Today: Ideal versus Real elements: Models for real elements

1

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

2


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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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

4


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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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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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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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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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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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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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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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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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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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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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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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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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.

Today: Ideal versus Real elements: Models for real elements

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