CAPACITANCE AND INDUCTANCEIntroduces two passive, energy storing devices: Capacitors and Inductors

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

CAPACITORS First 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 ε

=

284

12103141.6

10016.11085.855 mAAF ×=⇒×

×= −

−

PLATE SIZE FOR EQUIVALENT AIR-GAP CAPACITOR

gapinmaterialofconstant Dielectric ε

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 =

EXAMPLE Voltage 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 =

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

dtdVC

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 μ5=

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)( tdt

dqtqC

tp CCC =

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

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

−=

Charge stored at 3msec

)3()3( CC Cvq =

)0(21)6(

21)6,0( 22

CCC CvCvw −=

CVFqC μ60][12*][10*5)3( 6 == −

FC μ5=

EXAMPLE

VOLTAGETHE FIND .4 FC μ=

20 ≤≤ t

mst 42 ≤<][1082)( 3 Vttv −×+−=

0;)(1)0()(0

>+= ∫ tdxxiC

vtvt

2;)(1)2()(2

>+= ∫ tdxxiC

vtvt

0)0( =v

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

dtdvLφ

= 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

⇒= LLiφdtdiLv 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 =

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 L

LL = ⎟

⎠⎞

⎜⎝⎛= )(

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 =

∫ ⎟⎠⎞

⎜⎝⎛=

2

1

)(21),( 2

12

t

tLL dxxLi

dtdttw J Current in Amps, Inductance in Henrys

yield energy in Joules

EXAMPLE FIND THE TOTAL ENERGY STORED IN THE CIRCUIT

In 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 6

A AV VA A −− + + =

2

6 10.86 3C AV V V= =+

2 1.89

AL

VI A= =

1 2 13 1.2L L LI A I I A+ = ⇒ =−

1 1 19 6 16.2C L CV I V V= − ⇒ =

][581VVA =

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)/(10

20)/(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

EXAMPLE

CAPACITOR SPECIFICATIONS

VALUESSTANDARD INRANGE ECAPACITANC mFCFp 50≈≈

VV 5003.6 −RATINGS CAPACITOR STANDARD

%20%,10%,5 ±±±TOLERANCE STANDARD

VALUESSTANDARDINRANGESINDUCTANCE mHLnH 1001 ≤≈≤≈

AmA 1≈−≈RATINGSINDUCTOR STANDARD

%10%,5 ±±TOLERANCE STANDARD

INDUCTOR SPECIFICATIONS

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

Fμ6 Fμ3 =SCFμ2

PARALLEL CAPACITORS

)()( tdtdvCti kk =

)(ti

PC 4 6 2 3 15 Fμ= + + + =

SERIES INDUCTORS

)()( tdtdiLtv kk =

)()( tdtdiLtv S=

=eqL H7

PARALLEL INDUCTORS

)(ti

INDUCTORS COMBINE LIKE RESISTORSCAPACITORS COMBINE LIKE CONDUCTANCES

mH4 mH2∑=

=N

jj titi

100 )()( AAAAti 1263)( 0 −=+−=

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

−

−

××

=