capacitance and inductance
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
CAPACITANCE AND INDUCTANCE
Introduces two passive, energy storing devices: Capacitors and Inductors
LEARNING GOALS
CAPACITORSStore energy in their electric field (electrostatic energy)Model as circuit element
INDUCTORSStore energy in their magnetic fieldModel as circuit element
CAPACITOR AND INDUCTOR COMBINATIONSSeries/parallel combinations of elements
RC OP-AMP CIRCUITSIntegration and differentiation circuits
CAPACITORSFirst 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 faradsare not unusual
d
AC
284
12
103141.610016.1
1085.855 mA
AF
PLATE SIZE FOR EQUIVALENT AIR-GAP CAPACITOR
gapin material ofconstant Dielectric 電介質
Basic capacitance law )( CVfQ Linear capacitors obey Coulomb’s law CCVQ C is called the CAPACITANCE of the device and hasunits of
voltagecharge
One Farad(F)is the capacitance of a device that can store one Coulomb of charge at one Volt.
VoltCoulomb
Farad
EXAMPLEVoltage across a capacitor of 2 microFarads holding 10mC of charge
500010*1010*2
11 36
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 thereis a displacement current
CC CVQ Capacitance Law
One can also express the voltage across in terms of the current
QC
tVC1
)(
t
C dxxiC
)(1
Integral form of Capacitance law
dt
dVC
dt
dQi CC
… Or one can express the current throughin terms of the voltage across
Differential form of Capacitance law
The mathematicalimplication of the integralform is ...
ttVtV CC );()(
Voltage across a capacitorMUST be continuous
Implications of differential form??
0 CC iConstVDC 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
tCCC dxxi
Cdxxi
Ctv
t
t
CCC dxxiC
tvtv0
)(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
tCC dxxpttw
If t1 is minus infinity we talk about“energy stored at time t2.”
If both limits are infinity then we talkabout the “total energy stored.”
)(
2
1)( 2 tv
dt
dCtp CC
)(2
1)(
2
1),( 1
22
212 tCvtCvttw CCC
)(
2
11)( 2 tq
dt
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 Coulombsand capacitance in Faradsthen the energy is in ….
FC 5
LEARNING EXAMPLE
VOLTAGETHE FIND .4 FC
0 ) 0( v
20 t
mst 42 ][1082)( 3 Vttv
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 EXTENSIONCURRENT THE DETERMINE
FC 2)()( t
dt
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
tvt
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
tvtv0
)(1
)()( 0
)2(Cv 2
0
5.01)0( dxeC
v xC
2
0
5.06 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
tvt
CC )(1
)(
0;0)( ttvC t
xCC dxe
Cvtv
0
5.01)0()(
0;0
0);1(10)(
5.06
t
tetv
t
C VElectric 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 PROBLEMIf 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
66
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
236 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
30;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
INDUCTORS NOTICE USE OF PASSIVE SIGN CONVENTION
Circuit representation for an inductor
A TIME VARYING MAGNETIC FLUXINDUCES 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 ISPROPORTIONAL TO THE CURRENT
LLidt
diLv L
L DIFFERENTIAL FORM OF INDUCTION LAW
THE PROPORTIONALITY CONSTANT, L, ISCALLED THE INDUCTANCE OF THE COMPONENT
INDUCTANCE IS MEASURED IN UNITS OFhenry (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
titit
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
tLL dxxLi
dt
dttw J Current in Amps, Inductance in Henrys
yield energy in Joules
LEARNING EXAMPLEFIND 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 LW CV W LI
9@ : 3 0
9 6A A
V VA A
81[ ]
5AV V
2
610.8
6 3C AV V V
21.8
9A
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 ITSSLOPE
elsewhere
mstsA
mstsA
dt
di
0
42)/(10
20)/(10
mVVtvHL
sAtdt
di10010100)(
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 THEINDUCTOR IS SUPPLYING ENERGYPREVIOUSLY 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 COMPUTATIONSSAMPLE 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 ENERGYSTORED (AS FUNCTION OF TIME)
)(tv
FOR ENERGY STORED IN THE INDUCTOR
)(twL
NOTICE THAT ENERGY STORED ATANY 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 CHANGESSIGN
ENERGY IS NEVER NEGATIVE.THE DEVICE IS PASSIVE
FIND THE POWER
FIND THE ENERGY
L=5mHFIND 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 WAVEFORMDETERMINE 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 1RATINGS INDUCTOR STANDARD
%10%,5 TOLERANCE STANDARD
INDUCTOR SPECIFICATIONS
LEARNING EXAMPLE
%10100 HL
GIVEN THE CURRENT WAVEFORMDETERMINE THE VARIATIONS INVOLTAGE
)()( tdt
diLtv
S
AHv 6
36
1020
1020010100
vi
iv
LC
IDEAL AND PRACTICAL ELEMENTS
IDEAL ELEMENTSCAPACITOR/INDUCTOR MODELSINCLUDING 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 SIMILARITYWITH RESITORS INPARALLEL
21
21
CC
CCCs
Series Combination of twocapacitors
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 THEINITIAL 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 EXAMPLETwo uncharged capacitors are connected as shown.Find the unknown capacitance
1C FIND
CVFQCVQ 72)6)(12(
V18
FV
CC
418
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 CDETERMINE THE VARIOUS EQUIVALENT CAPACITANCES
SAMPLE PROBLEM
All capacitors are equalwith 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 RESISTORSCAPACITORS COMBINE LIKE CONDUCTANCES
LEARNING EXAMPLE
mH4 mH2
N
jj titi
100 )()(
AAAAti 1263)( 0
LEARNING EXTENSION
mH2
mH2WHEN 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 BETWEENPHYSICAL LAYOUT AND ELECTRICALCONNECTIONS
6||6||6
a
b
c
a
b
c
SELECTED LAYOUT
a
b
c
mH2
mH6
mH6
mH6
eqL
mHLeq 7
246
14
4866||)26(6
mHLeq 7
66
LEARNING EXTENSION
L-C
RC OPERATIONAL AMPLIFIER CIRCUITS
INTRODUCES TWO VERY IMPORTANT PRACTICAL CIRCUITSBASED 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
21
R
vi O
t
dxxiC
iRtv )(1
)( 11
111
KVL
)(2
1 R
vi oo1 vof terms in i replace
DIFFERENTIATE
)(111
111 t
dt
dvCi
dt
diCR
)(11211 tdt
dvCRv
dt
dvCR o
o
IF R1 COULD BE SET TO ZERO WE WOULD HAVEAN IDEAL DIFFERENTIATOR.IN PRACTICE AN IDEAL DIFFERENTIATOR AMPLIFIESELECTRIC NOISE AND DOES NOT OPERATE.THE RESISTOR INTRODUCES A FILTERINGACTION. ITS VALUE IS KEPT AS SMALL ASPOSSIBLE TO APPROXIMATE A DIFFERENTIATOR
ABOUT ELECTRIC NOISE
ALL ELECTRICAL SIGNALS ARE CORRUPTED BYEXTERNAL, UNCONTROLLABLE AND OFTENUNMEASURABLE, SIGNALS. THESE UNDESIREDSIGNALS ARE REFERRED TO AS NOISE
THE DERIVATIVE
)102cos(2000)120cos(120)( 9 ttdt
dy
SIMPLE MODEL FOR A NOISY 60Hz SINUSOIDCORRUPTED WITH ONE MICROVOLT OF 1GHzINTERFERENCE.
)102sin(10)120sin()( 96 ttty 610
amplitude signalamplitude noise
noisesignal
67.16120
2000
amplitude signalamplitude noise
noisesignal
LEARNING EXTENSION
)(112 tdt
dvCRvo
ATORDIFFERENTIIDEAL
FCkR 2,1 12 WITH ATORDIFFERENTIIDEAL TO INPUT
s
Vm 3105
10
sFCR 36312 102102101
sFV
QF
Q
sV
sQV
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
sQV
A
V
SL ANALYSIDIMENSIONA
DISCHARGEDINITIALLY IS CAPACITOR
sCR 321 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 WIRETO 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 EXAMPLESIMPLE CIRCUIT MODEL FOR DYNAMIC RANDOM ACCESSMEMORY CELL (DRAM)
REPRESENTS CHARGE LEAKAGEFROM CELL CAPACITOR
NOTICE THE VALUES OF THECAPACITANCES
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
312
15
105.11050
)(1050)(5.1
Ht/1 THAN HIGHERFREQUENCY A
AT”“REFRESHED BE MUSTCELL THE
THE ANALYSIS OF THE READ OPERATIONGIVES 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 CAPACITORSMUST HAVE THE SAME VOLTAGE
ASSUMING NO LOSS OF CHARGE THENTHE 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” EFFECT A SIMPLE MODEL CAN BE USED TO
DESCRIBE GROUND BOUNCE
MODELING THE GROUND BOUNCE EFFECT
)()( tdt
diLtV GballGB
IF ALL GATES IN A CHIP ARE CONNECTED TO A SINGLE GROUND THE CURRENTCAN BE QUITE HIGH AND THE BOUNCE MAY BECOME UNACCEPTABLE
USE SEVERAL GROUND CONNECTIONS (BALLS) AND ALLOCATE A FRACTION OFTHE GATES TO EACH BALL
nHLball 1.0 s
Am 9
3
1040
1040
LEARNING BY DESIGNPOWER OUTAGE “RIDE THROUGH” CIRCUIT
CAPACITOR MUST MAINTAIN AT LEAST 2.4V FORAT LEAST 1SEC.
v
DESIGN EQUATION
http://www.wiley.com/college/irwin/0470128690/animations/swf/6-21.swf
Proposed solution
DESIGN EXAMPLEDESIGN AN OP-AMP CIRCUIT TO REALIZE THE EQUATION
1 205 ( ) 2
t
Ov v d v
Needs integrator And adder
adder
Design equationsTwo 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 seemsreasonable
ANALYSIS OF THE SOLUTION
integrator