capacitance and inductance - الصفحات...
TRANSCRIPT
CAPACITANCE AND INDUCTANCE
Introduces two passive, energy storing devices: Capacitors and Inductors
LEARNING GOALS
CAPACITORS
Store energy in their electric field (electrostatic energy)
Model as circuit element
INDUCTORS
Store energy in their magnetic field
Model as circuit element
CAPACITOR AND INDUCTOR COMBINATIONS
Series/parallel combinations of elements
RC OP-AMP CIRCUITS
Integration and differentiation circuits
CAPACITORS First of the energy storage devices to be discussed
Basic parallel-plates capacitor
CIRCUIT REPRESENTATION
NOTICE USE OF PASSIVE SIGN CONVENTION
Typical Capacitors
Normal values of capacitance are small.
Microfarads is common.
For integrated circuits nano or pico farads
are not unusual
d
AC
28
4
12
103141.610016.1
1085.855 mA
AF
PLATE SIZE FOR EQUIVALENT AIR-GAP CAPACITOR
gap in material ofconstant Dielectric
Basic capacitance law )( CVfQ
Linear capacitors obey Coulomb’s law CCVQ C is called the CAPACITANCE of the device and hasunits of
voltage
charge
One Farad(F)is the capacitance of a device that can store one Coulomb of charge at one Volt.
Volt
CoulombFarad
EXAMPLE Voltage across a capacitor of 2 microFarads holding 10mC of charge
500010*1010*2
11 3
6
Q
CVC V
Capacitance in Farads, charge in Coulombsresult in voltage in Volts
Capacitors can be dangerous!!!
Linear capacitor circuit representation
The capacitor is a passive element and follows the passive sign convention
Capacitors only store and releaseELECTROSTATIC energy. They do not “create”
Linear capacitor circuit representation
)()( tdt
dvCti
LEARNING BY DOING
If the voltage varies the charge varies and there
is a displacement current
CC CVQ Capacitance Law
One can also express the voltage across
in terms of the current
QC
tVC
1)(
t
C dxxiC
)(1
Integral form of Capacitance law
dt
dVC
dt
dQi CC
… Or one can express the current through
in terms of the voltage across
Differential form of Capacitance law
The mathematical
implication of the integral
form is ...
ttVtV CC );()(
Voltage across a capacitor
MUST be continuous
Implications of differential form??
0 CC iConstV
DC or steady state behavior
A capacitor in steady state acts as an
OPEN CIRCUIT
CURRENT THE DETERMINE
FC 5
LEARNING EXAMPLECAPACITOR AS CIRCUIT ELEMENT
Cv
Ci
)()( tdt
dvCti c
C
t
CC dxxiC
tv )(1
)(
t
t
tt
0
0
0
0
)(1
)(1
)(
t t
t
CCC dxxiC
dxxiC
tv
t
t
CCC dxxiC
tvtv
0
)(1
)()( 0
The fact that the voltage is defined throughan integral has important implications...
RR
RR
Riv
vR
i
1
Ohm’s Law
)( Oc tv
elsewhereti 0)(
)()( tdt
dvCti
mAs
VFi 20
106
24][105
3
6
mA60
CAPACITOR AS ENERGY STORAGE DEVICE
)()()( titvtp CCC
Instantaneous power
)()( tdt
dvCti c
C
dt
dvtCvtp c
CC )()(
C
tqdxxi
Ctv C
t
CC
)()(
1)(
)()(1
)( tdt
dqtq
Ctp C
CC
Energy is the integral of power
2
1
)(),( 12
t
t
CC dxxpttw
If t1 is minus infinity we talk about
“energy stored at time t2.”
If both limits are infinity then we talk
about the “total energy stored.”
)(
2
1)( 2
tvdt
dCtp CC
)(2
1)(
2
1),( 1
22
212 tCvtCvttw CCC
)(
2
11)( 2
tqdt
d
Ctp cC
)(1
)(1
),( 12
22
12 tqC
tqC
ttw CCC
W
Cv
Ci
Energy stored in 0 - 6 msec
][)24(*][10*52
1)6,0( 226
VFwC
Charge stored at 3msec
)3()3( CC Cvq
)0(2
1)6(
2
1)6,0( 22
CCC CvCvw
CVFqC 60][12*][10*5)3( 6
“total energy stored?” ....
“total charge stored?” ...
If charge is in Coulombs
and capacitance in Farads
then the energy is in ….
FC 5
LEARNING EXAMPLE
VOLTAGETHE FIND .4 FC
0)0( v
20 t
mst 42
][1082)( 3Vttv
0;)(1
)0()(0
tdxxiC
vtvt
2;)(1
)2()(2
tdxxiC
vtvt
POWER THE FIND .4 FC
0)0( vtti
3108)(
mstttp 20,8)( 3
mst 42
elsewheretp ,0)(
elsewheretp ,0)(
FIND THE ENERGY
mstttp 20,8)( 3
mst 42
LEARNING EXTENSION
CURRENT THE DETERMINE
FC 2
)()( tdt
dvCti
s
VFi
3
6
102
12102
s
VFi
3
6
104
12102
Energy stored at a given time t )(2
1)( 2 tCvtE C )240/1(E
2sin130*][10*2
2
1 226 F J
Charge stored at a given time )()( tCvtq CC )120/1(Cq 0])[sin(*][10*2 6 VC C
Current through the capacitor )(tdt
dvCi C
C )120/1(Ci )cos(120*130*10*2 6 A
Electric power supplied to capacitor at a given time )()()( titvtp CCC
Energy stored over a given time interval
W
)(2
1)(
2
1),( 1
22
212 tCvtCvttw CC
J
FC 2
)120(sin130)( ttv
)(tv
WHAT VARIABLES CAN BECOMPUTED?
SAMPLE PROBLEM
Cv
Ci
C
FC 2
][0;0
0;)(
5.0
mAt
teti
t
C
Current through capacitor
Voltage at a given time t dxxiC
tv
t
CC )(1
)(
)0(Cv ][0 V
Voltage at a given time t when voltage at time to<t is also known
t
t
CCC dxxiC
tvtv
0
)(1
)()( 0
)2(Cv
2
0
5.01)0( dxe
Cv x
C
2
0
5.0
6 5.0
1
10*2
1
xe 61
610*6321.01
5.0
1
10*2
1
e V
Charge at a given time)()( tCvtq CC )2(Cq 6321.0*2 C
Voltage as a function of time dxxiC
tv
t
CC )(1
)(
0;0)( ttvC
t
x
CC dxeC
vtv0
5.01)0()(
0;0
0);1(10)(
5.06
t
tetv
t
CV
Electric power supplied to capacitor )()()( titvtp CCC
Energy stored in capacitor at a given time )(2
1)( 2
tCvtw C
W
J
“Total” energy stored in the capacitor )(2
1 2 CT Cvw 6266 10)10(*10*22
1
Tw J
SAMPLE PROBLEM If the current is known ...
SAMPLE PROBLEM
sec)(mt
5 10
Compute voltage as a function of time
At minus infinity everything is zero. Sincecurrent is zero for t<0 we have
sec50 mttsAt
s
At
ms
AtiC ]/[10*3
10
103
5
15)( 3
3
6
][10*4
10*3)(0)0(
06
3
VxdxtVVt
CC
][10*50];[8
10*3 323
stVt
In particular][
8
75][
8
)10*5(*10*3)5(
233
mVVmsVC
][10)(105 Atimst C
t
CC dxsAtVmVmsV310*5
6
6
3
]/)[10*10(10*4
1
8
10*75)(][
8
75)5(
][10*1010*5;][10*54
10
8
10*75)( 333
3
stVttVC
Charge stored at 5ms
)()( tCVtq CC
][8
10*75*][10*4)5(
36 VFmsq
][)2/75()5( nCmsq
Total energy stored
2
2
1CCVE
Total means at infinity. Hence
][8
10*2510*4*5.0
23
6 JET
Before looking into a formal way to describe the currentwe will look at additional questions that can be answered.
Now, for a formal way to represent piecewise functions....
Given current and capacitance
0;0)( ttVC
][10;8
25
][105;54
10
8
75
50;8
3
0;0
)(
2
mst
mstt
mstt
t
tVc][mV
Formal description of a piecewise analytical signal
Flux lines may extendbeyond inductor creatingstray inductance effects
A TIME VARYING FLUXCREATES A COUNTER EMFAND CAUSES A VOLTAGE TO APPEAR AT THETERMINALS OF THEDEVICE
INDUCTORSNOTICE USE OF
PASSIVE SIGN CONVENTION
Circuit representation for an inductor
A TIME VARYING MAGNETIC FLUX
INDUCES A VOLTAGE
dt
dvL
Induction law
INDUCTORS STORE ELECTROMAGNETIC ENERGY.
THEY MAY SUPPLY STORED ENERGY BACK TO
THE CIRCUIT BUT THEY CANNOT CREATE ENERGY.
THEY MUST ABIDE BY THE PASSIVE SIGN CONVENTION
FOR A LINEAR INDUCTOR THE FLUX IS
PROPORTIONAL TO THE CURRENT
LLi
dt
diLv L
L DIFFERENTIAL FORM
OF INDUCTION LAW
THE PROPORTIONALITY CONSTANT, L, IS
CALLED THE INDUCTANCE OF THE COMPONENT
INDUCTANCE IS MEASURED IN UNITS OF
henry (H). DIMENSIONALLY
secAmp
VoltHENRY
LEARNING by Doing
Follow passive sign convention
dt
diLv L
L Differential form of induction law
t
LL dxxvL
ti )(1
)(Integral form of induction law
00 ;)(1
)()(
0
ttdxxvL
titi
t
t
LLL
A direct consequence of integral form ttiti LL );()( Current MUST be continuous
A direct consequence of differential form 0. LL vConsti DC (steady state) behavior
Power and Energy stored
)()()( titvtp LLL W )()()( titdt
diLtp L
LL
)(
2
1 2 tLidt
dL
)(2
1)(
2
1),( 1
22
212 tLitLittw LL
Energy stored on the interval
Can be positive or negative
)(2
1)( 2
tLitw LL “Energy stored at time t”
Must be non-negative.
Passive element!!!
2
1
)(2
1),( 2
12
t
t
LL dxxLidt
dttw J
Current in Amps, Inductance in Henrys
yield energy in Joules
LEARNING EXAMPLE FIND THE TOTLA ENERGY STORED IN THE CIRCUIT
In steady state inductors act as
short circuits and capacitors act
as open circuits
2 21 1
2 2C C L L
W CV W LI
9@ : 3 0
9 6
A AV V
A A
81[ ]
5A
V V
2
610.8
6 3C A
V V V
21.8
9
A
L
VI A
1 2 13 1.2
L L LI A I I A
1 1 19 6 16.2
C L CV I V V
L=10mH. FIND THE VOLTAGE
s
A
s
Am 10
102
10203
3
s
Am 10
)()( tdt
diLtv
THE DERIVATIVE OF A STRAIGHT LINE IS ITS
SLOPE
elsewhere
mstsA
mstsA
dt
di
0
42)/(10
20)/(10
mVVtv
HL
sAtdt
di
10010100)(
1010
)/(10)( 3
3
ENERGY STORED BETWEEN 2 AND 4 ms
)2(2
1)4(
2
1)2,4( 22
LL LiLiw
233 )10*20(10*10*5.00)2,4( w J
THE VALUE IS NEGATIVE BECAUSE THE
INDUCTOR IS SUPPLYING ENERGY
PREVIOUSLY STORED
LEARNING EXAMPLE
2
)(Vv
2 )(st
L=0.1H, i(0)=2A. Find i(t), t>0
t
dxxvL
iti0
)(1
)0()(
20;2)(2)(0
ttdxxvxvt
stttiHL 20;202)(1.0
stititxv 2);2()(2;0)(
Initial energy stored in inductor
2)2]([1.0*5.0)0( AHw ][2.0 J
“Total energy stored in the inductor”
JAHw 2.88)42(*][1.0*5.0)( 2
Energy stored between 0 and 2 sec
)0(2
1)2(
2
1)0,2( 22
LL LiLiw
22 )2(*1.0*5.0)42(*1.0*5.0)0,2( w
][88)0,2( Jw
ENERGY COMPUTATIONS
SAMPLE PROBLEM
22 )(st
)(Ai42
)(2
1)(
2
1),( 1
22
212 tLitLittw LL
Energy stored on the interval
Can be positive or negative
FIND THE VOLTAGE ACROSS AND THE ENERGY
STORED (AS FUNCTION OF TIME)
)(tv
FOR ENERGY STORED IN THE INDUCTOR
)(twL
NOTICE THAT ENERGY STORED AT
ANY GIVEN TIME IS NON NEGATIVE
-THIS IS A PASSIVE ELEMENT-
LEARNING EXAMPLE
LEARNING EXAMPLE
VOLTAGETHE DETERMINE
mHL 10)()( t
dt
diLtv
mVv 100
s
AHv
3
33
102
1020][1010
L=200mH
FIND THE CURRENT
0)0(0;0)( ittv
0;)(1
)0()(0
tdxxvL
itit
)(ti
)(ti
LEARNING EXAMPLE
ENERGY
POWER
)(ti
L=200mH
)(tp
)(tw
NOTICE HOW POWER CHANGES
SIGN
ENERGY IS NEVER NEGATIVE.
THE DEVICE IS PASSIVE
FIND THE POWER
FIND THE ENERGY
L=5mH
FIND THE VOLTAGE
)()( tdt
diLtv
ms
mAm
1
20 )/(
12
2010sAm
Vv
m
0
0
)/(
34
100sAm
mVmstsAHv 10010);/(20)(105 3
mVv 50
mVv 50
LEARNING EXTENSION
CAPACITOR SPECIFICATIONS
VALUESSTANDARD IN
RANGE ECAPACITANC mFCFp 50
VV 5003.6
RATINGS CAPACITOR STANDARD
%20%,10%,5
TOLERANCE STANDARD
LEARNING EXAMPLE
GIVEN THE VOLTAGE WAVEFORM
DETERMINE THE VARIATIONS IN
CURRENT
%20100 nFC
)()( tdt
dvCti
nAs
VF 600
23
3)3(10100 9
current Nominal
nA300
nA300
CURRENT WAVEFORM
s
VALUESSTANDARD IN
RANGES INDUCTANCE mHLnH 1001
AmA 1
RATINGS INDUCTOR STANDARD
%10%,5
TOLERANCE STANDARD
INDUCTOR SPECIFICATIONS
LEARNING EXAMPLE
%10100 HL
GIVEN THE CURRENT WAVEFORM
DETERMINE THE VARIATIONS IN
VOLTAGE
)()( tdt
diLtv
S
AHv
6
36
1020
1020010100
vi
iv
LC
IDEAL AND PRACTICAL ELEMENTS
IDEAL ELEMENTS
CAPACITOR/INDUCTOR MODELS
INCLUDING LEAKAGE RESISTANCE
)(ti
)(tv
)()(
)( tdt
dvC
R
tvti
leak
MODEL FOR “LEAKY”
CAPACITOR
)(ti
)(tv
)()()( tdt
diLtiRtv leak
MODEL FOR “LEAKY”
INDUCTORS
)(tv
)(tv
)(ti )(ti
)()( tdt
dvCti )()( t
dt
diLtv
SERIES CAPACITORS
NOTICE SIMILARITY
WITH RESITORS IN
PARALLEL
21
21
CC
CCCs
Series Combination of two
capacitors
F6 F3 SC
F2
F2
F1
6
123
ALGEBRAIC SUM OF INITIAL VOLTAGES
POLARITY IS DICTATED BY THE REFERENCEDIRECTION FOR THE VOLTAGE
VVV 142
LEARNING EXAMPLE
DETERMINE EQUIVALENT
CAPACITANCE AND THE
INITIAL VOLTAGE
OR WE CAN REDUCE TWO AT A TIME
F30
C
+
-
V8
V12
SAME CURRENT. CONNECTED FOR THE SAME TIME PERIOD
SAME CHARGE ON BOTH CAPACITORS
CVFQ 240)8)(30(
V4
LEARNING EXAMPLE Two uncharged capacitors are connected as shown.
Find the unknown capacitance
1C FIND
CVFQCVQ 72)6)(12(
V18
FV
CC
4
18
721
PARALLEL CAPACITORS
)()( tdt
dvCti kk
)(ti
LEARNING EXAMPLE
PC 4 6 2 3 15 F
LEARNING EXTENSION
F6
F2
F3
F4
F12
eqC
F4
F3FCeq 2
3
F4 ARECAPACITORSALL
FIND EQUIVALENT CAPACITANCE
F8
F8
eqC F8
F8
F4
F12
32
3
328
3
8
SAMPLE PROBLEM
IF ALL CAPACITORS HAVE THE SAME CAPACITANCE VALUE C
DETERMINE THE VARIOUS EQUIVALENT CAPACITANCES
SAMPLE PROBLEM
All capacitors are equal
with C=8 microFaradsEQC
______ABC
Examples of interconnections
SERIES INDUCTORS
)()( tdt
diLtv kk
)()( tdt
diLtv S
LEARNING EXAMPLE
eqL H7
PARALLEL INDUCTORS
)(ti
INDUCTORS COMBINE LIKE RESISTORS
CAPACITORS COMBINE LIKE CONDUCTANCES
LEARNING EXAMPLE
mH4 mH2
N
jj titi
100 )()(
AAAAti 1263)( 0
LEARNING EXTENSION
mH2
mH2
WHEN IN
DOUBT…
REDRAW!
a
b
cd
mH4
mH2mH2
mH4
eqL
a
b
c
d
IDENTIFY ALL NODES
PLACE NODES IN CHOSEN LOCATIONS
a
b
cd
CONNECT COMPONENTS BETWEEN NODES
mH6
mHmHmHmHLeq 4.42)4||6(
ALL INDUCTORS ARE 4mH
ALL INDUCTORS ARE 6mH
NODES CAN HAVE COMPLICATED SHAPES.
KEEP IN MIND DIFFERENCE BETWEEN
PHYSICAL LAYOUT AND ELECTRICAL
CONNECTIONS
6||6||6
a
b
c
a
b
c
SELECTED LAYOUT
a
b
c
mH2
mH6
mH6
mH6
eqL
mHLeq7
246
14
4866||)26(6
mHLeq7
66
LEARNING EXTENSION
L-C
RC OPERATIONAL AMPLIFIER CIRCUITS
INTRODUCES TWO VERY IMPORTANT PRACTICAL CIRCUITS
BASED ON OPERATIONAL AMPLIFIERS
A
)(0 vvAvR OO
THE IDEAL OP-AMP
iR
ARR iO ,,0IDEAL
RC OPERATIONAL AMPLIFIER CIRCUITS -THE INTEGRATOR
0v
IDEAL OP-AMP ASSUMPTIONS
)(0
)(
_
_
iRi
Avv
RC OPERATIONAL AMPLIFIER CIRCUITS - THE DIFFERENTIATOR
1R
2i
1i
0v
iiiv 21:KCL@
IDEAL OP-AMP ASSUMPTIONS
)(0
)(
_
_
iRi
Avv 0
2
1 R
vi O
t
dxxiC
iRtv )(1
)( 1
1
111
KVL
)(2
1R
vi oo1 vof terms in i replace
DIFFERENTIATE
)(111
111 t
dt
dvCi
dt
diCR
)(11211 t
dt
dvCRv
dt
dvCR o
o
IF R1 COULD BE SET TO ZERO WE WOULD HAVE
AN IDEAL DIFFERENTIATOR.
IN PRACTICE AN IDEAL DIFFERENTIATOR AMPLIFIES
ELECTRIC NOISE AND DOES NOT OPERATE.
THE RESISTOR INTRODUCES A FILTERING
ACTION. ITS VALUE IS KEPT AS SMALL AS
POSSIBLE TO APPROXIMATE A DIFFERENTIATOR
ABOUT ELECTRIC NOISE
ALL ELECTRICAL SIGNALS ARE CORRUPTED BY
EXTERNAL, UNCONTROLLABLE AND OFTEN
UNMEASURABLE, SIGNALS. THESE UNDESIRED
SIGNALS ARE REFERRED TO AS NOISE
THE DERIVATIVE
)102cos(2000)120cos(120)( 9tt
dt
dy
SIMPLE MODEL FOR A NOISY 60Hz SINUSOID
CORRUPTED WITH ONE MICROVOLT OF 1GHz
INTERFERENCE.
)102sin(10)120sin()( 96ttty
610amplitude signal
amplitude noise
noisesignal
67.16120
2000
amplitude signal
amplitude noise
noisesignal
LEARNING EXTENSION
)(112 t
dt
dvCRvo
ATORDIFFERENTIIDEAL
FCkR 2,1 12 WITH ATORDIFFERENTIIDEAL TO INPUT
s
Vm
3105
10
sFCR363
12 102102101
sFV
QF
Q
sV
sQ
V
A
V
SL ANALYSIDIMENSIONA
LEARNING EXTENSION FCkR 2.0,5 21 WITH INTEGRATOR ANTO INPUT
t
ioo dxxvCR
vtv021
)(1
)0()(
INTEGRATOR
sFV
QF
Q
sV
sQ
V
A
V
SL ANALYSIDIMENSIONA
DISCHARGEDINITIALLY IS CAPACITOR
sCR3
21 10
31 1020)(:1.00 tvst
t
o sVtdxxvty0
31 1020)()( sVy 3102)1.0(
31 1020)(:2.01.0 tvst
t
oo sVtdxxvyty1.0
331 )1.0(1020102)()1.0()(
)(1
)(21
tyCR
tv oo
USING GROUND WIRE
TO REDUCE
CROSSTALK
APPLICATION EXAMPLE CROSS-TALK IN INTEGRATED CIRCUITS
REDUCE CROSSTALK BY
• Reducing C12
• Increasing C2
SMALLER
COST?
EXTRA SPACE BY
GROUND WIRE
Simplified Model
LEARNING EXAMPLE SIMPLE CIRCUIT MODEL FOR DYNAMIC RANDOM ACCESS
MEMORY CELL (DRAM)
REPRESENTS CHARGE LEAKAGE
FROM CELL CAPACITOR
NOTICE THE VALUES OF THE
CAPACITANCES
ONE LOGIC A OF
STORAGE CORRECT FOR VVcell 5.1
t
CCC dxxiC
vv0
)(1
)0(
VtC
It
C
IV
cell
leak
cell
leakcell 5.15.13
sA
FVtH
3
12
15
105.11050
)(1050)(5.1
Ht/1 THAN HIGHERFREQUENCY A
AT”“REFRESHED BE MUSTCELL THE
THE ANALYSIS OF THE READ OPERATION
GIVES FURTHER INSIGHT ON THE
REQUIREMENTS
SWITCHED CAPACITOR CIRCUIT
CELL AT THE BEGINNING OF A MEMORY
READ OPERATION
CELL READ OPERATION IF SWITCH IS CLOSED BOTH CAPACITORS
MUST HAVE THE SAME VOLTAGE
ASSUMING NO LOSS OF CHARGE THEN
THE CHARGE BEFORE CLOSING MUST BE
EQUAL TO CHARGE AFTER CLOSING
FVFVQbefore1515 10503104505.1
)10500( 15FVQ afterafter
VVafter 65.1
Even at full charge the voltage variation is small.
SENSOR amplifiers are required
After a READ operation the cell must be refreshed
LEARNING EXAMPLE
IC WITH WIREBONDS TO THE OUTSIDE
FLIP CHIP MOUNTING
GOAL: REDUCE INDUCTANCE IN
THE WIRING AND REDUCE THE
“GROUND BOUNCE” EFFECTA SIMPLE MODEL CAN BE USED TO
DESCRIBE GROUND BOUNCE
MODELING THE GROUND BOUNCE EFFECT
)()( tdt
diLtV G
ballGB
IF ALL GATES IN A CHIP ARE CONNECTED TO A SINGLE GROUND THE CURRENT
CAN BE QUITE HIGH AND THE BOUNCE MAY BECOME UNACCEPTABLE
USE SEVERAL GROUND CONNECTIONS (BALLS) AND ALLOCATE A FRACTION OF
THE GATES TO EACH BALL
nHLball 1.0 s
Am
9
3
1040
1040
LEARNING BY DESIGN POWER OUTAGE “RIDE THROUGH” CIRCUIT
CAPACITOR MUST MAINTAIN AT LEAST 2.4V FOR
AT LEAST 1SEC.
v
DESIGN EQUATION
http://www.wiley.com/college/irwin/0470128690/animations/swf/6-21.swf
Proposed solution
DESIGN EXAMPLE DESIGN AN OP-AMP CIRCUIT TO REALIZE THE EQUATION
1 205 ( ) 2
t
Ov v d v
Needs integrator And adder
adder
Design equations Two equations in five unknowns!!
Not too large. Not too small
Seems a reasonable value 310R k
220R k
If supply voltages are 10V (or less) all currents will be less than 1mA, which seems
reasonable
ANALYSIS OF THE SOLUTION
integrator