Part IA—Linear Circuits and Devices

Cambridge University EngineeringDepartment

Michaelmas Term 2001

Lecturer: Dr David HolburnEmail: dmh@eng.cam.ac.uk

Lectures 1–4

Contents.................................................................................................................... 1

Lecture 1

DC Circuits ................................................................................................. 1Ohm’s Law ........................................................................................... 1Electrical power.................................................................................... 1Series and parallel circuits .................................................................... 2

Design Example 1.1........................................................................ 3Design Example 1.2........................................................................ 5

Voltage sources........................................................................................... 7Ideal voltage source .............................................................................. 7Real voltage source............................................................................... 7

Current sources ........................................................................................... 8Ideal current source .............................................................................. 8Real current source ............................................................................... 8

Design Example 1.3........................................................................ 9

Lecture 2

Kirchhoff’s Laws I.................................................................................... 11The voltage law .................................................................................. 11The current law ................................................................................... 13

Design Example 2.1...................................................................... 14Design Example 2.2...................................................................... 18

Lecture 3

Kirchhoff’s Laws II .................................................................................. 21Mesh current loop analysis ................................................................. 21

Design Example 3.1...................................................................... 23Design Example 3.2...................................................................... 24

Superposition ............................................................................................ 27Design Example 3.3...................................................................... 29

Lecture 4

Equivalent Generators .............................................................................. 31Thévenin’s Theorem........................................................................... 31

Design Example 4.1...................................................................... 32Design Example 4.2...................................................................... 35

Norton’s Theorem............................................................................... 38Design Example 4.3...................................................................... 39

Circuit matching ....................................................................................... 41Power matching .................................................................................. 41Voltage matching................................................................................ 41Current matching ................................................................................ 42

A more difficult example.......................................................................... 43

Design Example 4.4...................................................................... 43

-1-

Lecture 1

DC Circuits

Ohm’s Law

Electrical power

Ohm’s law forms the basisfor all DC electrical circuitanalysis. It was first suggestedby Georg Simon Ohm (1787–1854) and was published as amemoir in 1826. Ohm waitedmany years for proper recog-nition for his work, eventuallyreceiving the Royal SocietyCopley Medal in 1841. Ohm’smemoir laid the foundationfor much subsequent work onelectrical theory in the secondquarter of the 19th century.

Note that by convention,current flows from high tolow potential, as though itwere composed of positivecharges. We know of coursethat electrical current is theflow of electrons. The reasonfor using the positive conven-tion dates back to BenjaminFranklin (1706–1790) whopostulated that only one kindof positive electrical chargeexisted. For largely historicalreasons we still use that con-vention.

The units of V are volts(V); the units of I are amps(A); and the units of R areohms (Ω).

–+

V R

I

V IR=

V

I R

Power IV=Power is measured in

Watts (W). The expression forpower may be written equiva-lently as I2R or V2 R .⁄

-2-

Series and parallel circuits

series resistors:

parallel resistors:

conductance

You will already have metthese results. They are left asa revision exercise for you toprove.

R1 RNR2

R R1 R2 … RN+ + +=

R1 R2 RN

1R--- 1

R1------ 1

R2------ … 1

RN------+ + +=

In the case of two resistorsin parallel this result takes aparticularly simple form,which is used so often thatyou should know it by heart:

.

“Product/Sum”

For parallel resistors it isoften more convenient towork in terms of conductance,defined as G=1/R. Then, forparallel resistors we just addthe conductances. The units ofconductance are , or alter-natively siemens.

RR1R2

R1 R2+------------------=

Ω 1– G1R---≡

G G1 G2 … GN+ + +=

-3-

Design Example 1.1A fuse can be modelled as a tempera-

tu re dependent res i s to r R where. The temperature rise

is assumed to be given by where P is the electrical power dissi-pated in R. Derive R as a function ofcurrent I.

We assume

then

But which gives us:

Solving for R

R R0 α∆T+=∆T ∆T kP=

Many electrical devices donot obey Ohm’s law, and inthis example we will study asimple model for the electricalfuse. The basic principle ofthe fuse is more complicatedthan many realise. As the cur-rent through the fuseincreases, it heats up, and itsresistance increases, whichcauses the voltage drop acrossthe fuse to increase. This inturn causes the current to risefurther, and thermal runawayresults. The fuse fails abruptlyat a critical current, say, IF.

We assume that the resist-ance of the fuse increases lin-early with temperature, andthat the temperature rise isproportional to the electricalpower dissipated in the fuse.We will also assume that theelectrical power dissipated bythe fuse is given by . Ineffect, this defines what wemean by resistance in thismodel. Strictly speaking, weshould define the resistance as

, and the power dissi-pated as . This would leadto a differential equation for Vwhich can be solved. How-ever, our simpler approachwill produce essentially thesame qualitative predictions.

I2R

V∂ I∂⁄IV

P I2R=

∆T kP kI2R= =

R R0 α∆T+=

R R0 αkI2R+=

RR0

1 αkI2–--------------------=

-4-

Define

then we can rewrite R as

fuse ‘blows’ when

i.e. when

IF1

αk-----------=We introduce so that wecan express R in terms of the

normalised current . Aswe will see shortly, definesthe critical current at whichthe fuse ‘blows’.

IF

I IF⁄IF

RR0

1I2

IF2

----–

--------------=

0 0.2 0.4 0.6 0.8 10

5

10

15

20

I IF⁄

R R0⁄As expected, the resist-

ance of the fuse depends non-linearly on the current. It doesnot obey Ohm’s law, except atlow currents. The expressionfor R is plotted opposite as afunction of the normalisedcurrent . Note that until Igets very close to , Rincreases only very slowly.Near the resistance risesvery rapidly, and in our modelbecomes infinite; that is thefuse suddenly ‘blows’, and inpractice would melt. In amore realistic model, where Ris defined as , the risewould is not infinite, but isstill very rapid as the currentapproaches .

I IF⁄IF

IF

V∂ I∂⁄

IF

R ∞→

I 1

αk-----------→

~

It is interesting to see thatin this model the critical cur-rent at which the fuse ‘blows’does not depend on factorssuch as the melting point ofthe fuse wire. Real fuses ofcourse are much more compli-cated. Nevertheless, our graphof R vs I is very similar quali-tatively to the experimentalbehaviour.

-5-

Design Example 1.2A long uniform cable AB is damaged

and has a leakage resistance to earth atan unknown point L. Find the positionof the leakage resistance using resist-ance measurements taken at end A only.

(i) short circuit B to earth

resistance measured at A is

(ii) open circuit B

resistance measured at A is now

yx a-x

A BL

leakage resistance

This is a famous problemof historical interest. Duringthe 19th century, when sub-marine telegraph cables cameinto wide spread use, therewas a need to be able to locatefaults along a cable quicklyand accurately. Most faultsconsisted of some sort ofdamage resulting in a leakageresistance to earth. Here, weconsider a method of faultlocation due to Oliver Heavi-side (1850–1925), which hepublished in the May 3rdissue of the Electrician in1901. The original accountwas found in his notebooks,dated January 16th 1871.

If we assume that thecable has a known constantresistance per unit length,then the problem reduces tofinding the resistance xbetween A and the leakagepoint at L. Assume that thetotal cable resistance betweenAB is a. We need to find twoequations so that we can elim-inate the leakage resistance y.We do this by measuring theresistance at A whilst the endB is short circuited, and thenopen circuited. When B isshort circuited, we measure xin series with the parallelcombination of y and (a-x).When open circuited, wemeasure x in series with y.

Rsc c≡ xy a x–( )y a x–+--------------------+=

Roc b≡ x y+=

-6-

eliminating y from these equations

solving for x gives

choosing the negative sign,

The leakage resistance y iseliminated from the abovetwo equations, leaving aquadratic in x. This is easy tosolve, but you do need tothink a little about which ofthe two possible solutions isvalid.

x2

2cx– ac bc ab–+ + 0=

x c a c–( ) b c–( )±=This is Heaviside’s result.

Unknown to Heaviside, it wasactually first derived byEdouard Blavier (1826–1887).

In any case, it turns outnot to be a good method.What do you think are itsmain disadvantages?

x c a c–( ) b c–( )–=

~

-7-

Voltage sources

Ideal voltage source

Voltage independent of the load

Real voltage source

Voltage depends on the load

–

+RLV V0=V0

In electronic circuit designwe frequently make use of theconcept of an ideal voltagesource. By this we mean atwo terminal device whoseoutput voltage is inde-pendent of any load resistor

connected across its out-put terminals. Since the cur-rent supplied by an idealvoltage source is given by

it is clear that it mustbe able, in principle, to pro-vide an unlimited amount ofcurrent. A good approxima-tion to an ideal voltage sourceis a newly charged car battery.

V0

RL

V0 RL⁄

RL

–

+Rs

RLV V0≠V0

A real voltage source canbe thought of as an ideal volt-age source together with aninternal series resistance .For example, a 12V newlycharged car battery has aninternal resistance of about0.006Ω , and a AA (penlight)1½ volt battery has an internalresistance of about 0.4Ω . Ageneral purpose, low voltageelectronic power supply mighthave an internal resistance ofaround 0.1Ω .

This means that the outputfrom a real voltage source willdepend on the size of the loadresistor . Voltage sources‘prefer’ to operate - that is,they operzte best - with theiroutput terminals opencircuited, and ‘dislike’ beingshort circuited.

Rs

RL

RL

-8-

Current sources

Ideal current source

Current independent of load .

Real current source

Current depends on load .

I0 I I0=RL

In addition to voltagesources we often employ asecond type of source, theideal current source. This is atwo terminal device whichdelivers a constant currentirrespective of the load con-nected across its terminals.This means that, in principle,it can sustain an arbitrary volt-age across a load, .Because of this, currentsources ‘like’ to operate(operate best) with their ter-minals short circuited. They‘dislike’ open circuits.

Whereas it is easy to thinkof approximations to voltagesources, there are no real anal-ogies to help us visualise cur-rent sources. They are no lessimportant, however, and arean important component inthe design of many electroniccircuits.

V I0RL=

RL

I0 RL I I0≠Rs

In practice, we deal withreal current sources, which wethink of as an ideal sourcewith an internal resistance Rsshunted across (i.e. in parallelwith) its output terminals.This means that the currentdelivered to a load willdepend on the size of the loadresistance . In a later lec-ture we will see how a currentsource can be approximatedusing a field effect transistorbiased in the following way:

RL

RL

+V

I0

RL

-9-

Design Example 1.3Measurements on a power supply pro-

duced the following results:(a) open circuit voltage ;(b) short circuit current ;Represent the power supply by (i) an

equivalent voltage source with seriesresistance; and (ii) an equivalent currentsource with parallel resistance.

(i) voltage source

open circuit voltage

short circuit current

This is somewhat con-trived, as you would not nor-mally short circuit a powersupply. A better methodwould be to connect a variableknown resistor across its out-put terminals and measure theresultant voltage and current.What this example does bringout however is the idea thatthe same two terminal devicecan be represented by eitheran equivalent voltage source,or an equivalent currentsource.

We shall meet this idea inmuch more detail when westudy Thévenin’s and Nor-ton’s theorems.

10 V=20 A=

–

+Rs

V0½

V0= 10 V= When the power supply isopen circuited no current isdrawn, so that its full voltageis seen across the output ter-minals. When it is shortcircuited, the voltage isdropped across its internalresistance.

V0 Rs⁄= 20 A=

Rs V0 20⁄=

Rs 0.5Ω=

-10-

(ii) current source

short circuit current

open circuit voltage

is often written as a conductance

For the above example:

.

½ I0 Rs

I0= 20 A=When a current generatoris short circuited, all of itscurrent flows via the outputterminals. This is because novoltage drop appears acrossits internal shunt resist-ance . When it is opencircuited, the entire currentflows through Rs and appearsas an output voltage .

Rs

I0Rs

10 V= I0Rs=

Rs 10 I0⁄=

Rs 0.5Ω=

Rs

Gs1Rs-----≡

Current generators areoften specified in terms of aninternal shunt conductance,defined by

.Gs1Rs-----=

Gs 2Ω1–

2 siemens= =

~

-11-

Lecture 2

Kirchhoff’s Laws I

The voltage law

let current I flow for time δt

work done by battery

= energy loss in resistors

Gustav Robert Kirchhoff(1824–1887) was born inKönigsberg. Although wellknown for his electrical cir-cuit laws, he is best known forhis work on the emission andabsorption of electromagneticradiation by black bodies.

His circuit laws were hisfirst piece of scientific work,carried out at the age of 21during the period 1845–46.He was also, in 1849, the firstperson to clarify Ohm’s rathervague ideas of electrical ten-sion, showing that it should beidentified with the electro-static potential, thus unifyingthe theories of electrostaticand voltaic currents.

–

+V0

R1

R2

V1

V2

II

Consider the simple cir-cuit drawn above. The voltagesource is driving a current Iaround the circuit, and indoing so performs work at therate of . This work mustgo somewhere by energy con-servation, and it results in theheating up of the two resis-tors. If the voltage drop across

is then it heats up atthe rate of . If the currentflows for a small time thenthe energy loss from the bat-tery will be , and thismust equal the energy dissi-pated in the two resistors

.

IV0

R1 V1IV1

δt

IV0δt

IV1δt IV2δt+

IV0( )δt IV1( )δt IV2( )δt+=

-12-

Now consider a more general circuit

Kirchhoff’s voltage law states:

V0 V1– V2– 0=We can cancel I and onboth sides of the energy bal-ance equation, to leave anequation relating the voltagedrops around the different ele-ments in the circuit. For thecircuit above, this says thatthe sum of the voltage dropsfor each component aroundthe entire circuit must add upto zero.

δt

Viloop∑ 0=

V1

V2

V3

VN

The result derived for thesimple circuit above can beshown to be true for the volt-ages around any closed cir-cuit, whatever voltage orcurrent sources, or other com-ponents are present, and isknown as Kirchhoff’s VoltageLaw. The circuit componentsdo not even need to be linear,or obey Ohm’s law; the volt-ages and currents may also bevarying with time, although atvery high frequencies (in theGHz) Kirchhoff’s voltage lawbreaks down. To prove all thiswould go a long way beyondthis course, and involves aknowledge of Maxwell’sequations which you willmeet in the second year.

Note that when applyingKirchhoff’s law you must bevery careful to get the alge-braic signs of the voltagedrops correct.

Viclosedloop

∑ 0=

-13-

The current law

Conservation of charge implies:

For a general node, we have

Kirchhoff’s current law:

–

+I1 I2 I3

I

V0 R1 R2 R3

We saw that Kirchhoff’sfirst law can be considered toresult from energy conserva-tion around a circuit. His sec-ond law is the result of chargeconservation. Consider a cir-cuit like the one shown oppo-site. We will assume that nocharge can accumulate on thecomponent leads or on thewires joining them. Then, anycharge entering a componentmust be balanced by an equalamount of charge leaving. Toput this another way, the alge-braic sum of the currentsentering a junction must bezero. This is Kirchhoff’s Cur-rent Law.

I I1– I2– I3– 0=

I1

I2 I3I4

INMore generally, Kirch-

hoff’s current law says thatthe algebraic sum of the cur-rents entering any node in acircuit must be zero. Again,care is often needed to get thesigns of the currents correct.

Ininto

a node

∑ 0=

-14-

Design Example 2.1Find the voltage across R in the follow-

ing circuit. Use (i) Kirchhoff’s VoltageLaw, and (ii) Kirchhoff’s Current Law.

(i) Using KVL (mesh current analysis)

Consider voltages around loop abef:

around loop bcde:

define x by

–

+

–

+e2e1 R

R1 R2

I1 I2

I1 I2+

a b c

def

The same circuit can besolved using either Kirch-hoff’s voltage law or currentlaw. We will use the circuitopposite to illustrate bothmethods of solution.

The circuit represents twobatteries, with voltages and , and internal resist-ances and respectively.The batteries are connected inparallel, the same way round,across a load of resistance R.We are to determine theresulting voltage across theload. We begin by usingKirchhoff’s voltage law, andthis method is often calledmesh current analysis. Welabel the currents through and as and , withdirections as shown on thediagram. You are, of course,free to label the current direc-tions as you wish. The currentthrough R is then given by

.Now consider the loop

abef. Kirchhoff’s law tells usthat the sum of voltagesaround this loop must be zero.This gives us one equationrelating unknown currents and . Note that for thevoltage drop is from a to b forthe current direction chosen.This is opposite to the voltage

, and is the reason for thenegative signs in our equa-tion. We get a second equa-tion by similarly applyingKirchhoff’s law to loop bcde.

To determine the voltageacross R we need , andin solving these two equationsit therefore makes sense todefine by a new varia-ble x.

e1e2R1 R2

R1R2 I1 I2

I1 I2+

I1I2 R1

e1

I1 I2+

I1 I2+

e1 I1R1– I1 I2+( )R– 0=

e2 I2R2– I1 I2+( )R– 0=

x I1 I2+=

-15-

eliminating we find

voltage across R is given by

using x from above,

we can rearrange this as

e1 I1R1– Rx– 0=We now rewrite our two

equations in terms of x and .It is not difficult to eliminate

to leave one equation for xwhich we then solve.

I1

I1

e2 x I1–( )R2– Rx– 0=

I1

xe1R2 e2R1+

R1R2 R R1 R2+( )+----------------------------------------------=

The voltage across R isgiven by xR.VR I1 I2+( )R xR= =

VR Re1R2 e2R1+

R1R2 R R1 R2+( )+----------------------------------------------⋅=

We can check that thisresult makes sense by seeingwhat happens if we make infinite, when the resistors and R form a simple potentialdivider. Our equation oppositethen reduces to

as expected. A similar thingshould happen if we make infinite, as you can check.

R2R1

VR e1R

R R1+----------------⋅=

R1

VR

e1R1------

e2R2------+

1R--- 1

R1------ 1

R2------+ +

-----------------------------=

-16-

(ii) Using KCL (nodal voltage analysis)

Write down currents into node at A

sum of currents into node = 0

–

+

–

+e2e1 R

R1 R2

I1 I2

I3

zero reference

A

VA

0

We will now see how thesame circuit can be solvedusing Kirchhoff’s current law.This is often referred to asnodal voltage analysis. Wechoose one node in the circuitas a reference node, and labelthe voltages at all the othernodes with respect to this ref-erence. In our case, there isone unknown nodal voltage,which we have labelled A.The voltage drops across eachcomponent are then calcu-lated. For example, the volt-age drop across is .The currents flowing at eachnode are then written down.The current into the node Athrough is

with similar equations for theother two currents. Kirch-hoff’s current law then saysthat the sum of the currentsflowing into each node mustbe zero. In our case, there isjust one node A, and this leadsto an equation for theunknown voltage across R.

R1 e1 VA–

R1

I1e1 VA–

R1-----------------=

I1e1 VA–

R1-----------------=

I2e2 VA–

R2-----------------=

I3VAR------=

I1 I2 I3–+ 0=

-17-

rearranging this

from which we find

e1 VA–

R1-----------------

e2 VA–

R2-----------------

VAR------–+ 0=

We obtain the same resultas before of course. But notethat the nodal voltage methodrequires rather less algebracompared to the mesh currentanalysis. This is no accident,and in many circuits onemethod will be quicker thanthe other. We will see whythis is so much more clearly ina later lecture, when we studyThévenin’s and Norton’s the-orems. In the present caserecall from the last lecturehow we saw that a voltagesource could also be repre-sented as a current sourcewith a shunt resistance. In ourpresent circuit, the voltagesource can be repre-sented as follows

with something similar for theother source. Then, the threeresistors , and appearin parallel, and the two cur-rent sources produce a totalcurrent of

The way in which we havewritten the final result clearlybrings this out.

e1 and R1

R1e1R1------

R R1 R2

e1R1------

e2R2------+

VA1R--- 1

R1------ 1

R2------+ +

e1R1------

e2R2------+=

VA

e1R1------

e2R2------+

1R--- 1

R1------ 1

R2------+ +

-----------------------------=

~

-18-

Design Example 2.2The following circuit represents a sim-

ple battery charger. Derive an expres-sion for the current through the battery.

(i) nodal voltage analysis (using KCL)

consider currents at node A

–+

RB

VBRsI0

VA

zero reference

IB

A battery charger worksby delivering a steady charg-ing current to the battery untilit reaches its fully chargedstate; depending on thedesign, the current mayremain constant throughoutthe charging period, or maybe reduced to a trickle chargewhen the battery is nearlyfully charged. The circuitopposite represents a simplebattery charger where the cur-rent is provided by the currentgenerator with internalshunt resistance . The bat-tery has internal resistance

. We will use Kirchhoff’slaws to find the battery charg-ing current as a function ofthe other component values.

I0Rs

RB

IB

There is only oneunknown nodal voltage, thatat A, which we shall call Vwith respect to the zero refer-ence node. The voltage across

is given by and thecurrent through this resistor isgiven by

Similarly, the current through is . The sum of the

currents at node A must bezero, and this gives us anequation for the unknownnodal voltage V.

RB V VB–

V VB–

RB----------------

Rs V Rs⁄

I0VRs-----–

V VB–

RB----------------– 0=

V

I0VBRB------+

1Rs----- 1

RB------+

-------------------=

-19-

battery current is given by

using V derived above, we find

.

(ii) mesh current analysis (using KVL)

consider voltages around loop abcd:

The charging currentthrough the battery is just thecurrent through . The volt-age drop across is .Then, using the expression forV found above, we obtain anexpression for the chargingcurrent.

RBRB V VB–

IBV VB–

RB----------------=

IBRs

RB Rs+------------------ I0

VBRs------–

⋅=

–+

RB

VBRsI0

IB

I0 IB–

a b

cd Consider loop abcd.Kirchhoff’s voltage law tellsus that the sum of the voltagesaround this loop must be zero.The voltage drop across is

, and the voltage dropacross is .Applying Kirchhoff’s lawthen gives us an equation forthe battery current .

RBIBRB

Rs I0 IB–( )Rs

IBIBRB VB I0 IB–( )Rs–+ 0=

-20-

solving for we again find

note: charging current positive only if

IB

IBRs

RB Rs+------------------ I0

VBRs------–

⋅=As expected, we get the

same result as using nodalvoltage analysis, but withrather less work. Also, thecharging current must be posi-tive for the battery charger towork, and this implies an ine-quality between , and

. How this is realised willdepend on the design of thecurrent generator. You willsee one way of implementingthis circuit using a standardintegrated circuit in the nextlecture.

I0 VBRs

I0VBRs------>

~

-21-

Lecture 3

Kirchhoff’s Laws II

Mesh current loop analysis

There are six currents to determine.

Current conservation:

–

+E1

R1 R2 R3

R4 R5

I1 I2 I3

I4 I5I6

R6With even moderately

complex circuits Kirchhoff’slaws quickly lead to a largenumber of equations. Theproblem with this is that someof these equations will beredundant, they repeat infor-mation already contained inother equations. The circuitabove illustrates this. Thereare six possible unknown cur-rents to determine, and apply-ing current conservation andKirchhoff’s voltage law leadsto seven equations for the sixunknowns. In principle, thereis no harm in this, as you willget the correct solution bysolving the equations. But itdoes become progressivelyworse as the circuit becomesmore complex. It also makesit very difficult to know whichequations to work with toavoid going round in circles.The purpose of this lecture isto introduce a more system-atic way of applying Kirch-hoff’s voltage law whichavoids the problem of redun-dant equations.

I4 I1 I2–=

I5 I2 I3–=

I6 I5 I3+=

I1 I4 I6+=

-22-

Apply Kirchhoff’s voltage law:

There are seven equations with sixunknowns.

Simpler method: use loop analysis:

Now there are only three equations.

E1 I1R1– I4R4– 0=

I2R2 I5R5 I4R4–+ 0=

I3 R3 R6+( ) I5R5– 0=

–

+E1

R1 R2 R3

R4 R5

I2 I3I1

R6

One way to ensure thatonly the minimum number ofcircuit equations needed aregenerated is to represent thecurrents as current loops, withone for each mesh of the cir-cuit. The reason that thisworks is that current conser-vation at the nodes is automat-ically enforced. For the circuitconsidered on the previouspage there are three independ-ent meshes, so we introducethree current loops, , ,and , as shown. This meansthat the current flowing downthrough resistor is givenby , and the voltagedrop across is .Similarly, the voltage dropacross is . These areboth in the opposite directionto that chosen for andaccounts for the sign differ-ence in the first circuit equa-tion opposite. It is easy to seethat applying Kirchhoff’svoltage law around each meshwill produce three equationsfor the three unknown cur-rents.

I1 I2I3

R4I1 I2–( )

R4 I1 I2–( )R4

R1 I1R1

E1E1 I1R1– R4 I1 I2–( )– 0=

I2R2 I2 I3–( )R5 I2 I1–( )R4+ + 0=

I3 R3 R6+( ) I3 I2–( )R5+ 0=

-23-

Design Example 3.1Determine the balance condition for

the Wheatstone bridge circuit below.

Apply Kirchhoff’s voltage law to thetwo current loops :

Eliminate and we get:

–+

G

i1

i1

i2

R1 R2

R X

The Wheatstone bridgeaffords a good example of theuse of current loops for circuitanalysis.

Charles Wheatstone camefrom a family of musicalinstrument makers, and hadno formal scientific training,starting out himself in thefamily music business. In1823 he patented the concer-tina. By 1834 he was Profes-sor of Physics at King’sCollege London. Rather lesswell known is the fact that hewas the uncle of the greatBritish electrical engineerOliver Heaviside.

In the circuit above, X isan unknown resistor to bemeasured, R is a precisionpotentiometer, and and are known standard resist-ances.

At balance, the currentthrough the galvanometer G iszero. This implies that thereshould be two equal currentloops around the twomeshes of the resistor bridgeas shown. There will be a sec-ond current loop aroundthe outer mesh containing thebattery.

Since the balance condi-tion will not depend on thebattery voltage we do notneed to apply Kirchhoff’s lawaround the loop including thebattery.

R1 R2

i1

i2

i1

i1R2 i1 i2–( )R1+ 0=

i1X i1 i2–( )R+ 0=

i1

XR2------ R

R1------ or X R

R2R1------⋅==

~

-24-

Design Example 3.2Determine the balance condition for

the Kelvin double bridge shown below.

Apply Kirchhoff’s voltage law:

Although the Wheatstonebridge is capable of highaccuracy it is not well suitedto measuring very low resist-ances, of the order 0.001Ω .The reason for this is that theleads to such a resistor willhave a comparable resistanceto what is being measured. Toovercome this, Sir WilliamThomson (1824–1907), laterknown as Lord Kelvin,invented a new bridge methodcalled the Kelvin doublebridge. This is one of the mostaccurate methods availablefor measuring very low resist-ances. The principle of thismethod is shown by the adja-cent circuit. The unknownresistor is X, and S is an accu-rate standard resistor compa-rable in value to X. Theresistors Q, M, and q, m, arestandard resistors of ordinaryvalues. Resistors Q and q arevariable resistors, and can beadjusted together so that theratio . Resistor ris a very low resistance madeby using a thick conductor.The reason this bridge worksis that the lead resistances ofX are either in series with amuch higher resistance, or ina part of the circuit (r) wheretheir resistance does not mat-ter.

The circuit has fourmeshes, so we introduce acurrent loop for each. At bal-ance two of these will beequal, labelled . We nowapply Kirchhoff’s law bysumming the voltages aroundeach mesh to zero. We do notworry about the mesh contain-ing the battery, as the balancecondition will not depend onthe battery voltage.

Q M⁄ q m⁄=

i1

Q M

q mX S

Gi1

i3

i1

i2r

i1Q i1 i2–( )q i1 i3–( )X+ + 0=

i1M i1 i3–( )S i1 i2–( )m+ + 0=

i1 i2–( )q i3 i2–( )r i1 i2–( )m+ + 0=

-25-

To solve these equations we define newvariables:

The circuit equations then become:

From equation (3)

Substitute into equations (1) and (2)

The circuit equations arenot difficult to solve, but arerather messy; a little thoughtwill simplify the analysis con-siderably. We note that thecurrent differences and occur severaltimes in the equations, and soit makes sense to define theseas new unknowns and .

i1 i2–( )i1 i3–( )

i12 i13

i12 i1 i2–=

i13 i1 i3–=

In terms of and theequations take the form givenopposite. Note that in writingthese equations we have madeuse of the following relation

which follows from the defi-nition of and .

Now it is much easier tosee what is going on. Equa-tion 3 has only twounknowns, and , so thisis probably the best one tostart with. We will use equa-tion 3 to express in termsof and then substitute thisin equations 1 and 3.

i12 i13

i3 i2– i12 i13–=

i12 i13

i12 i13

i12i13

i1Q i12q i13X+ + 0 (1)=

i1M i13S i12m+ + 0 (2)=

i12q i12 i13–( )r i12m+ + 0 (3)=

i12i13

r q m+ +---------------------- r⋅=

We now have just twoequations. It is straight for-ward to use one of them toexpress in terms of andto substitute for this in theother; will then cancel onboth sides, leaving an expres-sion for X.

i1 i13

i13

i1Qi13

r q m+ +---------------------- rq i13X+⋅+ 0=

i1M i13Si13

r q m+ +---------------------- rm⋅+ + 0=

-26-

Eliminating we find:

The bridge is adjusted so that:

Then the unknown resistor X is givenby:

i1The variable resistors Q

and M are adjusted so that theratios and areequal. In practice this isachieved by them being partof a double ratio arm resist-ance box. Then we see thatthe term on the right hand sidecontaining the resistance rvanishes, and with it any con-tribution from the connectingleads of the unknown resis-tor. The unknown resistor X isthen determined from Q, M,and S.

When measuring smallresistances thermo-electriceffects are important, but byreversing the battery and aver-aging the results they can beeliminated. If this is done, theKelvin double bridge can typ-ically measure resistances ofthe order 0.01–0.001Ω to anaccuracy of 0.02%.

Q M⁄ q m⁄ XQSM------- mr

r q m+ +----------------------

QM----- q

m----–

+=

QM----- q

m----=

X QM----- S⋅=

~

-27-

Superposition

The response of a linear network con-taining several sources is found by con-sidering each source separately and thenadding the individual responses.

Consider voltage source alone:

–+

RB

VBRsI0

IB

The idea of superpositionis a powerful one. It allowscircuits with many voltageand current sources to bereduced to a series of simplercircuits, each containing onlyone source.

This simplification mustbe offset against the largeramount of work required toobtain a solution.

The Superposition Theo-rem, quoted opposite, is validfor networks of linear ele-ments. These can includeideal resistors, capacitors andinductors. Such linear compo-nents are central to the ideascovered in this course, and itis convenient to define theterm linear at this point. Alinear element is one whoseproperties depend only on itsgeometry and the materials ofwhich it is made, not on thecurrent or voltage applied.The ideal resistors we haveseen already meet this specifi-cation.

The circuit opposite illus-trates the idea (we consideredit earlier, in Design Example2.2). We will calculate thebattery charging current using superposition. We con-sider the effect of each sourcealone, with all the othersremoved. If we remove thecurrent source we are left withan open circuit in its place.This is because an ideal cur-rent source has an infiniteimpedance. The circuit thenreduces to (a) opposite, wherewe denote the battery currentby . Note that in this casewe have kept the direction of

the same as , althoughwe are free to choose anydirection we wish.

IB

Ia

Ia IB

–+

RB

VBRs

Ia

(a)

-28-

The current is given by

Now consider the current source alone

Total battery current is then:

IaThe battery voltage

appears across the series com-bination of and ; thedirection chosen for the cur-rent results in the minussign.

VB

RB Rs

IaIa

VBRB Rs+------------------–=

RB

RsI0

Iba

V

(b)

Next, we remove the volt-age source and consider theeffect of the current sourcealone. An ideal voltage sourcehas zero impedance, so wereplace it by a short circuit.The resulting current is .The circuit has now reducedto a current generator con-nected to the parallel combi-nation of and .Consider the voltage V at thenode labelled a. This is givenby

where denotes theparallel combination of and , so that

The current is then givenby

and this leads to the expres-sion given opposite. Note thatthis is just the current equiva-lent of the potential divider,with the current dividingbetween and in inverseproportion to the resistances.

The total current through the battery is thesuperposition of and and results in the same answeras we obtained in WorkedExample 2.2.

Ib

I0

RB Rs

V I0 RB Rs||⋅=

RB Rs||RB

Rs

V I0RBRs

RB Rs+------------------⋅=

Ib

IbV

RB------=

I0RB Rs

IB

Ia Ib

IbRs

RB Rs+------------------ I0⋅=

IB Ia Ib+=

RsRB Rs+------------------ I0

VBRs------–

⋅=

-29-

Design Example 3.3Determine the response of the voltage

adding circuit below.

Consider alone:

–

+

–

+

R

R

R'V1 V2 Vout

The circuit opposite isknown as a voltage adder, andis a simple way of producingan output proportional to thesum of two independent volt-age sources and . Infact we have already met thiscircuit in a different context inDesign Example 2.1. We willuse superposition, consideringthe contribution to fromeach source separately, andadding the two results.

V1 V2

Vout

V1

–

+

R

R

R'V1

Va

Consider the effect of alone. We remove byreplacing it with a short cir-cuit. The resultant output volt-age is denoted by in thecircuit opposite. It is notimmediately obvious from theway the circuit is drawn, butthe circuit now consists of connected across a potentialdivider, consisting of R inseries with the parallel combi-nation of and , which wedenote by . We can seethis more clearly by redraw-ing the circuit below

V1V2

Va

V1

R R'R R'||

V1

+

-

R

R R'|| Va

VaR R'||

R R R'||+----------------------- V1⋅=

where R R'||RR'

R R'+--------------=

-30-

This gives:

Now consider alone:

By superposition

VaR'

R 2R'+------------------ V1⋅=

V2

–

+

R

R

R'V2 Vb

You should be getting theidea by now. This time wereplace by a short circuit,and consider the effect of alone. The contribution to theoutput voltage is . The cir-cuit is drawn opposite. In fact,it is the same as the previouscircuit used to determine .This can be seen by noticingthat and are connectedin parallel, and that is con-nected to them through , sowe have the same voltagedivider as before, and we havea similar expression to the onederived for above.

V1V2

Vb

Va

R R'V2

R

Va

VbR'

R 2R'+------------------ V2⋅=

By superposition, the out-put voltage is , and wefind that the result is propor-tional to the sum of and

. It is not difficult to gener-alise the circuit to add morethan two voltages. When wemeet operational amplifierslater in this course we willmeet other methods for sum-ming voltages. Note that ourfinal result does not appear tobe valid when we set the resis-tors to zero. Can youexplain what is going on? Dis-cuss this with your supervisorif you are unsure.

Va Vb+

V1V2

R

Vout Va Vb+=

R'R 2R'+------------------ V1 V2+( )=

~

-31-

Lecture 4

Equivalent Generators

Thévenin’s Theorem

Any two terminal linear network maybe replaced by a voltage source inseries with a resistance .

We will now consideranother very useful techniquefor the analysis of linear cir-cuits. This is the method ofequivalent circuits, morecommonly known as theThévenin and Norton theo-rems.

We deal first withThévenin’s theorem. Theidea here is that any two ter-minal network containingvoltage and current sources,together with other linear cir-cuit elements, such as resis-tors, has exactly the sameelectrical characteristics asan equivalent circuit consist-ing of a single voltage source

in series with a singleresistor . We shall notgive a general proof ofThévenin’s theorem here, butit follows on from superposi-tion discussed in Lecture 3.The Thévenin voltage isequal to the open circuit volt-age seen at the output termi-nals of the linear circuit (seediagram opposite). TheThévenin resistance is givenin terms of the open circuitvoltage and the short circuitcurrent. Another way ofexpressing this, which isoften used in practice, is tosay that is equal to theresistance seen across theoutput terminals with allvoltage sources shortcircuited and all currentsources open circuited.

VThRTh

VTh

RTh

VThRTh

linearcircuit

–

+

RTh

VTh

VTh V open circuit( )=

RThV open circuit( )I short circuit( )-------------------------------------=

-32-

Note the following equivalence:

Design Example 4.1Investigate the loading effect of an ana-

logue voltmeter (200kΩ resistance) on a20 volt divide-by-two potential divider.

–

+I0 R0

RTh R0=

VTh I0R0=

Before looking at someexamples of Thévenin’s theo-rem I will mention one veryuseful result, the Théveninequivalent of a current gener-ator. Consider an ideal cur-rent generator with shuntresistance . The open cir-cuit voltage across the outputterminals is , and theresistance seen across theoutput terminals with the cur-rent source open circuited isjust . This gives us theThévenin equivalent shownopposite.

I0R0

I0R0

R0

When you connect a volt-meter across a circuit themeter’s internal resistancewill draw some current whichmay result in a false reading.In the case of digital meters,which have very high internalresistances, this loadingeffect can usually be ignored.In analogue moving coilinstruments however theinternal resistance is muchlower. It is usually specifiedby a figure like 20,000 ohmsper volt, meaning that whenthe meter is switched to theten volt scale, for example, itwill have an internal resist-ance of 200kΩ . In thisexample we will study theloading effect of such a meteron a potential divider circuit.The general situation is rep-resented by the circuit oppo-site, where we will eventuallyset for a divide bytwo potential divider. Wewill use Thévenin’s theoremto analyse the circuit,although any of the methodswe have met previouslycould also be used.

R1 R2=

R1

R2 RmVm

V0A

A'

general circuit

meter

-33-

Consider the Thévenin equivalent forthe circuit to the left of

The Thévenin parameters are:

With , and we have:

Our first step is to find theThévenin circuit for the partof the circuit to the left of

on the previous diagram.In doing so, we ignore for themoment the presence of themeter. The Thévenin voltageis the open circuit voltageseen at , assuming nometer is connected. This willjust be the voltage producedby the unloaded potentialdivider and , whichgives us

.

Next, we consider theThévenin resistance. The eas-iest way to get this is to notethat it is the resistance seenlooking to the left of with all voltage sources shortcircuited, and all currentsources open circuited. In ourcase there is only the voltagesource and shortcircuiting this we see that theresistors and are thenin parallel. This gives us

.

The result is the Théveninequivalent circuit drawnabove, with the meter shownconnected. The advantage ofusing Thévenin’s theorem forthis example is that theequivalent circuit is just apotential divider formed by

and . This makes itvery easy to visualise theloading effect of the meter.Let us now insert some com-ponent values. The meterresistance is 200kΩ, the volt-age source is 20V, and sincewe want a divide by two cir-cuit we set .Then, using the results justderived, the Thévenin voltageis 10V, and the Théveninresistance is , giving usthe circuit opposite.

AA'

AA'

R1 R2

VThR2

R1 R2+------------------ V0⋅=

AA'

V0

R1 R2

RThR1R2

R1 R2+------------------=

RTh Rm

R1 R2 R= =

R 2⁄

AA'

–

+RmVm

A

A'

RTh

VTh

RThR1R2

R1 R2+------------------=

VThR2

R1 R2+------------------ V0⋅=

V0 20 V= Rm 200 kΩ=R1 R2 R= =

–

+Vm

A

A'

10V

R 2⁄

200k

-34-

Voltage across meter is then given by:

Calculate for various values of

R ohms 1M 100k 10k 1k

volts 2.86 8.0 9.76 9.97

The Thévenin equivalentcircuit is a simple potentialdivider, so it is now straightforward to write down thevoltage across the meter. Thegeneral expression is

which reduces to the equationon the right when we insertthe component values givenin the previous circuit. In theexpression for we haveexpressed the resistor valuesin kΩ.

The table opposite showsthe meter readings for vari-ous values of the dividerresistors . Thecorrect meter reading (withno circuit loading) should be10 V. The loading effect ofthe meter is clearly apparentfrom this table.

Only when is thereading close to 10V. This isa general principle for allinstruments. Their internalresistance must be muchgreater than that of the circuitthey are connected to if accu-rate readings are to beobtained.

VmRm

RTh Rm+----------------------

·V0⋅=

Vm

R1 R2 R= =

Rm R»

Vm200

200 R 2⁄+-------------------------- 10 volts⋅=

Vm R

Vm

~

-35-

Design Example 4.2Use Thévenin’s theorem to find the

current I in the circuit below.

(i) Replace current generator withThévenin equivalent

+

10k10k

8mA

4k

15k75V

I

A B

CD

It is not immediatelyapparent how to applyThévenin’s theorem in thiscircuit. We are asked to findthe current through the resis-tor BC , which suggests thatwe find the Thévenin equiva-lent for the circuit to the leftof nodes B and C. A first stepwould be to make everythinginto a voltage source, whichsuggests that we start by find-ing the Thévenin equivalentfor the current generator and10k resistor to the left ofnodes AD.

10k

8mA

–

+80V

10kA

D

We have already seenhow to replace a current gen-erator with a Thévenin equiv-alent circuit. In this case, theopen circuit voltage is 8mAtimes 10kΩ, giving 80V forthe Thévenin voltage. Theresistance with the currentgenerator replaced by anopen circuit is 10k, which isthe required Thévenin resis-tor. This gives us the equiva-lent circuit shown opposite.

-36-

(ii) Replace voltage source and resistorswith Thévenin equivalent:

(iii) Redraw circuit diagram with theThévenin equivalents:

+

10k

15k75V+

30 V

6 kA B

Next we consider theThévenin equivalent for all ofthe circuit above the nodeslabelled AB, reproducedopposite. The open circuitvoltage across AB is theresult of the 75V generatorconnected across the seriescombination of the 10k and15k resistors, which there-fore act as a potential divider.The open circuit voltageacross AB is then given by

which gives us our Théveninvoltage of 30V. TheThévenin resistance is theresistance seen across ABwith the 75V generator shortcircuited. This is just the par-allel combination of the 10kand 15k resistors, giving

.

This gives us the 6k resistorshown in the equivalent cir-cuit.

Voc10

10 15+------------------ 75 volts⋅=

RTh10 15×10 15+------------------ kΩ=

–

+

80V

10k

+

30 V

6 k

4k I

B

C

A

D

We now redraw the origi-nal circuit with the currentand voltage generatorsreplaced by their Théveninequivalents derived above.The circuit is shown oppo-site. The original nodesABCD are shown so that it isclear where the Théveninequivalents are placed. Ourfinal step is to replace all ofthe circuit to the left of BC bya Thévenin equivalent. Theopen circuit voltage at BC isjust (80-30) = 50V, (note thenegative sign because thegenerators have oppositesign). The Thévenin resist-ance is (10+6) = 16kΩ . Thisproduces the equivalent cir-cuit shown on the next page.

-37-

(iv) Replace circuit to left of BC byThévenin equivalent:

The current I is given by:

–

+50 V 4 k

16 k

I

B

C

By applying Thévenin’stheorem three times we havereduced the original circuit tothe one shown on the left.The current through the 4kresistor is now easily calcu-lated.

Although we chose to useThévenin’s theorem for thisexample we could have ana-lysed the original circuit byany of the other techniquesthat we have met so far usingKirchhoff’s laws or superpo-sition. In many cases there isno one best method and it isoften a matter of preferenceas to which method of analy-sis is employed.

I 5016 4+( )

--------------------mA=

2.5mA=

~

-38-

Norton’s Theorem

Any two terminal linear network maybe replaced by a current source inparallel with a shunt resistance .

Note the following equivalence:

The second equivalentcircuit theorem that we shallmeet is called Norton’s theo-rem and, like Thévenin’s the-orem, applies to any twoterminal circuit consisting ofvoltage and current sourcestogether with other linearcomponents. Norton’s theo-rem tells us that such a circuitcan be replaced by a singlecurrent generator in paral-lel with a shunt resistance

. The Norton current isequal to the current seenwhen the output of the twoterminal circuit is shortcircuited. The Norton resist-ance is defined in thesame way as the Théveninresistance, and is also equalto the resistance seen look-ing back into the circuit withall voltage sources shortcircuited and all currentsources open circuited. LikeThévenin’s theorem, Nor-ton’s theorem is a conse-quence of the linearity of thecircuit.

IN

RN IN

RN

INRN

linearcircuit

IN RN

IN I short circuit( )=

RNV open circuit( )I short circuit( )-------------------------------------=

–

+V

R

VR--- R

A

B

An important conse-quence of Norton’s theoremis that a voltage source can berepresented by an equivalentcurrent source. Referring tothe diagram on the right, thecurrent with AB shortcircuited is , and theresistance seen from the ter-minals AB with the voltagesource short circuited is R.This gives us the equivalentNorton circuit shown.

V R⁄

-39-

Design Example 4.3Use Norton’s theorem to find the volt-

age across R in the circuit below.

(i) Apply Norton’s theorem to eachvoltage source:

–

+

–

+e2e1 R

R1 R2

A

B

We met this circuit previ-ously in Design Example 2.1.

We will see that the appli-cation of Norton’s theoremleads to a direct and simplesolution. We use the result onthe previous page to replacethe voltage sources to the leftand right of nodes AB withequivalent current sources.Consequently, for source we use the following Nortonequivalent, with a similar cir-cuit for .

This then gives us theequivalent circuit shownadjacent. Note that every-thing, current sources andresistors, are now all in paral-lel. This makes it very easy tosimplify the circuit further.The parallel resistors can allbe combined into a singleresistor of value

.

The current sources, being inparallel, can be replaced by asingle current source of value

All of this then leads to thecircuit on the next page.

e1

e2

e1

R1

R1e1R1------

+

-

11

R1------ 1

R--- 1

R2------+ +

-----------------------------

e1R1------

e2R2------+

R1 R R2e2R2------

e1R1------

-40-

(ii) Combine the current sources, andthe parallel resistors:

Voltage across resistor is then given by:

e1R1------

e2R2------+

11

R1------ 1

R2------ 1

R---+ +

-----------------------------

Here is the result of ourlabours. The circuit has sim-plified to a single currentsource connected across aresistor. It is a simple matternow to write down the volt-age. The result, given below,is of course the same as wederived in Design Example2.1. I have written it in thisform, without further simpli-fication, as the reductionusing the Norton equivalentsis then very clear. You shouldhave no trouble generalisingthis result to a circuit contain-ing N sources connected inparallel across a resistor R. Ileave it to you to show thatthe voltage in this case isgiven by

.

eiRi-----

1

N

∑

1R--- 1

Ri-----

1

N

∑+-----------------------

V

e1R1------

e2R2------+

1R1------ 1

R2------ 1

R---+ +

-----------------------------=

~

-41-

Circuit matching

Power matching

Maximum power in load when

Voltage matching

Voltage across load is given by:

For maximum voltage we need:

What value of load resis-tor should we choose to max-imise the power in the load inthis circuit? The currentthrough the load is

Then the power delivered tothe load is given by

and this is a maximum when

.

It follows from this that thecondition for maximumpower to be dissipated in theload is , a result youwill have met previously atA-level; we refer to this aspower matching.

iV

Rs RL+------------------=

RL

PL i2RL

RL

Rs RL+( )2

-------------------------- V2⋅= =

RLd

dPL0=

RL Rs=

–

+RLV load

Rs

RL Rs=

In many situations wewant to maximise the volt-age delivered to a load. Forexample, in the above circuitthe load might representthe input resistance of anamplifier, being fed by atransducer or an audio pick-up, represented by a voltagesource of internal resistance

. In this case, to maximisethe voltage delivered to we need to make muchlarger than the source resist-ance . Although manybooks emphasise powermatching, it is of less practi-cal importance than maximis-ing input voltage, and mostcircuits are designed so thattheir input resistance is muchlarger than the source driv-ing them.

RL

RsRL

RL

Rs

VLRL

Rs RL+------------------ V⋅=

RL Rs»

-42-

Current matching

Current through load is

For maximum current we need

Rs RLI

This situation is met lessoften than the previous two.Sometimes we may need tomaximise the current deliv-ered to a load from a currentsource. In this case, it is easyto see from the expression forthe current through the loadthat we need to make the loadresistance much smaller thanthe resistance of the currentsource.

ILRs

Rs RL+------------------ I⋅=

RL Rs«

-43-

A more difficult example

Design Example 4.4Use Thévenin’s theorem to find the

charging current in the following cir-cuit of a battery charger.

IB

VinAdj

Rs

R1

R2 VB

IB

I 0≈

Vout

Vref 1.25 V=Rs 0.2=R1 240=R2 2.4 k=

LM317

Vref

A

A'

X

This is a more difficultexample. It is difficult notbecause of harder mathemat-ics, but because you need tothink more carefully aboutthe circuit.

The circuit is of a 12Vbattery charger and is takenfrom a National Semiconduc-tors data sheet, who alsomake the LM317 used in thecircuit. The LM317 is adevice known as a voltageregulator, and is typical of thewide range of integrated cir-cuits available today. It isessentially a three terminaldevice which takes a voltageat the input Vin of between4–40V and converts it to avery stable reference voltageVref of 1.25V between theterminal Vout and theadjuster terminal Adj. Thedata sheet suggests the com-ponent values shown, andtells us that the batterycharger will gradually reducethe charging current as thebattery gets close to its fullycharged 12V state. No otherexplanation of the circuit isgiven. Our task, which is typ-ical of real life electronicdesign, is to analyse the cir-cuit and discover how itworks.

-44-

We will find Thévenin equivalent of thecircuit to the left of

(i) Find the open circuit voltage

Current through and :

(ii) Find short circuit current

With short circuited no currentflows through , .

Hence

AA'

We begin by making an open circuit, that is wecompletely remove the bat-tery . The voltage across

is then the same as thevoltage drop across and

since the current I flow-ing from the Adj terminal isnegligible. This means thatwe need the current i flowingthrough these resistors. Thisis also the same as the currentthrough and . The volt-age drop across and is

, so that the current ithrough these resistors is

.

Therefore, the voltage dropacross and is

, giving us theexpression for the open cir-cuit voltage opposite.

AA'

VBAA'

R1R2

Rs R1Rs R1

Vref

iVref

Rs R1+------------------=

R1 R2i R1 R2+( )

Voc

Rs R1

iVref

Rs R1+------------------=

Voc voltage drop across R1 and R2=

Voc i R1 R2+( )=

R1 R2+

Rs R1+------------------ Vref⋅=

iscThis is conceptually more

difficult. When is shortcircuited all the current willflow through , and therewill be no current through and . In particular, thevoltage drop across willbe zero. This means that thenode marked X is at earthpotential (zero volts) and that

is effectively droppedacross , giving us a shortcircuit current of .

AA'

RsR1

R2R2

VrefRs

Vref Rs⁄

AA'R1 R2

iscVrefRs

---------=

-45-

(iii) Thévenin parameters:

(iv) Draw the Thévenin equivalentcircuit:

We are now in a positionto construct the Théveninequivalent circuit. TheThévenin voltage is the opencircuit voltage derived above,and the Thévenin resistanceis defined as . This isan example of a situationwhere we have to use the for-mal definition of theThévenin resistance. With-out knowing the details of theinternal construction of thevoltage regulator we cannoteasily use the method of cal-culating the resistance seenlooking to the left of with voltage sources shortcircuited and current sourcesopen circuited as we did withprevious examples.

Voc isc⁄

AA'

VTh Voc=

R1 R2+

Rs R1+------------------ Vref⋅=

RThVocisc--------=

RsRs R1+------------------ R1 R2+( )⋅=

–

+VTh

RTh

VB

IB

And this is our Théveninequivalent circuit. The com-plexities of the original cir-cuit have been swept away toleave a very simple equiva-lent circuit from which it isnow easy to write down anexpression for the batterycharging current . Notethat this circuit contains eve-rything needed to predict thebehaviour of the original cir-cuit as far as connecting any-thing to the right of isconcerned. This is the realpower of the Thévenin equiv-alent circuit approach.

IB

AA'

IBVTh VB–

RTh---------------------=

-46-

Inserting values for and :

Inserting component values, we find

.

Consider how varies with

Charging current decreases as battery voltage increases.

(volts) 8 10 12 13.9

(amps) 2.65 1.75 0.85 0

VTh RTh

IBVrefRs

---------VBRs------

R1 Rs+

R1 R2+------------------

⋅–=

IB 6.25 0.45VB amps–=

We can now calculate as a function of using thecomponent values suggestedby National Semiconductors.In fact, the 240Ω value for

is suggested in the datasheet as the best choicebetween Vout and Adj for awide variety of applicationsfor this device. We see that,as expected, the charging cur-rent gradually reduces asthe battery voltage builds up.The table opposite showshow the current starts atabout 2.6A for a battery volt-age of 8V, and falls to zero ataround 13.9V. Most batter-ies will charge to a littleabove their nominal workingvoltage, so this is probably anacceptable design. Modifyingthe component values wouldallow the charging rate to beadjusted, but bear in mindthat the values of resistorschosen must fall within therange of those manufactured,which will place some con-straints on the design.

IBVB

R1

IBIB VB

VB

IB

~