Today: Ideal versus Real elements: Models for real elements
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Transcript of Today: Ideal versus Real elements: Models for real elements
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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
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
MODELING NON-IDEAL VOLTAGE AND CURRENT SOURCES
VBB
i
+
v
+
Real battery
Voltage drops if large
current
i
vVBB
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).
What combination of voltage sources,
current sources and resistors has this I-V
characteristic?
Example: A real battery
Two circuits are equivalent if they have the same I-V characteristic.
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
MODELING NON-IDEAL VOLTAGE AND CURRENT SOURCES
Current-voltage characteristic of a real battery
Approximation over range where i > 0: v = VBB iR
i = (VBB v)/R (straight line with slope of -1/R)
VBB
i
+
v
+
Real battery
Voltage drops if large
current
i
vVBB Model of battery
+
-
+
v
i
R
Simple resistor in series with ideal voltage source “models” real battery
VBB
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
REAL VOLTMETERSConcept of “Loading” as Application of Parallel
ResistorsHow 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 M
Real Voltmeter
Ideal Voltmeter
Rin
Model
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
REAL VOLTMETERSConcept of “Loading” as Application of Parallel
Resistors
21
2SS2 RR
RVV
1in2
in2SS2 RR||R
R||RVV
Example: V1VK900R ,K100R ,V10V 212SS
+VSS
R1
R2
-
+
V2
But if ,V991.0V ,M10R 2in a 1% error
Computation of voltage (uses ideal Voltmeter)
-+
VSS
R1
R2 Rin 2V
Measurement of voltage (including loading by real VM)
-
+
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
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 1.
Real Ammeter
Ideal Ammeter
Rin?
Model
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
MEASURING CURRENT
Potential measurement error due to non-zero input resistance:
R in_+
V
Imeas
R1
R 2
ammeter
with ammeter
_+
V
I
R1
R 2
undisturbed circuit
Example: V = 1 V: R1 + R2 = 1 K , Rin = 1
21
1RR
VI
in21
1meas RRR
VI
mA 999.01K1
1I ,mA1I meas
error) %1.0(
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
IDEAL AND NON-IDEAL METERS
DMMamps
MODEL OF REAL DIGITAL AMMETER
C +
Rin
Note: Rin may depend on range Note: Rin usually depends on current range
Rin typically > 10 M Rin typically < 1
C +IDEAL
DMMamps
C +IDEAL
DMMvolts
DMMvolts
MODEL OF REAL DIGITAL VOLTMETER
C +
Rin
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
CAPACITORS IN SERIES
, Ci
dtdV
, Ci
dtdV
So2
2
1
1
dtdV
Cdt
dVCi 2
21
1
dt)Vd(V
Cdt
dVCi and VVV 21
eqeq
eq21eq
Clearly, SERIES IN CAPACITORS CC
CC
C1
C1
1C
21
21
21
eq
C1
V1
i(t)C2
| ( | (V2+ +
Ce
qi(t)
| (Veq
+
Equivalent to
eq21
eq
Ci
)C1
C1
i(dt
dV so
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
CAPACITORS IN PARALLEL
C1i(t) C2
| ( | (
+
V
dtdV
CdtdV
C)t(i 21
Equivalent capacitance defined by
dtdV
Ci eqCeqi(t)
| ( +
V(t)
Clearly, PARALLEL IN CAPACITORS CCC 21eq
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
CHARGE REDISTRIBUTIONPre-charged capacitor CA is connected to CB at t = 0 Let CA = CB = 1mF
CA CB
t = 0
Find vA(t = 0+).
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 CBvB
2(t>0)
=½ [½ CA vA2 (0)] so
vA2(t>0) = [½ VA
2 (0)]
Or vA2(t>0) =½
These answers are inconsistent. What is wrong with this circuit?
Hint : We set up a paradox : Capacitor V jumps
(infinite current so we dare not ingore the wire resistance)
Initial Voltage = 1V
Initial Voltage = 0
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
DIGITAL CIRCUIT EXAMPLE(Memory cell is read like this in
DRAM)For simplicity, let CC = CB. If VC = V0, t < 0.
Find VC(t), i(t), energy dissipated in R.
)CC for Q of ion(conservat V21
)( VV)(0V BC0C0C
BCC
BC
BC CC if 2
C R)
CC
CCR(
t/
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5
VC/ V
0
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5
cu
rre
nt
(fra
cti
on
of
Vo
/R)
t/
)(eRV t
0
0t
CV CC BC
R+
initially uncharged
-
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
ENERGY DISSIPATION IN R
R
i
R)t(i)t(P 2 R )(eR
V2t
0
R eR
VP(t)
2t20
dteR
VRE
t22
0
0R
2R
VR
20
2
2CR
R
RVE
C
2
20
R
20CVC
41
TWO FACTS:
(1) 1/2 of initial E lost (for CC = CB)
(2) Independent of R!
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
Simple Proof of Energy Division
For simplicity, let CC = CB. If VC = V0, t < 0.
Find VC(t), i(t), energy dissipated in R.
)CC for Q of ion(conservat V21
)( VV)(0V BC0C0C
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).
0t
CV CC BC
R+
initially uncharged
-
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
THE BASIC INDUCTOR CIRCUIT
+vi(t) R
LvX
i
KVL: R iv But R idt
i dLv Xii
V1
v(t)
tt=0
XX
i vdtv d
RL
v
i
t
RV1
RL
)e1(RV
i R/Lt1
0t
1V
RL
Xv
0
)e1(Vv R/Lt
1X
=L/R Solution has same form as RC!
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EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham
TRANSIENTS IN SINGLE-INDUCTOR OR SINGLE-CAPACITOR 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.