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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
−
−
××
=
