Transcript Linear capacitor circuit representation The is a

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
CAPACITOR AND INDUCTOR COMBINATIONS
Series/parallel combinations of elements
RC OP-AMP CIRCUITS
Integration and differentiation circuits
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 電介質 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
LEARNING BY DOING
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  1
C
C
… Or one can express the current through
in terms of the voltage across
t
 iC ( x)dx

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

vC
LEARNING EXAMPLE
C  5F
DETERMINE THE CURRENT

iC (t )  C
i (t )  C
dvc
(t )
dt
1
iR  v R
R
vR  Ri R
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

dvc
iC (t )  C
(t )
dt
dvc
pC (t )  CvC (t )
dt
t
q (t )
1
vC (t )   iC ( x)dx  C
C 
C
dq
1
pC (t )  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 2 1
C
c
 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 (t 2 )  qC (t1 )
C
C
Energy stored in 0 - 6 msec
C  5F
1
1
wC (0,6)  CvC2 (6)  CvC2 (0)
2
2
1
wC (0,6)  5 *106 [ F ] * (24) 2 [V 2 ]
2
Charge stored at 3msec
qC (3)  CvC (3)
LEARNING EXAMPLE
qC (3)  5 *10 6 [ F ] *12[V ]  60C
“total energy stored?” ....
“total charge stored?” ...
If charge is in Coulombs
and capacitance in Farads
then the energy is in ….
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
C  4 F . FIND THE POWER
i (t )  8 103 t
v (0)  0
p(t )  8t 3 , 0  t  2ms
p(t )  0, elsewhere
2  t  4ms
FIND THE ENERGY
p(t )  8t 3 , 0  t  2ms
p(t )  0, elsewhere
2  t  4ms
LEARNING EXTENSION
C  2 F
DETERMINE THE CURRENT
i  2 106 F 
i  2 106 F 
12 V 

3  
2 10  s 
 12 V 
4 103  s 
i (t )  C
dv
(t )
dt
SAMPLE PROBLEM

C  2 F
v (t )

v ( t )  130 sin (120 t )
WHAT VARIABLES CAN BE
COMPUTED?
Charge stored at a given time
1
E (t )  CvC2 (t ) E (1 / 240)  1 2 *10 6 [ F ] *1302 sin 2   
2
2
2
qC (1 / 120)  2 *106 [C ] * sin(  )[V ]  0 C
qC (t )  CvC (t )
Current through the capacitor
iC  C
Energy stored at a given time t
dvC
(t )
dt
Electric power supplied to capacitor at a given time
Energy stored over a given time interval
iC (1 / 120)  2 *106 *130 *120 cos( )
pC (t )  vC (t )iC (t )
W
1
1
w (t 2 , t1 )  CvC2 (t 2 )  CvC2 (t1 )
2
2
J
J
A
If the current is known ...
SAMPLE PROBLEM
iC
C
Current through capacitor

vC
e 0.5t ; t  0
iC (t )  
[mA]
 0; t  0

C  2F
t
Voltage at a given time t
1
vC (t )   iC ( x)dx
C 
vC (0)  0[V ]
Voltage at a given time t when voltage at time to<t is also known
2
1
1 0.5 x

v
(
0
)

e
dx
vC (2)  C
2 *106
C 0
Charge at a given time
2
1
1
 1 0.5 x 
1
6



1

e

0
.
6321
*
10

e
V

6
 0.5
 2 *10 0.5
0
qC (t )  CvC (t )
qC (2)  2 * 0.6321
t
Voltage as a function of time
vC (t ) 
Electric power supplied to capacitor
t
1
vC (t )  vC (t0 )   iC ( x)dx
C t0
1
iC ( x)dx

C 
vC (t )  0; t  0
pC (t )  vC (t )iC (t )
Energy stored in capacitor at a given time w (t ) 
W
1 2
CvC (t ) J
2
“Total” energy stored in the capacitor wT  1 CvC2 ()
2
C
t
1
vC (t )  vC (0)   e 0.5 x dx
C0
106 (1  e 0.5t ); t  0
vC (t )  
0; t  0

1
wT  2 *106 * (106 ) 2  106
2
J
V
SAMPLE PROBLEM
Given current and capacitance
Compute voltage as a function of time
At minus infinity everything is zero. Since
current is zero for t<0 we have
0  t  5m sec 
VC (0)  0  VC (t ) 
5
10
VC (t )  0; t  0
15 A
106 A
iC (t ) 
t  3 3 t  3 *103[ A / s] t
5 ms
10 s
3 t
3
3 *10
xdx [V ]  3 *10 t 2 [V ]; 0  t  5 *103[ s ]
6 
4 *10 0
8
In particular
3 *103 * (5 *103 ) 2
75
VC (5ms) 
[V ]  [mV ]
8
8
t (m sec)
5  t  10 ms  iC (t )  10 [A]
t
75
75 *103
1
VC (5ms )  [mV ]  VC (t ) 

(10 *106 )[ A / s]dx
6 
8
8
4 *10 5*103
Charge stored at 5ms
qC (t )  CVC (t )
75 *103
q(5ms)  4 *10 [ F ] *
[V ]
8
6
q(5ms)  (75 / 2) [nC ]
75 *103 10
VC (t ) 

t  5 *103 [V ]; 5 *103  t  10 *103 [ s]
8
4

Total energy stored
1
E  CVC2
2
Total means at infinity. Hence

Before looking into a formal way to describe the current
we will look at additional questions that can be answered.
2
3

 6  25 *10
 [ J ]
ET  0.5 * 4 *10 
8


Now, for a formal way to represent piecewise functions....
Formal description of a piecewise analytical signal
0;

 3 2
t ;

8

Vc (t )   75 10
 t  5;
8 4

25
 ;

8

t0
0  t  5ms
5  t  10 [ms]
t  10[ms]
[mV ]
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
LEARNING by Doing
Induction law
FOR A LINEAR INDUCTOR THE FLUX IS
PROPORTIONAL TO THE CURRENT
  LiL 
diL
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
p L (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
diL
d 1

(t )iL (t )   LiL2 (t ) 
dt  2
dt

Current in Amps, Inductance in Henrys
yield energy in Joules
Energy stored on the interval
Can be positive or negative
“Energy stored at time t”
Must be non-negative.
Passive element!!!
LEARNING EXAMPLE
FIND THE TOTLA 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 WL  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
LEARNING 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
SAMPLE PROBLEM L=0.1H, i(0)=2A. Find i(t), t>0
ENERGY COMPUTATIONS
v (V )
1 2
1 2
w (t 2 , t1 )  LiL (t 2 )  LiL (t1 )
2
2
2
1t
i (t )  i (0)   v ( x )dx
L0
2 t (s)
Energy stored on the interval
Can be positive or negative
Initial energy stored in inductor
w(0)  0.5 * 0.1[ H ](2 A)2  0.2[ J ]
t
v ( x )  2   v ( x )dx  2t ; 0  t  2
0
L  0.1H  i (t )  2  20t ; 0  t  2 s
“Total energy stored in the inductor”
w()  0.5 * 0.1[ H ] * (42 A)2  88.2J
v ( x )  0; t  2  i (t )  i (2); t  2 s
Energy stored between 0 and 2 sec
42
1
1
w (2,0)  LiL2 (2)  LiL2 (0)
2
2
i (A)
w(2,0)  0.5 * 0.1* (42)2  0.5 * 0.1* (2)2
w (2,0)  88[ J ]
2
2
t (s)
LEARNING EXAMPLE
FIND THE VOLTAGE ACROSS AND THE ENERGY
STORED (AS FUNCTION OF TIME)
v (t )
FOR ENERGY STORED IN THE INDUCTOR
w L (t )
NOTICE THAT ENERGY STORED AT
ANY GIVEN TIME IS NON NEGATIVE
-THIS IS A PASSIVE ELEMENT-
LEARNING EXAMPLE
L  10mH
DETERMINE THE VOLTAGE
v  100mV
20 103  A 
v  10 10 [ H ] 
2 103  s 
3
v (t )  L
di
(t )
dt
LEARNING EXAMPLE
FIND THE CURRENT
i (t )
L=200mH
1t
i (t )  i (0)   v ( x )dx; t  0
L0
v (t )  0; t  0  i (0)  0
i (t )
L=200mH
FIND THE POWER
NOTICE HOW POWER CHANGES
SIGN
i (t )
POWER
p(t )
ENERGY
w (t )
FIND THE ENERGY
ENERGY IS NEVER NEGATIVE.
THE DEVICE IS PASSIVE
LEARNING EXTENSION
L=5mH
FIND THE VOLTAGE
m
20mA
1ms
m
10  20
( A / s)
2 1
v  50mV
m0
v  0V
v  5 103 ( H )  20( A / s); 0  t  1ms  100mV
v (t )  L
m
di
(t )
dt
0  10
( A / s)
43
v  50mV
CAPACITOR SPECIFICATIONS
CAPACITANC E RANGE p F  C  50mF
IN STANDARD VALUES
STANDARD CAPACITOR RATINGS
6.3V  500V
STANDARD TOLERANCE
 5%,  10%,  20%
Nominal current
300nA
100  109 F
(3)  3 V 
 600nA
3  2  s 
300nA
LEARNING EXAMPLE
C  100nF  20%
i (t )  C
dv
(t )
dt
GIVEN THE VOLTAGE WAVEFORM
DETERMINE THE VARIATIONS IN
CURRENT
INDUCTOR SPECIFICATIONS
CURRENT WAVEFORM
INDUCTANCE RANGES  1nH  L  100mH
IN STANDARD VALUES
200 103  A 
v  100 10 H 
20 106  S 
6
STANDARD INDUCTOR RATINGS
 mA   1A
STANDARD TOLERANCE
 5%,  10%
s
LEARNING EXAMPLE
L  100 H  10%
v (t )  L
di
(t )
dt
GIVEN THE CURRENT WAVEFORM
DETERMINE THE VARIATIONS IN
VOLTAGE
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 )  Rleak i (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
LEARNING EXAMPLE
1 F
DETERMINE EQUIVALENT
CAPACITANCE AND THE
INITIAL VOLTAGE
2 F

3  2 1
6
OR WE CAN REDUCE TWO AT A TIME
 2V  4V  1V
ALGEBRAIC SUM OF INITIAL VOLTAGES
POLARITY IS DICTATED BY THE REFERENCE
DIRECTION FOR THE VOLTAGE
LEARNING EXAMPLE

8V
+
-
12V
Two uncharged capacitors are connected as shown.
Find the unknown capacitance
FIND C1
30  F


4V


18V

C
SAME CURRENT. CONNECTED FOR THE SAME TIME PERIOD
SAME CHARGE ON BOTH CAPACITORS
Q  (30 F )(8V )  240C
Q  CV  Q  (12F )(6V )  72C
C1 
72C
 4 F
18V
PARALLEL CAPACITORS
ik ( t )  C k
dv
(t )
dt
i (t )
LEARNING EXAMPLE
C P  4  6  2  3  15 F
LEARNING EXTENSION
6 F
2 F
3 F
C eq 
3
C eq   F
2
4 F
12  F
3 F
4 F
SAMPLE PROBLEM
FIND EQUIVALENT CAPACITANCE
ALL CAPACITORS ARE 4 F
8 F
8 F
4 F
C eq
8
32
8 
3
3
32
F
12
8 F
8 F
SAMPLE PROBLEM
IF ALL CAPACITORS HAVE THE SAME CAPACITANCE VALUE C
DETERMINE THE VARIOUS EQUIVALENT CAPACITANCES
Examples of interconnections
C EQ
CAB  ______
All capacitors are equal
with C=8 microFarads
SERIES INDUCTORS
v (t )  LS
vk (t )  Lk
di
(t )
dt
LEARNING EXAMPLE
Leq 
7H
di
(t )
dt
PARALLEL INDUCTORS
i (t )
LEARNING EXAMPLE
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 EXTENSION
ALL INDUCTORS ARE 4mH
a
CONNECT COMPONENTS BETWEEN NODES
d
6mH
a 4mH
2mH
WHEN IN
DOUBT…
REDRAW!
c
4mH
Leq
c
d
2mH
2mH
b
2mH
IDENTIFY ALL NODES
PLACE NODES IN CHOSEN LOCATIONS
a
Leq  (6mH || 4mH )  2mH  4.4mH
d
b
b
c
LEARNING EXTENSION
ALL INDUCTORS ARE 6mH
a
a
2mH
6 || 6 || 6
Leq
b
b
6mH
6mH
c
6mH
c
NODES CAN HAVE COMPLICATED SHAPES.
KEEP IN MIND DIFFERENCE BETWEEN
PHYSICAL LAYOUT AND ELECTRICAL
CONNECTIONS
a
b
c
SELECTED LAYOUT
Leq  6  (6  2) || 6  6 
Leq 
48
24
 6 mH
14
7
66
mH
7
L-C
RC OPERATIONAL AMPLIFIER CIRCUITS
INTRODUCES TWO VERY IMPORTANT PRACTICAL CIRCUITS
BASED ON OPERATIONAL AMPLIFIERS
THE IDEAL OP-AMP
IDEAL  RO  0, Ri  , A  
RO  0  vO  A(v  v )
Ri   
A
RC OPERATIONAL AMPLIFIER CIRCUITS -THE INTEGRATOR
v  0
IDEAL OP-AMP ASSUMPTIONS
v _  v  ( A  )
i _  0 ( Ri  )
RC OPERATIONAL AMPLIFIER CIRCUITS - THE DIFFERENTIATOR
i2
R1
i1
KVL
v  0
KCL@ v : i1  i2  i
IDEAL OP-AMP ASSUMPTIONS
v _  v  ( A  )
v
i1  O  0
R2
i _  0 ( Ri  )
t
1
i1 ( x )dx
C1 
di
dv
R1C1 1  i1  C1 1 (t )
dt
dt
v1 (t )  R1i1 
DIFFERENTIATE
replace i1 in terms of v o (i1  
R1C1
vo
)
R2
dvo
dv
 vo   R2C1 1 (t )
dt
dt
IF R1 COULD BE SET TO ZERO WE WOULD HAVE
AN IDEAL DIFFERENTIATOR.
IN PRACTICE AN IDEAL DIFFERENTIATOR AMPLIFIES
ELECTRIC NOISE AND DOES NOT OPERATE.
THE RESISTOR INTRODUCES A FILTERING
ACTION. ITS VALUE IS KEPT AS SMALL AS
POSSIBLE TO APPROXIMATE A DIFFERENTIATOR
ABOUT ELECTRIC NOISE
ALL ELECTRICAL SIGNALS ARE CORRUPTED BY
EXTERNAL, UNCONTROLLABLE AND OFTEN
UNMEASURABLE, SIGNALS. THESE UNDESIRED
SIGNALS ARE REFERRED TO AS NOISE
SIMPLE MODEL FOR A NOISY 60Hz SINUSOID
CORRUPTED WITH ONE MICROVOLT OF 1GHz
INTERFERENCE.
6
y (t )  sin(120 t )  10 sin( 2 10  t )
signal
9
noise amplitude
 106
signal amplitude
noise
THE DERIVATIVE
noise amplitude 2000
dy
9

 16.67
(t )  120 cos(120 )  2000 cos(2 10  t )
signal amplitude 120
dt
signal
noise
LEARNING EXTENSION
INPUT TO IDEAL DIFFERENTIATOR WITH R2  1k, C1  2 F
m
10 V
5 103 s
IDEAL DIFFERENTIATOR
dv
vo   R2C1 1 (t )
dt
DIMENSIONAL ANALYSIS
V
V
V s
 

A Q
Q
s
Q
F    F  s
V
R2C1  1103   2 106 F  2 103 s
INPUT TO AN INTEGRATOR WITH R1  5k, C2  0.2 F
CAPACITOR IS INITIALLY DISCHARGED
INTEGRATOR
LEARNING EXTENSION
R1C2  103 s
1 t
vo (t )  vo (0) 
vi ( x )dx

R1C 2 0
DIMENSIONAL ANALYSIS
V
V
V s
 

A Q
Q
s
Q
F    F  s
V
t
3
3
0  t  0.1s :v1 (t )  20 103  yo (t )   v1 ( x )dx  20 10 t V  s   y(0.1)  2 10 V  s 
0
t
0.1  t  0.2 s : v1 (t )  20 103  y (t )  y (0.1)  v ( x )dx  2 103  20 103 (t  0.1)V  s 
o
o
1
0.1
vo ( t ) 
1
yo (t )
R1C 2
APPLICATION EXAMPLE
CROSS-TALK IN INTEGRATED CIRCUITS
Simplified Model
REDUCE CROSSTALK BY
• Reducing C12
• Increasing C2
COST?
EXTRA SPACE BY
GROUND WIRE

USING GROUND WIRE
TO REDUCE
CROSSTALK
SMALLER
LEARNING EXAMPLE
SIMPLE CIRCUIT MODEL FOR DYNAMIC RANDOM ACCESS
MEMORY CELL (DRAM)
REPRESENTS CHARGE LEAKAGE
FROM CELL CAPACITOR
NOTICE THE VALUES OF THE
CAPACITANCES
Vcell  1.5V FOR CORRECT STORAGE
OF A LOGIC ONE
1t
vC  vC (0)   iC ( x )dx
C0
I leak
I
t  1.5  leak t  1.5V
Ccell
Ccell
1.5(V )  50  1015 ( F )
3
tH 

1
.
5

10
s
12
50  10 A
THE CELL MUST BE “REFRESHED ” AT
A FREQUENCY HIGHER THAN 1 / t H
Vcell  3 
CELL AT THE BEGINNING OF A MEMORY
READ OPERATION
SWITCHED CAPACITOR CIRCUIT
THE ANALYSIS OF THE READ OPERATION
GIVES FURTHER INSIGHT ON THE
REQUIREMENTS
CELL READ OPERATION
IF SWITCH IS CLOSED BOTH CAPACITORS
MUST HAVE THE SAME VOLTAGE
ASSUMING NO LOSS OF CHARGE THEN
THE CHARGE BEFORE CLOSING MUST BE
EQUAL TO CHARGE AFTER CLOSING
Qbefore  1.5V  450 1015 F  3V  50 1015 F
Qafter  Vafter  (500 1015 F )
Vafter  1.65V
Even at full charge the voltage variation is small.
SENSOR amplifiers are required
After a READ operation the cell must be refreshed
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
LEARNING BY DESIGN
POWER OUTAGE “RIDE THROUGH” CIRCUIT
CAPACITOR MUST MAINTAIN AT LEAST 2.4V FOR
AT LEAST 1SEC.

v

DESIGN EQUATION
http://www.wiley.com/college/irwin/0470128690/animations/swf/6-21.swf
DESIGN EXAMPLE
DESIGN AN OP-AMP CIRCUIT TO REALIZE THE EQUATION
vO  50 v1 ( )d  2v2
t
Proposed solution
Needs integrator
And adder
adder
integrator
Design equations
Two equations in five unknowns!!
Not too large. Not too small
Seems a reasonable value
 R3  10k 
 R2  20k 
ANALYSIS OF THE SOLUTION
If supply voltages are 10V (or less) all currents will be less than 1mA, which seems
reasonable