CAPACITANCE AND INDUCTANCEIntroduces 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 REPRESENTATIONNOTICE 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

dAC

284

12

103141.610016.11085.855 mAAF

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

voltagecharge

One Farad(F)is the capacitance of a device that can store one Coulomb of charge at one Volt.

VoltCoulombFarad

EXAMPLEVoltage across a capacitor of 2 microFarads holding 10mC of charge

500010*1010*211 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

)()( tdtdvCti

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 lawdtdVC

dtdQi C

C

… 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 DETERMINEFC 5

LEARNING EXAMPLECAPACITOR AS CIRCUIT ELEMENT

Cv

Ci

)()( tdtdvCti c

C

t

CC dxxiC

tv )(1)(

t

t

tt

0

0

0

0

)(1)(1)(t t

tCCC dxxi

Cdxxi

Ctv

t

tCCC dxxi

Ctvtv

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)(

)()( tdtdvCti

mAsVFi 20

10624][105 3

6

mA60

CAPACITOR AS ENERGY STORAGE DEVICE

)()()( titvtp CCC Instantaneous power

)()( tdtdvCti c

C

dtdvtCvtp c

CC )()(

Ctqdxxi

Ctv C

t

CC)()(1)(

)()(1)( tdtdqtq

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.”

)(21)( 2 tv

dtdCtp CC

)(21)(

21),( 1

22

212 tCvtCvttw CCC

)(211)( 2 tq

dtd

Ctp cC

)(1)(1),( 12

22

12 tqC

tqC

ttw CCC

W

Cv

Ci

Energy stored in 0 - 6 msec

][)24(*][10*521)6,0( 226 VFwC

Charge stored at 3msec

)3()3( CC Cvq

)0(21)6(

21)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

dtdvCti

sVFi 3

6

10212102

sVFi 3

6

10412102

Energy stored at a given time t )(21)( 2 tCvtE C )240/1(E

2sin130*][10*2

21 226 F J

Charge stored at a given time )()( tCvtq CC )120/1(Cq 0])[sin(*][10*2 6 VC C

Current through the capacitor )(tdtdvCi 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

)(21)(

21),( 1

22

212 tCvtCvttw CC J

FC 2

)120(sin130)( ttv

)(tv

WHAT VARIABLES CAN BECOMPUTED?

SAMPLE PROBLEM

Cv

Ci

C

FC 2

][0;00;

)(5.0

mAtte

tit

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

tCCC dxxi

Ctvtv

0

)(1)()( 0

)2(Cv 2

0

5.01)0( dxeC

v xC

2

0

5.06 5.0

110*21

xe 61

6 10*6321.015.01

10*21

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;00);1(10

)(5.06

tte

tvt

C VElectric power supplied to capacitor )()()( titvtp CCC

Energy stored in capacitor at a given time )(21)( 2 tCvtw C

W

J

“Total” energy stored in the capacitor )(21 2 CT Cvw 6266 10)10(*10*2

21

Tw J

SAMPLE PROBLEMIf the current is known ...

SAMPLE PROBLEM

sec)(mt

5 10

Compute voltage as a function of timeAt minus infinity everything is zero. Sincecurrent is zero for t<0 we have

sec50 mttsAt

sAt

msAtiC ]/[10*3

10103

515)( 3

3

6

][10*410*3)(0)0(

06

3

VxdxtVVt

CC

][10*50];[810*3 323

stVt

In particular ][875][

8)10*5(*10*3)5(233

mVVmsVC

][10)(105 Atimst C

t

CC dxsAtVmVmsV310*5

66

3]/)[10*10(

10*41

810*75)(][

875)5(

][10*1010*5;][10*5410

810*75)( 333

3stVttVC

Charge stored at 5ms)()( tCVtq CC

][810*75*][10*4)5(

36 VFmsq

][)2/75()5( nCmsq

Total energy stored2

21

CCVE

Total means at infinity. Hence

][810*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;825

][105;5410

875

50;83

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

INDUCTORS NOTICE USE OF PASSIVE SIGN CONVENTION

Circuit representation for an inductor

A TIME VARYING MAGNETIC FLUXINDUCES A VOLTAGE

dtdvL

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

LLidtdiLv 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

dtdiLv L

L Differential form of induction law

t

LL dxxvL

ti )(1)(Integral form of induction law

00 ;)(1)()(0

ttdxxvL

titit

tLLL

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 )()()( titdtdiLtp LL

L

)(21 2 tLi

dtd

L

)(21)(

21),( 1

22

212 tLitLittw LL Energy stored on the interval

Can be positive or negative

)(21)( 2 tLitw LL

“Energy stored at time t”Must be non-negative. Passive element!!!

2

1

)(21),( 2

12

t

tLL dxxLi

dtdttw J Current in Amps, Inductance in Henrys

yield energy in Joules

LEARNING EXAMPLEFIND THE TOTLA ENERGY STORED IN THE CIRCUITIn steady state inductors act as short circuits and capacitors act as open circuits

2 21 12 2C C L LW CV W LI

9@ : 3 09 6A AV V

A A

81[ ]5AV V

2

6 10.86 3C AV V V

2 1.89A

L

VI A

1 2 13 1.2L L LI A I I A

1 1 19 6 16.2C L CV I V V

L=10mH. FIND THE VOLTAGE

sA

sAm 10

1021020

3

3

sAm 10

)()( tdtdiLtv

THE DERIVATIVE OF A STRAIGHT LINE IS ITSSLOPE

elsewhere

mstsAmstsA

dtdi

042)/(1020)/(10

mVVtvHL

sAtdtdi

10010100)(1010

)/(10)( 3

3

ENERGY STORED BETWEEN 2 AND 4 ms

)2(21)4(

21)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(21)2(

21)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

)(21)(

21),( 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 DETERMINEmHL 10 )()( t

dtdiLtv

mVv 100

sAHv 3

33

1021020][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

)()( tdtdiLtv

msmAm120

)/(122010 sAm

Vvm00

)/(34100 sAm

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

)()( tdtdvCti

nAsVF 600

233)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

)()( tdtdiLtv

SAHv 6

36

10201020010100

viivLC

IDEAL AND PRACTICAL ELEMENTS

IDEAL ELEMENTSCAPACITOR/INDUCTOR MODELSINCLUDING LEAKAGE RESISTANCE

)(ti

)(tv

)()()( tdtdvC

Rtvti

leak

MODEL FOR “LEAKY”CAPACITOR

)(ti

)(tv

)()()( tdtdiLtiRtv leak

MODEL FOR “LEAKY”INDUCTORS

)(tv

)(tv

)(ti )(ti

)()( tdtdvCti )()( t

dtdiLtv

SERIES CAPACITORS

NOTICE SIMILARITYWITH RESITORS INPARALLEL

21

21

CCCCCs

Series Combination of twocapacitors

F6 F3 SCF2

F2

F1

6123

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 EXAMPLE Two uncharged capacitors are connected as shown.Find the unknown capacitance

1C FIND

CVFQCVQ 72)6)(12(

V18

FVCC 4

1872

1

PARALLEL CAPACITORS

)()( tdtdvCti kk

)(tiLEARNING EXAMPLE

PC 4 6 2 3 15 F

LEARNING EXTENSION

F6

F2

F3

F4

F12

eqC

F4

F3FCeq 23

F4 ARECAPACITORSALL FIND EQUIVALENT CAPACITANCE

F8

F8

eqC F8

F8

F4

F1232

3328

38

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

)()( tdtdiLtv kk

)()( tdtdiLtv 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 NODESPLACE 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

mH6mH6

mH6

eqL

mHLeq 7246

144866||)26(6

mHLeq 766

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

Rvi O

t

dxxiC

iRtv )(1)( 11

111

KVL

)(2

1 Rvi oo1 vof terms in i replace

DIFFERENTIATE

)(1111

11 tdtdvCi

dtdiCR

)(11211 tdtdvCRv

dtdvCR 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 ttdtdy

SIMPLE MODEL FOR A NOISY 60Hz SINUSOIDCORRUPTED WITH ONE MICROVOLT OF 1GHzINTERFERENCE.

)102sin(10)120sin()( 96 ttty 610amplitude signal

amplitude noise

noisesignal

67.161202000 amplitude signal

amplitude noise

noisesignal

LEARNING EXTENSION

)(112 tdtdvCRvo

ATORDIFFERENTIIDEAL

FCkR 2,1 12 WITH ATORDIFFERENTIIDEAL TO INPUT

sVm 3105

10

sFCR 36312 102102101

sFVQF

QsV

sQV

AV

SL ANALYSIDIMENSIONA

LEARNING EXTENSION FCkR 2.0,5 21 WITH INTEGRATOR ANTO INPUT

t

ioo dxxvCR

vtv021

)(1)0()(

INTEGRATOR

sFVQF

QsV

sQV

AV

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(

VtCIt

CIV

cell

leak

cell

leakcell 5.15.13

sA

FVtH3

12

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

)()( tdtdiLtV G

ballGB

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 sAm 9

3

10401040

LEARNING BY DESIGNPOWER OUTAGE “RIDE THROUGH” CIRCUITCAPACITOR 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 ( ) 2t

Ov v d v Needs integrator And adder

adder

Design equationsTwo equations in five unknowns!!

Not too large. Not too smallSeems a reasonable value 3 10R k

2 20R k

If supply voltages are 10V (or less) all currents will be less than 1mA, which seemsreasonable

ANALYSIS OF THE SOLUTION

integrator