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EECS 40 Fall 2002 Lecture 13
Copyright, Regents University of California
S. Ross and W. G. Oldham
Today:
• Ideal versus Real elements: Models for real
elements
• Non-ideal voltages sources, real voltmeters,
ammeters
• Series and parallel capacitors
• Charge sharing among capacitors, the paradox
1
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
MODELING NON-IDEAL VOLTAGE AND CURRENT SOURCES
Two circuits are equivalent if they have the same I-V
characteristic.
A model of a device is a collection of ideal circuit elements
that has the same I vs V characteristic as the actual (real)
device (and is therefore equivalent).
i
+
VBB
Example: A real battery
i
+
v

What combination of
voltage sources,
current sources and
resistors has this I-V
characteristic?
Voltage
drops if
large
current
v
Real battery
VBB
2
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
MODELING NON-IDEAL VOLTAGE AND CURRENT SOURCES
Current-voltage characteristic of a
real battery
i
i
+
VBB
Voltage
+ drops if
large
v current

Simple resistor
in series with
ideal voltage
source “models”
real battery
R
+
VBB
-
i
+
v

v
Real battery
Model of battery
VBB
Approximation over range where i > 0:
v = VBB  iR
i = (VBB  v)/R (straight line with slope of -1/R)
3
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
REAL VOLTMETERS
Concept of “Loading” as Application of Parallel Resistors
How is voltage measured? Modern answer: Digital multimeter (DMM)
Problem: Connecting leads from a real voltmeter across two nodes
changes the circuit. The voltmeter may be modeled by an ideal
voltmeter (open circuit) in parallel with a resistance:
“voltmeter input resistance,” Rin. Typical value: 10 MW
Real
Voltmeter
Ideal
Voltmeter
Rin
Model
4
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
REAL VOLTMETERS
Concept of “Loading” as Application of Parallel Resistors
Computation of voltage
(uses ideal Voltmeter)
Measurement of voltage
(including loading by real VM)
R1
VSS
R1
+
+
R2
V2
VSS
+
+
-
R2
Rin
 R2 
V2  VSS 

R

R
 1
2
V2
-
 R 2 || Rin 
V2  VSS 

R 2 || Rin  R1 
Example: VSS  10V, R2  100K, R1  900K  V2  1V
But if Rin  10M, V2  0.991V, a 1% error
5
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
MEASURING CURRENT
Insert DMM (in current measurement mode) into circuit. But ammeters
disturb the circuit. Ammeters are characterized by their “ammeter input
resistance,” Rin. Ideally this should be very low. Typical value 1W.
Real
Ammeter
?
Ideal
Ammeter
Rin
Model
6
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
MEASURING CURRENT
Potential measurement error due to non-zero input resistance:
Imeas
I
R1
ammeter
R1
+
V _
V
Rin
+
_
R2
R2
undisturbed circuit
I
V1
R1  R2
with ammeter
Imeas 
Example: V = 1 V: R1 + R2 = 1 KW , Rin = 1W
1
I  1mA, Imeas 
 0.999 mA
1K  1W
V1
R1  R2  Rin
(0.1% error)
7
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
IDEAL AND NON-IDEAL METERS
DMM
volts
DMM
amps
+
C
+
C
IDEAL
IDEAL
DMM
volts
DMM
amps
Rin
Rin
C
+
MODEL OF REAL
DIGITAL VOLTMETER
Note: Rin may depend on range
Rin typically > 10 MW
C
+
MODEL OF REAL
DIGITAL AMMETER
Note: Rin usually depends on
current range
Rin typically < 1 W
8
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
CAPACITORS IN SERIES
+ V1  + V2 
i(t)
|(
|(
C1
C2
+ Veq 
|(
Equivalent to
Ce
i(t)
q
dV1
dV
 C2 2
dt
dt
dVeq
d(V1  V2 )
Veq  V1  V2 and i  Ceq
 Ceq
dt
dt
i  C1
So
dV1
i

,
dt
C1
Clearly, Ceq 
dV2
i

,
dt
C2
1
1
1

C1 C2

C1C2
C1  C2
so
dVeq
dt
 i(
1
1
i

)
C1 C2
Ceq
CAPACITORS IN SERIES
9
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
CAPACITORS IN PARALLEL
+
C2
|(
C1
|(
i(t)
V
i( t )  C1

Clearly,
Ceq  C1  C2
i(t)
|(
Equivalent capacitance defined by
dV
i  Ceq
dt
dV
dV
 C2
dt
dt
Ceq
+
V(t)

CAPACITORS IN PARALLEL
10
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
CHARGE REDISTRIBUTION
Pre-charged capacitor CA is connected to CB at t = 0 Let CA = CB = 1mF
t=0
Initial Voltage
= 1V
CA
Find vA(t = 0+).
Initial
Voltage = 0
CB
These answers are inconsistent.
What is wrong with this circuit?
From conservation of charge:
QA (t>0) = QB (t>0) = ½ QA (0)
Thus vA (t>0) = ½ V
From conservation of energy:
½ CA vA2(t>0) = ½ CBvB2(t>0)
=½ [½ CA vA2 (0)] so
vA2(t>0) = [½ VA2 (0)]
Or vA2(t>0) =½
Hint : We set up a paradox : Capacitor V jumps
(infinite current so we dare not ingore the wire resistance)
11
EECS 40 Fall 2002 Lecture 13
t0
S. Ross and W. G. Oldham
Copyright, Regents University of California
DIGITAL CIRCUIT EXAMPLE
(Memory cell is read like this in DRAM)
R
For simplicity, let CC = CB.
If VC = V0, t < 0.
initially
uncharged
+
Find VC(t), i(t), energy
CB
dissipated in R.
C C
R CC
  R( C B ) 
if CC  CB
CC  CB
2
VC CC
VC (0 )  V0
VC () 
1
V0 (conservation of Q for CC  CB )
2
current (fraction of Vo/R)
1
VC/ V0
0.8
0.6
0.4
0.2
0
1
V0  t 

(e
)
R
0.8
0.6
0.4
0.2
0
0
1
2
3
t/
4
5
0
1
2
t/
3
4
5
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EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
ENERGY DISSIPATION IN R
R
i
P( t )  i2 ( t )R
2
2
 V0  t 
  (e  )  R
 R

 V0   2t 
P(t)  
R
 e
R 

2
 V0  2t 
ER   R
dt
 e
R

0 
TWO FACTS:
 CC 
2
R
RV0 
2 
ER 
2
R2

2
V  
 R 0 
R  2
1
CC V02
4
(1) 1/2 of initial E lost (for CC = CB)
(2) Independent of R!
13
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
Simple Proof of Energy Division
t0
R
+
VC CC
CB
VC (0 )  V0
initially
uncharged
VC () 
For simplicity, let CC = CB.
If VC = V0, t < 0.
Find VC(t), i(t), energy
dissipated in R.
1
V0 (conservation of Q for CC  CB )
2
Thus initial Energy Stored in Capacitors is 1/2CCV02 + 0
Final Energy is 1/4 CCV02 so clearly the resistor dissipated
the rest, independent of the value of the resistance.
So even if the resistance is very, very, very small, it still
dissipates half the energy in this example (where CC =CB).
14
EECS 40 Fall 2002 Lecture 13
S. Ross and W. G. Oldham
Copyright, Regents University of California
THE BASIC INDUCTOR CIRCUIT
v(t)
L
i
vX
V1
vi(t) +
R
t
t=0
KVL: v i  Li
vi 
di
iR
dt
L d vX
 vX
R dt
i
=L/R
Solution has same form as RC!
t
V
i  1 (1  e  t L / R )
R
V1
R
0
But v X  i R
L
R
t
v X  V1(1  e L / R )
vX
V1
0
L
R
t
15
EECS 40 Fall 2002 Lecture 13
Copyright, Regents University of California
S. Ross and W. G. Oldham
TRANSIENTS IN SINGLE-INDUCTOR OR SINGLECAPACITOR CIRCUITS - THE EASY WAY
1) Find Resistance seen from terminals of L or C
(short voltage sources, open current sources).
2) The circuit time constant is L/R or RC (for every node,
every current, every voltage).
3) Use initial conditions and inductor/capacitor rules to find
initial values of all transient variables. (Capacitor voltage and
inductor current must be continuous.)
4) Find t= value of all variables by setting all time
derivatives to zero.
5) Sketch the time-behavior of all transient variables,
based on initial and final values and known time constant.
6) Write the equation for each transient variable by inspection. 16