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CAPACITANCE AND INDUCTANCE
Introduces two passive, energy storing devices: Capacitors and Inductors
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
CAPACITORS
First of the energy storage devices to be discussed
Typical Capacitors
Basic parallel-plates capacitor
CIRCUIT REPRESENTATION
NOTICE USE OF PASSIVE SIGN CONVENTION
C
A
d
 Dielectric constant of material in gap
PLATE SIZE FOR EQUIVALENT AIR-GAP CAPACITOR
8.85  1012 A
8 2
55F 

A

6
.
3141

10
m
4
1.016  10
Normal values of capacitance are small.
Microfarads is common.
For integrated circuits nano or pico farads
are not unusual
Basic capacitance law
Q  f (VC )
Linear capacitors obey Coulomb’s law
Q  CVC
C is called the CAPACITANCE of the device and has
units of
charge
voltage
One Farad(F)is the capacitance of a device that can
store one Coulomb of charge at one Volt.
Coulomb
Farad 
Volt
Linear capacitor circuit representation
EXAMPLE Voltage across a capacitor of 2 micro
Farads holding 10mC of charge
VC 
1
1
3
Q
10
*
10
 5000
6
C
2 *10
V
Capacitance in Farads, charge in Coulombs
result in voltage in Volts
Capacitors can be dangerous!!!
Capacitors only store and release
ELECTROSTATIC energy. They do not “create”
The capacitor is a passive element
and follows the passive sign convention
Linear capacitor circuit representation
i (t )  C
dv
(t )
dt
QC  CVC
Capacitance Law
If the voltage varies the charge varies and there
is a displacement current
One can also express the voltage across
in terms of the current
1
VC (t )  Q
C
… Or one can express the current through
in terms of the voltage across
t
1
  iC ( x)dx
C 
Integral form of Capacitance law
The mathematical
implication of the integral
form is ...
VC (t )  VC (t ); t
Voltage across a capacitor
MUST be continuous
iC 
dV
dQ
C C
dt
dt
Differential form of Capacitance law
Implications of differential form??
VC  Const  iC  0
DC or steady state behavior
A capacitor in steady state acts as an
OPEN CIRCUIT
CAPACITOR AS CIRCUIT ELEMENT
iC

LEARNING EXAMPLE
C  5F
DETERMINE THE CURRENT
vC

iC (t )  C
i (t )  C
dvc
(t )
dt
1
iR  v R
R
v R  RiR
t
1
vC (t )   iC ( x)dx
C 
t



t0


Ohm’s Law
t

t0
vc ( t O )
t
1 0
1 t
vC (t )   iC ( x )dx   iC ( x )dx
C 
C t0
t
1
vC (t )  vC (t0 )   iC ( x)dx
C t0
The fact that the voltage is defined through
an integral has important implications...
dv
(t )
dt
 60mA
i  5  106 [ F ] 
24 V 
 20mA
6  103  s 
i ( t )  0 elsewhere
CAPACITOR AS ENERGY STORAGE DEVICE
iC

vC
Instantaneous power
pC (t )  vC (t )iC (t )
W

dv c
iC (t )  C
(t )
dt
dvc
pC (t )  CvC (t )
dt
t
q (t )
1
vC (t )   iC ( x)dx  C
C 
C
pC (t ) 
dq
1
qC (t ) C (t )
C
dt
d  1 2  Energy is the integral of power
t
pC ( t )  C  vC ( t ) 
dt  2
 w ( t , t )  p ( x )dx p ( t )  1 d  1 q 2 ( t ) 
C
c
C 2 1
 C
C
dt
2


t
2
1
If t1 is minus infinity we talk about
“energy stored at time t2.”
1
1
wC (t 2 , t1 )  CvC2 (t 2 )  CvC2 (t1 )
2
2
If both limits are infinity then we talk
about the “total energy stored.”
1 2
1 2
wC (t 2 , t1 )  qC (t2 )  qC (t1 )
C
C
Energy stored in 0 - 6 msec
C  5F
1
1
wC (0,6)  CvC2 (6)  CvC2 (0)
2
2
wC (0,6) 
1
5 *10 6 [ F ] * (6) 2 [V 2 ]
2
Charge stored at 3msec
qC (3)  CvC (3)
EXAMPLE
qC (3)  5 *106[F ]*12[V ]  60C
C  4 F . FIND THE VOLTAGE
v(0)  0
1t
v (t )  v (0)   i ( x )dx; t  0
C0
0t 2
v (t )  v (2) 
t
1
i ( x )dx; t  2

C2
3
v (t )  2t  8 10 [V ]
2  t  4ms
INDUCTORS
Flux lines may extend
beyond inductor creating
stray inductance effects
NOTICE USE OF
PASSIVE SIGN CONVENTION
Circuit representation
for an inductor
A TIME VARYING FLUX
CREATES A COUNTER EMF
AND CAUSES A VOLTAGE
TO APPEAR AT THE
TERMINALS OF THE
DEVICE
A TIME VARYING MAGNETIC FLUX
INDUCES A VOLTAGE
vL 
d
dt
Induction law
FOR A LINEAR INDUCTOR THE FLUX IS
PROPORTIONAL TO THE CURRENT
  LiL 
di L
vL  L
dt
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
HENRY 
Volt
Amp
sec
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
Follow passive sign convention
di
vL  L L
dt
Differential form of induction law
t
1
iL (t )   vL ( x)dx
L 
Integral form of induction law
t
1
iL (t )  iL (t0 )   vL ( x)dx; t  t0
L t0
A direct consequence of integral form i
L
(t )  iL (t ); t
A direct consequence of differential form i
L
Current MUST be continuous
 Const.  vL  0
DC (steady state) behavior
Power and Energy stored
pL (t )  vL (t )iL (t )
t2
w L (t 2 , t1 )  
t1
W
pL (t )  L
d 1 2

 LiL ( x ) dx
dt  2

1 2
1 2
w (t 2 , t1 )  LiL (t 2 )  LiL (t1 )
2
2
1
w L (t )  LiL2 (t )
2
J
di L
d 1

(t )iL (t )   LiL2 (t ) 
dt
dt  2

Current in Amps, Inductance in Henrys
yield energy in Joules
Energy stored on the interval
Can be positive or negative
EXAMPLE
FIND THE TOTAL ENERGY STORED IN THE CIRCUIT
In steady state inductors act as
short circuits and capacitors act
as open circuits
1
1 2
2
WC  CVC W L  LI L
2
2
@ A : 3 A 
VA VA  9

0
9
6
VA 
I L1  3 A  I L 2  I L1  1.2 A V  6 V  10.8V
VC 1  9  6 I L1  VC 1  16.2V
C2
6 3
A
VA
I L2 
 1.8 A
9
81
[V ]
5
EXAMPLE
L=10mH. FIND THE VOLTAGE
v (t )  L
20  103 A
 A
m

10
 s 
2  103 s
di
(t )
dt
 A
m  10  
s
THE DERIVATIVE OF A STRAIGHT LINE IS ITS
SLOPE
 10( A / s) 0  t  2ms
di 
  10( A / s) 2  t  4ms
dt 
0 elsewhere

di

(t )  10( A / s)
3
dt
  v (t )  100 10 V  100mV
L  10 103 H 

ENERGY STORED BETWEEN 2 AND 4 ms
1
1
w (4,2)  LiL2 (4)  LiL2 (2)
2
2
w(4,2)  0  0.5 *10 *103 (20 *103 )2
THE VALUE IS NEGATIVE BECAUSE THE
INDUCTOR IS SUPPLYING ENERGY
PREVIOUSLY STORED
J
CAPACITOR SPECIFICATIONS
CAPACITANCE RANGE p F  C  50mF
IN STANDARDVALUES
STANDARD CAPACITOR RATINGS
6.3V  500V
STANDARD TOLERANCE
 5%,  10%,  20%
INDUCTOR SPECIFICATIONS
INDUCTANCERANGES  1nH  L  100mH
IN STANDARDVALUES
STANDARD INDUCTORRATINGS
 mA   1A
STANDARD TOLERANCE
 5%,  10%
CL
vi
iv
IDEAL AND PRACTICAL ELEMENTS

i (t )
i (t )
i (t )
i (t )



v (t )
v (t )


v (t )
v (t )

IDEAL ELEMENTS
i (t )  C
dv
(t )
dt
v (t )  L
CAPACITOR/INDUCTOR MODELS
INCLUDING LEAKAGE RESISTANCE
di
(t )
dt
i (t ) 
v (t )
dv
 C (t )
Rleak
dt
MODEL FOR “LEAKY”
CAPACITOR

v (t )  Rleaki (t )  L
di
(t )
dt
MODEL FOR “LEAKY”
INDUCTORS
SERIES CAPACITORS
C1C2
Cs 
C1  C2
Series Combination of two
capacitors
6F
3F
CS 
2 F
NOTICE SIMILARITY
WITH RESITORS IN
PARALLEL
PARALLEL CAPACITORS
ik ( t )  C k
dv
(t )
dt
i (t )
CP
 4  6  2  3  15 F
SERIES INDUCTORS
v (t )  LS
vk (t )  Lk
di
(t )
dt
Leq 
7H
di
(t )
dt
PARALLEL INDUCTORS
i (t )
N
i (t0 )   i j (t0 )
j 1
4mH
2mH
i (t0 )  3 A  6 A  2 A  1A
INDUCTORS COMBINE LIKE RESISTORS
CAPACITORS COMBINE LIKE CONDUCTANCES
LEARNING EXAMPLE
FLIP CHIP MOUNTING
IC WITH WIREBONDS TO THE OUTSIDE
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
VGB (t )  Lball
Lball  0.1nH
diG
(t )
dt
40  103 A
m
40  109 s
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