin)( 0ttXtx Mlead

0t

“Lead by t0”

EMLAB

27

AC (Alternating Current)• Easy to generate ( 교류 전압은 만들기 쉽다 .)

• Easy to change voltage levels. ( 전압을 변화하기도 쉽다 .)

• Less damage on human compared with DC ( 직류에 비해 덜 위험하다 .)

rmsV 220 Hz,60

EMLAB

28

0)0(),()()( :KVL itVtRitdt

diL S

)0(0

)0(cos)(V 0

S t

ttVt

0)()( tRit

dt

diL h

h

)()()( tVtRitdt

diL Sp

p

Solution of Differential Eq.

To solve a differential equation, initial conditions must be specified.

solution Particular:)(

solution sHomogeneou:)(

)()()(

ti

ti

tititi

p

h

ph

)()]()([)]()([

tVtitiRdt

titidL Sph

ph

EMLAB

29

0)()( tRitdt

diL h

h

)()()( tVtRitdt

diL Sp

p

tL

Rst

h AeAeti

)(

)cos()( tBtip

tVtBLRtBLR

tVtRBtLB

cossin)cossin(cos)sincos(

cos)cos()sin(

0

0

R

L

LR

VB

tan,sincos

0

R

Lt

LR

Vtip

1

22

0 tan),cos()(

)(

R

Lt

LR

VAeti

tL

R

1

22

0 tan),cos()(

)(

Solution method #1

For a particular solution, choose a trial function that might produce VS(t) on entering the differential Eq.

EMLAB

30Simpler method for sinusoidal source case

)()()( tVtRitdt

diL Sp

p }Re{}Re{)cos()( tjtjjp eeBetBti I

}Re{Re

}Re{}Re{}Re{

0

0

tjtjtj

tjtjtj

eVeRedt

dL

eVeRedt

dL

II

II

0Re 0

tjtjtj eVeRe

dt

dL II

LjR

eVeti

RLj

V

eVeReLj

eVeRedt

dL

tjtj

p

tjtjtj

tjtjtj

00

0

0

Re}Re{)(,

0

II

II

II

)cos()()(

ReRe)(22

0

22

00

t

LR

V

LR

eeV

LjR

eVti

tjjtj

p

tjte tj sincos

sin,cos

tan, 122

ryrxx

yyxr

rejyx

PolarComplexj

EMLAB

31

tjtjjtj

eeLR

eV

LjR

eVti

IRe

)(ReRe)(

22

00

Phasor

22

0

22

00

)()(I:Phasor

LR

V

LR

eV

LjR

V j

• With sinusoidal source function, it is simpler to use a trial solution ~ Re{I ejwt}.

• The complex coefficient of the exponential function is called as a phasor.

EMLAB

32

VoutRR1R=2 Ohm

ItSineSRC1

Phase=0Damping=0Delay=0 nsecFreq=0.16 HzAmplitude=1 AIdc=0 mA

I_ProbeI_Probe1

Phasor - resistor

2 4 6 8 10 12 14 16 180 20

-1

0

1

-2

2

time, sec

Vou

t, V

I_P

robe

1.i,

A

(t)(t) SViR

)(t

t

)(ti

Relationship between sinusoids

EMLAB

33

ILjV

Iedt

dLVe

dt

tdiLtv

tjtj

}Re{}Re{

)()(

Relationship between sinusoids

)2/cos(}Re{

}Re{}Re{)(

}Re{}Re{

)()(

)2/(

tLIIeL

IeLjVetv

ILjV

Iedt

dLVe

dt

tdiLtv

tj

tjtj

tjtj

Phasor - inductor

)(t

t

)(ti

0

2

EMLAB

34

Relationship between sinusoids

Cj

Cj

edt

dCe

dt

tdvCti

tjtj

IV

VI

VI

}Re{}Re{

)()(

Phasor - capacitor

)(t

t

)(ti

0

2

EMLAB

35Examples

inductor theacross voltage theFind

60),(304,05.0 HzfAIHL

1202 fLIjV

30490105.0120V 6024V

)60120cos(24)( tv

inductor theacross voltage theFind

60,1456.3,150 HzfIFC

1202 f

Cj

IVCVjI

90110150120

1456.36

V

235200

V

)235120cos(200

)( ttv

EMLAB

36Impedance and Admittance

jXRZI

V

I

VzMiv

M

M

iM

vM

)(I

VZ

AC circuitvMV

iMI

zZ

Impedance :

jBGY YY V

IYAdmittance :

Element Impedance Admittance

R

L

C

LjL Z

CjC 1

Z

LjYL

1

CjYC

RR ZR

YR

1

EMLAB

37

1Z

2Z

3Z

321 ZZZZeq

321 YYYYeq

11

1

ZY

22

1

ZY

33

1

ZY

321

1111

ZZZ

Zeq

321

1111

YYY

Yeq

Series / parallel combination

EMLAB

38Example

Find the current i(t) in the network in Fig. E8.8.

25.90323.00052.00319.021 jYYYeq

0189.01050377 61 jjY

024.00319.0104037720

132 j

jY

1Y

2Y

25.3988.33012025.90323.0VYI eq

)25.39377cos(88.3)( tti

EMLAB

39Example 8.15

0621 VV

011

0211

221

j

VV

j

V

21111

6 222

j

VV

j

V

11

62

1

11

11

12 jjj

V

5.15.22 jV

][5.15.21

20 Aj

VI

Super node