Part IA—Linear Circuits and Devices
Cambridge University Engineering
Department
Michaelmas Term 2001
Lecturer: Dr David Holburn
Email: dmh@eng.cam.ac.uk
Lectures 1–4
Contents
.................................................................................................................... 1
Lecture 1
DC Circuits .................................................................................................
Ohm’s Law ...........................................................................................
Electrical power....................................................................................
Series and parallel circuits....................................................................
Design Example 1.1........................................................................
Design Example 1.2........................................................................
Voltage sources...........................................................................................
Ideal voltage source ..............................................................................
Real voltage source...............................................................................
Current sources ...........................................................................................
Ideal current source ..............................................................................
Real current source ...............................................................................
Design Example 1.3........................................................................
1
1
1
2
3
5
7
7
7
8
8
8
9
Lecture 2
Kirchhoff’s Laws I....................................................................................
The voltage law ..................................................................................
The current law ...................................................................................
Design Example 2.1......................................................................
Design Example 2.2......................................................................
11
11
13
14
18
Lecture 3
Kirchhoff’s Laws II ..................................................................................
Mesh current loop analysis .................................................................
Design Example 3.1......................................................................
Design Example 3.2......................................................................
Superposition ............................................................................................
Design Example 3.3......................................................................
21
21
23
24
27
29
Lecture 4
Equivalent Generators ..............................................................................
Thévenin’s Theorem...........................................................................
Design Example 4.1......................................................................
Design Example 4.2......................................................................
Norton’s Theorem...............................................................................
Design Example 4.3......................................................................
Circuit matching .......................................................................................
Power matching ..................................................................................
Voltage matching................................................................................
Current matching ................................................................................
A more difficult example..........................................................................
31
31
32
35
38
39
41
41
41
42
43
Design Example 4.4...................................................................... 43
Lecture 1
DC Circuits
Ohm’s Law
I
+
V
–
R
V = IR
Electrical power
R
I
Ohm’s law forms the basis
for all DC electrical circuit
analysis. It was first suggested
by Georg Simon Ohm (1787–
1854) and was published as a
memoir in 1826. Ohm waited
many years for proper recognition for his work, eventually
receiving the Royal Society
Copley Medal in 1841. Ohm’s
memoir laid the foundation
for much subsequent work on
electrical theory in the second
quarter of the 19th century.
Note that by convention,
current flows from high to
low potential, as though it
were composed of positive
charges. We know of course
that electrical current is the
flow of electrons. The reason
for using the positive convention dates back to Benjamin
Franklin (1706–1790) who
postulated that only one kind
of positive electrical charge
existed. For largely historical
reasons we still use that convention.
The units of V are volts
(V); the units of I are amps
(A); and the units of R are
ohms (Ω).
V
Power = IV
-1-
Power is measured in
Watts (W). The expression for
power may be written equiva2
2
lently as I R or V ⁄ R .
-2-
Series and parallel circuits
series resistors:
You will already have met
these results. They are left as
a revision exercise for you to
prove.
R1
R2
RN
R = R1 + R2 + … + RN
parallel resistors:
R1
R2
RN
In the case of two resistors
in parallel this result takes a
particularly simple form,
which is used so often that
you should know it by heart:
R1 R2
R = ------------------ .
R1 + R2
“Product/Sum”
For parallel resistors it is
often more convenient to
work in terms of conductance,
defined as G=1/R. Then, for
parallel resistors we just add
the conductances. The units of
–1
conductance are Ω , or alternatively siemens.
1 = ----1- + ----1- + … + -----1
--R
R1 R 2
RN
conductance G ≡ -1-R
G = G1 + G2 + … + GN
-3Design Example 1.1
A fuse can be modelled as a temperature dependent resistor R where
R = R 0 + α∆T . The temperature rise
∆T is assumed to be given by ∆T = kP
where P is the electrical power dissipated in R. Derive R as a function of
current I.
We assume
2
P = I R
then
2
∆T = kP = kI R
But R = R 0 + α∆T which gives us:
2
R = R 0 + αkI R
Solving for R
R0
R = -------------------21 – αkI
Many electrical devices do
not obey Ohm’s law, and in
this example we will study a
simple model for the electrical
fuse. The basic principle of
the fuse is more complicated
than many realise. As the current
through
the
fuse
increases, it heats up, and its
resistance increases, which
causes the voltage drop across
the fuse to increase. This in
turn causes the current to rise
further, and thermal runaway
results. The fuse fails abruptly
at a critical current, say, IF.
We assume that the resistance of the fuse increases linearly with temperature, and
that the temperature rise is
proportional to the electrical
power dissipated in the fuse.
We will also assume that the
electrical power dissipated by
2
the fuse is given by I R . In
effect, this defines what we
mean by resistance in this
model. Strictly speaking, we
should define the resistance as
∂V ⁄ ∂I , and the power dissipated as IV . This would lead
to a differential equation for V
which can be solved. However, our simpler approach
will produce essentially the
same qualitative predictions.
-4We introduce I F so that we
can express R in terms of the
normalised current I ⁄ I F . As
we will see shortly, IF defines
the critical current at which
the fuse ‘blows’.
1---------Define I F =
αk
then we can rewrite R as
R0
R = -------------2I
1 – ----2
IF
As expected, the resistance of the fuse depends nonlinearly on the current. It does
not obey Ohm’s law, except at
low currents. The expression
for R is plotted opposite as a
function of the normalised
current I ⁄ I F . Note that until I
gets very close to I F , R
increases only very slowly.
Near I F the resistance rises
very rapidly, and in our model
becomes infinite; that is the
fuse suddenly ‘blows’, and in
practice would melt. In a
more realistic model, where R
is defined as ∂V ⁄ ∂I , the rise
would is not infinite, but is
still very rapid as the current
approaches I F .
It is interesting to see that
in this model the critical current at which the fuse ‘blows’
does not depend on factors
such as the melting point of
the fuse wire. Real fuses of
course are much more complicated. Nevertheless, our graph
of R vs I is very similar qualitatively to the experimental
behaviour.
R ⁄ R0
20
15
10
5
0
0
0.2
0.4
0.6
0.8
1
I ⁄ IF
fuse ‘blows’ when R → ∞
1i.e. when I → ---------αk
~
-5Design Example 1.2
A long uniform cable AB is damaged
and has a leakage resistance to earth at
an unknown point L. Find the position
of the leakage resistance using resistance measurements taken at end A only.
A
L
B
x
a-x
y
leakage resistance
This is a famous problem
of historical interest. During
the 19th century, when submarine telegraph cables came
into wide spread use, there
was a need to be able to locate
faults along a cable quickly
and accurately. Most faults
consisted of some sort of
damage resulting in a leakage
resistance to earth. Here, we
consider a method of fault
location due to Oliver Heaviside (1850–1925), which he
published in the May 3rd
issue of the Electrician in
1901. The original account
was found in his notebooks,
dated January 16th 1871.
(i) short circuit B to earth
resistance measured at A is
y(a – x )
-------------------R sc ≡ c = x +
y+a –x
(ii) open circuit B
resistance measured at A is now
Roc ≡ b = x + y
If we assume that the
cable has a known constant
resistance per unit length,
then the problem reduces to
finding the resistance x
between A and the leakage
point at L. Assume that the
total cable resistance between
AB is a. We need to find two
equations so that we can eliminate the leakage resistance y.
We do this by measuring the
resistance at A whilst the end
B is short circuited, and then
open circuited. When B is
short circuited, we measure x
in series with the parallel
combination of y and (a-x).
When open circuited, we
measure x in series with y.
-6The leakage resistance y is
eliminated from the above
two equations, leaving a
quadratic in x. This is easy to
solve, but you do need to
think a little about which of
the two possible solutions is
valid.
eliminating y from these equations
2
x – 2cx + ac + bc – ab = 0
solving for x gives
x = c ± ( a – c )(b – c )
This is Heaviside’s result.
Unknown to Heaviside, it was
actually first derived by
Edouard
Blavier
(1826–
1887).
In any case, it turns out
not to be a good method.
What do you think are its
main disadvantages?
choosing the negative sign,
x = c – ( a – c)(b – c )
~
-7-
Voltage sources
Ideal voltage source
In electronic circuit design
we frequently make use of the
concept of an ideal voltage
source. By this we mean a
two terminal device whose
output voltage V 0 is independent of any load resistor
R L connected across its output terminals. Since the current supplied by an ideal
voltage source is given by
V 0 ⁄ R L it is clear that it must
be able, in principle, to provide an unlimited amount of
current. A good approximation to an ideal voltage source
is a newly charged car battery.
+
V0
V = V0
–
RL
Voltage independent of the load R L
Real voltage source
+
V0
–
Rs
V ≠ V0
Voltage depends on the load R L
RL
A real voltage source can
be thought of as an ideal voltage source together with an
internal series resistance R s .
For example, a 12V newly
charged car battery has an
internal resistance of about
0.006Ω , and a AA (penlight)
1½ volt battery has an internal
resistance of about 0.4 Ω . A
general purpose, low voltage
electronic power supply might
have an internal resistance of
around 0.1Ω .
This means that the output
from a real voltage source will
depend on the size of the load
resistor R L . Voltage sources
‘prefer’ to operate - that is,
they operzte best - with their
output
terminals
open
circuited, and ‘dislike’ being
short circuited.
-8-
Current sources
In addition to voltage
sources we often employ a
second type of source, the
ideal current source. This is a
two terminal device which
delivers a constant current
irrespective of the load connected across its terminals.
This means that, in principle,
it can sustain an arbitrary voltage across a load, V = I 0 R L .
Because of this, current
sources ‘like’ to operate
(operate best) with their terminals short circuited. They
‘dislike’ open circuits.
Whereas it is easy to think
of approximations to voltage
sources, there are no real analogies to help us visualise current sources. They are no less
important, however, and are
an important component in
the design of many electronic
circuits.
In practice, we deal with
real current sources, which we
think of as an ideal source
with an internal resistance Rs
shunted across (i.e. in parallel
with) its output terminals.
This means that the current
delivered to a load will
depend on the size of the load
resistance R L . In a later lecture we will see how a current
source can be approximated
using a field effect transistor
biased in the following way:
Ideal current source
I0
RL
I = I0
Current independent of load R L .
Real current source
I0
Rs
RL
I ≠ I0
+V
I0
RL
Current depends on load R L .
-9Design Example 1.3
Measurements on a power supply produced the following results:
(a) open circuit voltage = 10 V ;
(b) short circuit current = 20 A ;
Represent the power supply by (i) an
equivalent voltage source with series
conresistance; and (ii) an equivalent current trived,Thisas isyousomewhat
would not normally short circuit a power
source with parallel resistance.
supply. A better method
(i) voltage source
+
½ V0
Rs
would be to connect a variable
known resistor across its output terminals and measure the
resultant voltage and current.
What this example does bring
out however is the idea that
the same two terminal device
can be represented by either
an equivalent voltage source,
or an equivalent current
source.
We shall meet this idea in
much more detail when we
study Thévenin’s and Norton’s theorems.
–
open circuit voltage = V 0 = 10 V
short circuit current = V0 ⁄ Rs = 20 A
R s = V 0 ⁄ 20
R s = 0.5Ω
When the power supply is
open circuited no current is
drawn, so that its full voltage
is seen across the output terminals. When it is short
circuited, the voltage is
dropped across its internal
resistance.
-10(ii) current source
When a current generator
is short circuited, all of its
current flows via the output
terminals. This is because no
voltage drop appears across
its internal shunt resistance R s . When it is open
circuited, the entire current
flows through Rs and appears
as an output voltage I0 R s .
Rs
I0
½
short circuit current = I 0 = 20 A
open circuit voltage = 10 V = I 0 R s
R s = 10 ⁄ I 0
R s = 0.5Ω
R s is often written as a conductance
1G s ≡ ---Rs
For the above example:
Current generators are
often specified in terms of an
internal shunt conductance,
defined by
1
G s = ----- .
Rs
G s = 2Ω
–1
= 2 siemens .
~
Lecture 2
Kirchhoff’s Laws I
The voltage law
V1
I
R1
+
V0
I
R2
–
V2
let current I flow for time δt
work done by
battery
=
energy loss in
resistors
( IV 0 )δt = ( IV 1 )δt + ( IV 2 )δt
-11-
Gustav Robert Kirchhoff
(1824–1887) was born in
Königsberg. Although well
known for his electrical circuit laws, he is best known for
his work on the emission and
absorption of electromagnetic
radiation by black bodies.
His circuit laws were his
first piece of scientific work,
carried out at the age of 21
during the period 1845–46.
He was also, in 1849, the first
person to clarify Ohm’s rather
vague ideas of electrical tension, showing that it should be
identified with the electrostatic potential, thus unifying
the theories of electrostatic
and voltaic currents.
Consider the simple circuit drawn above. The voltage
source is driving a current I
around the circuit, and in
doing so performs work at the
rate of IV 0 . This work must
go somewhere by energy conservation, and it results in the
heating up of the two resistors. If the voltage drop across
R 1 is V 1 then it heats up at
the rate of IV 1 . If the current
flows for a small time δt then
the energy loss from the battery will be IV 0 δt , and this
must equal the energy dissipated in the two resistors
IV 1 δt + IV2 δt .
-12V0 – V 1 – V 2 = 0
We can cancel I and δt on
both sides of the energy balance equation, to leave an
equation relating the voltage
drops around the different elements in the circuit. For the
circuit above, this says that
the sum of the voltage drops
for each component around
the entire circuit must add up
to zero.
∑ Vi
= 0
loop
Now consider a more general circuit
The result derived for the
simple circuit above can be
shown to be true for the voltages around any closed circuit, whatever voltage or
current sources, or other components are present, and is
known as Kirchhoff’s Voltage
Law. The circuit components
do not even need to be linear,
or obey Ohm’s law; the voltages and currents may also be
varying with time, although at
very high frequencies (in the
GHz) Kirchhoff’s voltage law
breaks down. To prove all this
would go a long way beyond
this course, and involves a
knowledge of Maxwell’s
equations which you will
meet in the second year.
Note that when applying
Kirchhoff’s law you must be
very careful to get the algebraic signs of the voltage
drops correct.
V2
V1
V3
VN
Kirchhoff’s voltage law states:
∑
closed
loop
Vi = 0
-13-
The current law
I
I1
I2
I3
R1
R2
R3
+
V0
–
We saw that Kirchhoff’s
first law can be considered to
result from energy conservation around a circuit. His second law is the result of charge
conservation. Consider a circuit like the one shown opposite. We will assume that no
charge can accumulate on the
component leads or on the
wires joining them. Then, any
charge entering a component
must be balanced by an equal
amount of charge leaving. To
put this another way, the algebraic sum of the currents
entering a junction must be
zero. This is Kirchhoff’s Current Law.
Conservation of charge implies:
I – I1 – I2 – I3 = 0
For a general node, we have
IN
I1
I2
Kirchhoff’s current law:
I3
I4
∑
into
a node
In = 0
More generally, Kirchhoff’s current law says that
the algebraic sum of the currents entering any node in a
circuit must be zero. Again,
care is often needed to get the
signs of the currents correct.
-14Design Example 2.1
Find the voltage across R in the following circuit. Use (i) Kirchhoff’s Voltage
Law, and (ii) Kirchhoff’s Current Law.
The same circuit can be
solved using either Kirchhoff’s voltage law or current
law. We will use the circuit
opposite to illustrate both
methods of solution.
The circuit represents two
batteries, with voltages e 1
and e 2 , and internal resistances R 1 and R 2 respectively.
The batteries are connected in
parallel, the same way round,
across a load of resistance R.
We are to determine the
resulting voltage across the
load. We begin by using
Kirchhoff’s voltage law, and
this method is often called
mesh current analysis. We
label the currents through R 1
and R 2 as I 1 and I 2 , with
directions as shown on the
diagram. You are, of course,
free to label the current directions as you wish. The current
through R is then given by
I1 + I2 .
Now consider the loop
abef. Kirchhoff’s law tells us
that the sum of voltages
around this loop must be zero.
This gives us one equation
relating unknown currents I 1
and I 2 . Note that for R 1 the
voltage drop is from a to b for
the current direction chosen.
This is opposite to the voltage
e 1 , and is the reason for the
negative signs in our equation. We get a second equation by similarly applying
Kirchhoff’s law to loop bcde.
To determine the voltage
across R we need I 1 + I 2 , and
in solving these two equations
it therefore makes sense to
define I 1 + I 2 by a new variable x.
a
+
e1
–
I1
I2
b
R1
c
R2
R
f
I1 + I2
e
+
e2
–
d
(i) Using KVL (mesh current analysis)
Consider voltages around loop abef:
e 1 – I 1 R 1 – ( I 1 + I 2 )R = 0
around loop bcde:
e 2 – I 2 R 2 – ( I 1 + I 2 )R = 0
define x by
x = I1 + I2
-15e 1 – I 1 R 1 – Rx = 0
e 2 – ( x – I 1 )R 2 – Rx = 0
We now rewrite our two
equations in terms of x and I 1 .
It is not difficult to eliminate
I 1 to leave one equation for x
which we then solve.
eliminating I 1 we find
e1 R2 + e2 R 1
x = ---------------------------------------------R 1 R 2 + R ( R1 + R 2 )
voltage across R is given by
V R = ( I 1 + I 2 )R = xR
The voltage across R is
given by xR.
using x from above,
e1 R2 + e2 R1
V R = R ⋅ ---------------------------------------------R1 R2 + R ( R1 + R2 )
we can rearrange this as
e1 e2
------ + -----R1 R2
V R = ----------------------------1 ----1
1
--+ - + -----R R1 R2
We can check that this
result makes sense by seeing
what happens if we make R 2
infinite, when the resistors R 1
and R form a simple potential
divider. Our equation opposite
then reduces to
R
V R = e 1 ⋅ ---------------R + R1
as expected. A similar thing
should happen if we make R 1
infinite, as you can check.
-16(ii) Using KCL (nodal voltage analysis)
I1
We will now see how the
same circuit can be solved
using Kirchhoff’s current law.
This is often referred to as
nodal voltage analysis. We
choose one node in the circuit
as a reference node, and label
the voltages at all the other
nodes with respect to this reference. In our case, there is
one unknown nodal voltage,
which we have labelled A.
The voltage drops across each
component are then calculated. For example, the voltage drop across R 1 is e 1 – V A .
The currents flowing at each
node are then written down.
The current into the node A
through R 1 is
e 1 – VA
I 1 = ----------------R1
with similar equations for the
other two currents. Kirchhoff’s current law then says
that the sum of the currents
flowing into each node must
be zero. In our case, there is
just one node A, and this leads
to an equation for the
unknown voltage across R.
+
e1
–
R1
I2
VA
A
R2
I3
R
+
e2
–
0
zero reference
Write down currents into node at A
e1 – VA
I 1 = ----------------R1
e2 – VA
I 2 = ----------------R2
VA
I 3 = -----R
sum of currents into node = 0
I1 + I2 – I3 = 0
-17e1 – VA e2 – VA VA
----------------- + ----------------- – ------ = 0
R1
R2
R
rearranging this
VA
e1 e2
1
1
1
--- + ------ + ------ = ------ + -----R R1 R 2
R1 R2
from which we find
e1 e2
------ + -----R1 R2
V A = ----------------------------1 ----1
1
--+ - + -----R R1 R2
~
We obtain the same result
as before of course. But note
that the nodal voltage method
requires rather less algebra
compared to the mesh current
analysis. This is no accident,
and in many circuits one
method will be quicker than
the other. We will see why
this is so much more clearly in
a later lecture, when we study
Thévenin’s and Norton’s theorems. In the present case
recall from the last lecture
how we saw that a voltage
source could also be represented as a current source
with a shunt resistance. In our
present circuit, the voltage
source e 1 and R 1 can be represented as follows
e
-----1R1
R1
with something similar for the
other source. Then, the three
resistors R , R 1 and R 2 appear
in parallel, and the two current sources produce a total
current of
e
e
-----1- + -----2R1 R2
The way in which we have
written the final result clearly
brings this out.
-18Design Example 2.2
The following circuit represents a simple battery charger. Derive an expression for the current through the battery.
A battery charger works
by delivering a steady charging current to the battery until
it reaches its fully charged
state; depending on the
design, the current may
remain constant throughout
the charging period, or may
be reduced to a trickle charge
when the battery is nearly
fully charged. The circuit
opposite represents a simple
battery charger where the current is provided by the current
generator I0 with internal
shunt resistance R s . The battery has internal resistance
R B . We will use Kirchhoff’s
laws to find the battery charging current I B as a function of
the other component values.
RB
V
A
I0
IB
+
–
Rs
zero reference
(i) nodal voltage analysis (using KCL)
consider currents at node A
There
is
only
one
unknown nodal voltage, that
at A, which we shall call V
with respect to the zero reference node. The voltage across
R B is given by V – V B and the
current through this resistor is
given by
V–V
----------------B
RB
Similarly, the current through
R s is V ⁄ R s . The sum of the
currents at node A must be
zero, and this gives us an
equation for the unknown
nodal voltage V.
VB
V – VB
V
I 0 – ----- – ---------------- = 0
Rs
RB
VB
I 0 + -----RB
V = ------------------1- ----1---+
Rs RB
-19battery current is given by
The charging current
through the battery is just the
current through R B . The voltage drop across R B is V – V B .
Then, using the expression for
V found above, we obtain an
expression for the charging
current.
V – VB
I B = ---------------RB
using V derived above, we find
VB
Rs

I B = ------------------ ⋅ I 0 – ------ .
Rs 
RB + Rs 
(ii) mesh current analysis (using KVL)
a IB
RB
b
I0 – IB
I0
Rs
d
+
–
VB
c
consider voltages around loop abcd:
I B R B + V B – ( I 0 – I B )R s = 0
Consider
loop
abcd.
Kirchhoff’s voltage law tells
us that the sum of the voltages
around this loop must be zero.
The voltage drop across R B is
I B R B , and the voltage drop
across R s is ( I 0 – I B )R s .
Applying Kirchhoff’s law
then gives us an equation for
the battery current I B .
-20solving for I B we again find
As expected, we get the
same result as using nodal
voltage analysis, but with
rather less work. Also, the
charging current must be positive for the battery charger to
work, and this implies an inequality between I 0 , V B and
R s . How this is realised will
depend on the design of the
current generator. You will
see one way of implementing
this circuit using a standard
integrated circuit in the next
lecture.
VB
Rs

I B = ------------------ ⋅ I 0 – ------
Rs 
RB + Rs 
note: charging current positive only if
VB
I 0 > -----Rs
~
Lecture 3
Kirchhoff’s Laws II
Mesh current loop analysis
+
E1
–
R1
R2
R3
I1
I2
I3
I4
R4
R5
I5 R 6
I6
There are six currents to determine.
Current conservation:
I4 = I1 – I2
I5 = I2 – I3
I6 = I5 + I3
I1 = I4 + I6
-21-
With even moderately
complex circuits Kirchhoff’s
laws quickly lead to a large
number of equations. The
problem with this is that some
of these equations will be
redundant, they repeat information already contained in
other equations. The circuit
above illustrates this. There
are six possible unknown currents to determine, and applying current conservation and
Kirchhoff’s voltage law leads
to seven equations for the six
unknowns. In principle, there
is no harm in this, as you will
get the correct solution by
solving the equations. But it
does become progressively
worse as the circuit becomes
more complex. It also makes
it very difficult to know which
equations to work with to
avoid going round in circles.
The purpose of this lecture is
to introduce a more systematic way of applying Kirchhoff’s voltage law which
avoids the problem of redundant equations.
-22Apply Kirchhoff’s voltage law:
E1 – I1 R1 – I4 R4 = 0
I2 R2 + I5 R5 – I4 R4 = 0
I3 ( R3 + R6 ) – I5 R5 = 0
There are seven equations with six
unknowns.
Simpler method: use loop analysis:
One way to ensure that
only the minimum number of
circuit equations needed are
generated is to represent the
currents as current loops, with
one for each mesh of the circuit. The reason that this
works is that current conservation at the nodes is automatically enforced. For the circuit
considered on the previous
page there are three independent meshes, so we introduce
three current loops, I 1 , I 2 ,
and I 3 , as shown. This means
that the current flowing down
through resistor R 4 is given
by ( I 1 – I 2 ) , and the voltage
drop across R 4 is ( I 1 – I 2 )R 4 .
Similarly, the voltage drop
across R 1 is I 1 R 1 . These are
both in the opposite direction
to that chosen for E 1 and
accounts for the sign difference in the first circuit equation opposite. It is easy to see
that applying Kirchhoff’s
voltage law around each mesh
will produce three equations
for the three unknown currents.
+
E1
–
R1
R2
R3
I1
I2
I3
R4
R5
R6
E1 – I1 R1 – R4 ( I1 – I2 ) = 0
I 2 R 2 + ( I 2 – I 3 )R 5 + ( I 2 – I 1 )R 4 = 0
I 3 ( R 3 + R 6 ) + ( I 3 – I 2 )R 5 = 0
Now there are only three equations.
-23Design Example 3.1
Determine the balance condition for
the Wheatstone bridge circuit below.
R1
+
–
i2
i1
R2
G
R
i1
X
Apply Kirchhoff’s voltage law to the
two current loops i 1 :
i 1 R 2 + ( i 1 – i 2 )R1 = 0
i 1 X + ( i 1 – i 2 )R = 0
Eliminate i 1 and we get:
R2
XR
----= ------ or X = R ⋅ -----R2
R1
R1
~
The Wheatstone bridge
affords a good example of the
use of current loops for circuit
analysis.
Charles Wheatstone came
from a family of musical
instrument makers, and had
no formal scientific training,
starting out himself in the
family music business. In
1823 he patented the concertina. By 1834 he was Professor of Physics at King’s
College London. Rather less
well known is the fact that he
was the uncle of the great
British electrical engineer
Oliver Heaviside.
In the circuit above, X is
an unknown resistor to be
measured, R is a precision
potentiometer, and R 1 and R 2
are known standard resistances.
At balance, the current
through the galvanometer G is
zero. This implies that there
should be two equal current
loops i 1 around the two
meshes of the resistor bridge
as shown. There will be a second current loop i 2 around
the outer mesh containing the
battery.
Since the balance condition will not depend on the
battery voltage we do not
need to apply Kirchhoff’s law
around the loop including the
battery.
-24Design Example 3.2
Determine the balance condition for
the Kelvin double bridge shown below.
Although the Wheatstone
bridge is capable of high
accuracy it is not well suited
to measuring very low resistances, of the order 0.001Ω .
The reason for this is that the
leads to such a resistor will
have a comparable resistance
to what is being measured. To
overcome this, Sir William
Thomson (1824–1907), later
known as Lord Kelvin,
invented a new bridge method
called the Kelvin double
bridge. This is one of the most
accurate methods available
for measuring very low resistances. The principle of this
method is shown by the adjacent circuit. The unknown
resistor is X, and S is an accurate standard resistor comparable in value to X. The
resistors Q, M, and q, m, are
standard resistors of ordinary
values. Resistors Q and q are
variable resistors, and can be
adjusted together so that the
ratio Q ⁄ M = q ⁄ m . Resistor r
is a very low resistance made
by using a thick conductor.
The reason this bridge works
is that the lead resistances of
X are either in series with a
much higher resistance, or in
a part of the circuit (r) where
their resistance does not matter.
The circuit has four
meshes, so we introduce a
current loop for each. At balance two of these will be
equal, labelled i1 . We now
apply Kirchhoff’s law by
summing the voltages around
each mesh to zero. We do not
worry about the mesh containing the battery, as the balance
condition will not depend on
the battery voltage.
Q
M
G
i1
i1
q
X
m
S
i2
i3
r
Apply Kirchhoff’s voltage law:
i 1 Q + ( i 1 – i 2 )q + ( i 1 – i 3 )X = 0
i 1 M + ( i 1 – i 3 )S + ( i 1 – i 2 )m = 0
( i 1 – i 2 )q + ( i 3 – i 2 )r + ( i 1 – i 2 )m = 0
-25To solve these equations we define new
The circuit equations are
variables:
not difficult to solve, but are
rather messy; a little thought
will simplify the analysis considerably. We note that the
current differences ( i 1 – i 2 )
and ( i1 – i 3 ) occur several
times in the equations, and so
it makes sense to define these
as new unknowns i 12 and i 13 .
i 12 = i 1 – i 2
i 13 = i 1 – i 3
The circuit equations then become:
i 1 Q + i 12q + i 13X = 0
(1)
In terms of i 12 and i 13 the
equations take the form given
opposite. Note that in writing
these equations we have made
use of the following relation
i 1 M + i 13S + i 12m = 0
(2)
i 3 – i 2 = i 12 – i13
i 12q + ( i 12 – i 13 )r + i 12m = 0
(3)
From equation (3)
i 12
i 13
= ---------------------- ⋅ r
r+q+m
which follows from the definition of i 12 and i 13 .
Now it is much easier to
see what is going on. Equation 3 has only two
unknowns, i 12 and i 13 , so this
is probably the best one to
start with. We will use equation 3 to express i 12 in terms
of i 13 and then substitute this
in equations 1 and 3.
Substitute into equations (1) and (2)
i 13
i 1 Q + ---------------------- ⋅ rq + i 13X = 0
r+q+m
i 13
i 1 M + i 13S + ---------------------- ⋅ rm = 0
r+q+m
We now have just two
equations. It is straight forward to use one of them to
express i1 in terms of i 13 and
to substitute for this in the
other; i 13 will then cancel on
both sides, leaving an expression for X.
-26Eliminating i 1 we find:
The variable resistors Q
and M are adjusted so that the
ratios Q ⁄ M and q ⁄ m are
equal. In practice this is
achieved by them being part
of a double ratio arm resistance box. Then we see that
the term on the right hand side
containing the resistance r
vanishes, and with it any contribution from the connecting
leads of the unknown resistor. The unknown resistor X is
then determined from Q, M,
and S.
When measuring small
resistances
thermo-electric
effects are important, but by
reversing the battery and averaging the results they can be
eliminated. If this is done, the
Kelvin double bridge can typically measure resistances of
the order 0.01–0.001Ω to an
accuracy of 0.02%.
QSQ- --q-
mr

--------X =
+ ----------------------  – 
M r+q+m M m
The bridge is adjusted so that:
Q
q
----- = ---M
m
Then the unknown resistor X is given
by:
Q- ⋅ S
X = ---M
~
-27-
Superposition
The response of a linear network con- The idea of superposition
is a powerful one. It allows
with many voltage
taining several sources is found by con- circuits
and current sources to be
to a series of simpler
sidering each source separately and then reduced
circuits, each containing only
one source.
This simplification must
adding the individual responses.
be offset against the larger
amount of work required to
IB RB
obtain a solution.
I0
Rs
+
–
VB
Consider voltage source alone:
Ia
RB
(a)
Rs
+
–
VB
The Superposition Theorem, quoted opposite, is valid
for networks of linear elements. These can include
ideal resistors, capacitors and
inductors. Such linear components are central to the ideas
covered in this course, and it
is convenient to define the
term linear at this point. A
linear element is one whose
properties depend only on its
geometry and the materials of
which it is made, not on the
current or voltage applied.
The ideal resistors we have
seen already meet this specification.
The circuit opposite illustrates the idea (we considered
it earlier, in Design Example
2.2). We will calculate the
battery charging current I B
using superposition. We consider the effect of each source
alone, with all the others
removed. If we remove the
current source we are left with
an open circuit in its place.
This is because an ideal current source has an infinite
impedance. The circuit then
reduces to (a) opposite, where
we denote the battery current
by I a . Note that in this case
we have kept the direction of
I a the same as I B , although
we are free to choose any
direction we wish.
-28The current I a is given by
The battery voltage V B
appears across the series combination of R B and R s ; the
direction chosen for the current I a results in the minus
sign.
VB
I a = – -----------------RB + Rs
Now consider the current source alone
Next, we remove the voltage source and consider the
effect of the current source
alone. An ideal voltage source
has zero impedance, so we
replace it by a short circuit.
The resulting current is I b .
The circuit has now reduced
to a current generator I 0 connected to the parallel combination of R B and R s .
Consider the voltage V at the
node labelled a. This is given
by
V Ib
a
I0
RB
Rs
V = I 0 ⋅ R B || R s
where R B || R s denotes the
parallel combination of R B
and R s , so that
RB R s
V = I 0 ⋅ -----------------R B + Rs
The current I b is then given
by
V
I b = -----RB
and this leads to the expression given opposite. Note that
this is just the current equivalent of the potential divider,
with the current I 0 dividing
between R B and R s in inverse
proportion to the resistances.
The total current I B
through the battery is the
superposition of I a and I b
and results in the same answer
as we obtained in Worked
Example 2.2.
Rs
I b = ------------------ ⋅ I 0
RB + Rs
Total battery current is then:
IB = Ia + Ib
VB
Rs

= ------------------ ⋅  I 0 – ------
Rs
RB + Rs
(b)
-29Design Example 3.3
Determine the response of the voltage
adding circuit below.
R
R
+
V1
–
+
V2
–
R' V out
Consider V 1 alone:
R
R
+
V1
–
R'
Va
The circuit opposite is
known as a voltage adder, and
is a simple way of producing
an output proportional to the
sum of two independent voltage sources V 1 and V 2 . In
fact we have already met this
circuit in a different context in
Design Example 2.1. We will
use superposition, considering
the contribution to V out from
each source separately, and
adding the two results.
Consider the effect of V 1
alone. We remove V 2 by
replacing it with a short circuit. The resultant output voltage is denoted by V a in the
circuit opposite. It is not
immediately obvious from the
way the circuit is drawn, but
the circuit now consists of V 1
connected across a potential
divider, consisting of R in
series with the parallel combination of R and R' , which we
denote by R || R' . We can see
this more clearly by redrawing the circuit below
R
R || R' ⋅ V
V a = ----------------------R + R || R' 1
RR'
where R || R' = -------------R + R'
+
V1
-
R || R'
Va
-30This gives:
R' - ⋅ V
V a = ----------------R + 2R' 1
Now consider V 2 alone:
You should be getting the
idea by now. This time we
replace V 1 by a short circuit,
and consider the effect of V 2
alone. The contribution to the
output voltage is V b . The circuit is drawn opposite. In fact,
it is the same as the previous
circuit used to determine V a .
This can be seen by noticing
that R and R' are connected
in parallel, and that V 2 is connected to them through R , so
we have the same voltage
divider as before, and we have
a similar expression to the one
derived for V a above.
R
R
+
V2
–
R'
Vb
R' - ⋅ V
V b = ----------------R + 2R' 2
By superposition, the output voltage is V a + V b , and we
find that the result is proportional to the sum of V 1 and
V 2 . It is not difficult to generalise the circuit to add more
than two voltages. When we
meet operational amplifiers
later in this course we will
meet other methods for summing voltages. Note that our
final result does not appear to
be valid when we set the resistors R to zero. Can you
explain what is going on? Discuss this with your supervisor
if you are unsure.
By superposition
V out = V a + V b
R' - ( V + V )
= ----------------2
R + 2R' 1
~
Lecture 4
Equivalent Generators
Thévenin’s Theorem
Any two terminal linear network may
be replaced by a voltage source V Th in
series with a resistance R Th .
R Th
linear
circuit
V Th
+
–
V Th = V ( open circuit )
R Th
V
( open circuit )-----------------------------------=
I ( short circuit )
-31-
We will now consider
another very useful technique
for the analysis of linear circuits. This is the method of
equivalent circuits, more
commonly known as the
Thévenin and Norton theorems.
We deal first with
Thévenin’s theorem. The
idea here is that any two terminal network containing
voltage and current sources,
together with other linear circuit elements, such as resistors, has exactly the same
electrical characteristics as
an equivalent circuit consisting of a single voltage source
V Th in series with a single
resistor R Th . We shall not
give a general proof of
Thévenin’s theorem here, but
it follows on from superposition discussed in Lecture 3.
The Thévenin voltage V Th is
equal to the open circuit voltage seen at the output terminals of the linear circuit (see
diagram
opposite).
The
Thévenin resistance is given
in terms of the open circuit
voltage and the short circuit
current. Another way of
expressing this, which is
often used in practice, is to
say that R Th is equal to the
resistance seen across the
output terminals with all
voltage
sources
short
circuited and all current
sources open circuited.
-32Before looking at some
examples of Thévenin’s theorem I will mention one very
useful result, the Thévenin
equivalent of a current generator. Consider an ideal current generator I 0 with shunt
resistance R 0 . The open circuit voltage across the output
terminals is I 0 R 0 , and the
resistance seen across the
output terminals with the current source open circuited is
just R 0 . This gives us the
Thévenin equivalent shown
opposite.
Note the following equivalence:
+
I0
R0
R Th = R 0
–V = I R
Th
0 0
Design Example 4.1
When you connect a voltmeter across a circuit the
Investigate the loading effect of an anameter’s internal resistance
will draw some current which
logue voltmeter (200kΩ resistance) on a
may result in a false reading.
In the case of digital meters,
which have very high internal 20 volt divide-by-two potential divider.
resistances, this loading
effect can usually be ignored.
In analogue moving coil
instruments however the
internal resistance is much
lower. It is usually specified
by a figure like 20,000 ohms
per volt, meaning that when
the meter is switched to the
ten volt scale, for example, it
will have an internal resistance of 200kΩ . In this
example we will study the
loading effect of such a meter
on a potential divider circuit.
The general situation is represented by the circuit opposite, where we will eventually
set R 1 = R 2 for a divide by
two potential divider. We
will use Thévenin’s theorem
to analyse the circuit,
although any of the methods
we have met previously
could also be used.
general circuit
R1
A
V0
R2
Vm
A'
Rm
meter
-33Consider the Thévenin equivalent for
Our first step is to find the
the circuit to the left of AA'
Thévenin circuit for the part
of the circuit to the left of
on the previous diagram.
R Th
In doing so, we ignore for the
moment the presence of the
A
meter. The Thévenin voltage
is the open circuit voltage
+
seen at
, assuming no
meter is connected. This will
Rm
V Th
Vm
just be the voltage produced
by the unloaded potential
divider
and
, which
–
gives us
AA'
AA'
R1
R2
⋅V .
V Th = -----------------R 1 + R2 0
A'
Next, we consider the
Thévenin resistance. The easiest way to get this is to note
that it is the resistance seen
looking to the left of AA'
with all voltage sources short
circuited, and all current
sources open circuited. In our
case there is only the voltage
source
and
short
V0
circuiting this we see that the
resistors R 1 and R 2 are then
in parallel. This gives us
The Thévenin parameters are:
R Th
R1 R2
= -----------------R1 + R2
V Th
R2
= ------------------ ⋅ V 0
R1 + R2
R1 R2
R Th = ------------------ .
R1 + R2
With V0 = 20 V , R m = 200 kΩ and
R 1 = R 2 = R we have:
R⁄2
A
+
Vm
10V
–
A'
R2
200k
The result is the Thévenin
equivalent circuit drawn
above, with the meter shown
connected. The advantage of
using Thévenin’s theorem for
this example is that the
equivalent circuit is just a
potential divider formed by
R Th and R m . This makes it
very easy to visualise the
loading effect of the meter.
Let us now insert some component values. The meter
resistance is 200kΩ, the voltage source is 20V, and since
we want a divide by two circuit we set R 1 = R 2 = R .
Then, using the results just
derived, the Thévenin voltage
is 10V, and the Thévenin
resistance is R ⁄ 2 , giving us
the circuit opposite.
-34The Thévenin equivalent
circuit is a simple potential
divider, so it is now straight
forward to write down the
voltage across the meter. The
general expression is
·
Rm
⋅V
V m = ---------------------R Th + R m 0
which reduces to the equation
on the right when we insert
the component values given
in the previous circuit. In the
expression for V m we have
expressed the resistor values
in kΩ.
The table opposite shows
the meter readings for various values of the divider
resistors R1 = R2 = R . The
correct meter reading (with
no circuit loading) should be
10 V. The loading effect of
the meter is clearly apparent
from this table.
Only when R m » R is the
reading close to 10V. This is
a general principle for all
instruments. Their internal
resistance must be much
greater than that of the circuit
they are connected to if accurate readings are to be
obtained.
Voltage across meter is then given by:
200 - ⋅ 10 volts
V m = ------------------------200 + R ⁄ 2
Calculate V m for various values of R
R ohms
1M
V m volts 2.86
100k
8.0
~
10k
9.76
1k
9.97
-35Design Example 4.2
Use Thévenin’s theorem to find the
current I in the circuit below.
75V
+
15k
A
8mA
B
10k
10k
I
4k
D
It is not immediately
apparent how to apply
Thévenin’s theorem in this
circuit. We are asked to find
the current through the resistor BC , which suggests that
we find the Thévenin equivalent for the circuit to the left
of nodes B and C. A first step
would be to make everything
into a voltage source, which
suggests that we start by finding the Thévenin equivalent
for the current generator and
10k resistor to the left of
nodes AD.
C
(i) Replace current generator with
Thévenin equivalent
10k
A
8mA
+
10k
80V
–
D
We have already seen
how to replace a current generator with a Thévenin equivalent circuit. In this case, the
open circuit voltage is 8mA
times 10kΩ, giving 80V for
the Thévenin voltage. The
resistance with the current
generator replaced by an
open circuit is 10 k, which is
the required Thévenin resistor. This gives us the equivalent circuit shown opposite.
-36-
which gives us our Thévenin
voltage
of
30V.
The
Thévenin resistance is the
resistance seen across AB
with the 75V generator short
circuited. This is just the parallel combination of the 10k
and 15 k resistors, giving
10 × 15
R Th = ------------------ kΩ .
10 + 15
This gives us the 6k resistor
shown in the equivalent circuit.
30 V
15k
+
V oc
10
= ------------------ ⋅ 75 volts
10 + 15
75V
+
Next we consider the
Thévenin equivalent for all of
the circuit above the nodes
labelled AB, reproduced
opposite. The open circuit
voltage across AB is the
result of the 75V generator
connected across the series
combination of the 10k and
15k resistors, which therefore act as a potential divider.
The open circuit voltage
across AB is then given by
(ii) Replace voltage source and resistors
with Thévenin equivalent:
6k
A
10k
B
(iii) Redraw circuit diagram with the
Thévenin equivalents:
30 V
+
We now redraw the original circuit with the current
and
voltage
generators
replaced by their Thévenin
equivalents derived above.
The circuit is shown opposite. The original nodes
ABCD are shown so that it is
clear where the Thévenin
equivalents are placed. Our
final step is to replace all of
the circuit to the left of BC by
a Thévenin equivalent. The
open circuit voltage at BC is
just (80-30) = 50V, (note the
negative sign because the
generators have opposite
sign). The Thévenin resistance is (10+6) = 16kΩ . This
produces the equivalent circuit shown on the next page.
10k
6k
A
B
+
80V
4k
–
D
C
I
-37(iv) Replace circuit to left of BC by
Thévenin equivalent:
16 k
B
+
50 V
I
4k
–
C
The current I is given by:
50 - mA
I = ------------------( 16 + 4 )
= 2.5mA
~
By applying Thévenin’s
theorem three times we have
reduced the original circuit to
the one shown on the left.
The current through the 4k
resistor is now easily calculated.
Although we chose to use
Thévenin’s theorem for this
example we could have analysed the original circuit by
any of the other techniques
that we have met so far using
Kirchhoff’s laws or superposition. In many cases there is
no one best method and it is
often a matter of preference
as to which method of analysis is employed.
-38-
Norton’s Theorem
The second equivalent
circuit theorem that we shall
meet is called Norton’s theorem and, like Thévenin’s theorem, applies to any two
terminal circuit consisting of
voltage and current sources
together with other linear
components. Norton’s theorem tells us that such a circuit
can be replaced by a single
current generator I N in parallel with a shunt resistance
R N . The Norton current I N is
equal to the current seen
when the output of the two
terminal circuit is short
circuited. The Norton resistance R N is defined in the
same way as the Thévenin
resistance, and is also equal
to the resistance seen looking back into the circuit with
all voltage sources short
circuited and all current
sources open circuited. Like
Thévenin’s theorem, Norton’s theorem is a consequence of the linearity of the
circuit.
Any two terminal linear network may
be replaced by a current source I N in
parallel with a shunt resistance R N .
linear
circuit
IN
RN
I N = I ( short circuit )
RN
V
( open circuit )-----------------------------------=
I ( short circuit )
Note the following equivalence:
An important consequence of Norton’s theorem
is that a voltage source can be
represented by an equivalent
current source. Referring to
the diagram on the right, the
current with AB short
circuited is V ⁄ R , and the
resistance seen from the terminals AB with the voltage
source short circuited is R.
This gives us the equivalent
Norton circuit shown.
R
A
+
V
--R
V
–
B
R
-39Design Example 4.3
Use Norton’s theorem to find the voltage across R in the circuit below.
We met this circuit previously in Design Example 2.1.
We will see that the application of Norton’s theorem
leads to a direct and simple
solution. We use the result on
the previous page to replace
the voltage sources to the left
and right of nodes AB with
equivalent current sources.
Consequently, for source e 1
we use the following Norton
equivalent, with a similar circuit for e 2 .
A
+
e1
–
R1
R2
+
e2
–
R
R1
B
e1
+
-
e
-----1R1
R1
(i) Apply Norton’s theorem to each This then gives us the
equivalent circuit shown
adjacent. Note that everyvoltage source:
thing, current sources and
R1
e1
-----R1
R
resistors, are now all in parallel. This makes it very easy to
simplify the circuit further.
The parallel resistors can all
be combined into a single
resistor of value
R2
e2
-----R2
1
----------------------------.
1
1 ----1------ + --+
R1 R R2
The current sources, being in
parallel, can be replaced by a
single current source of value
e
e
-----1- + -----2R 1 R2
All of this then leads to the
circuit on the next page.
-40(ii) Combine the current sources, and
the parallel resistors:
Here is the result of our
labours. The circuit has simplified to a single current
source connected across a
resistor. It is a simple matter
now to write down the voltage. The result, given below,
is of course the same as we
derived in Design Example
2.1. I have written it in this
form, without further simplification, as the reduction
using the Norton equivalents
is then very clear. You should
have no trouble generalising
this result to a circuit containing N sources connected in
parallel across a resistor R. I
leave it to you to show that
the voltage in this case is
given by
N
e
-i
∑ ---R
1
1
---------------------------1- ----1- -1-----+
+
R 1 R2 R
e1 e2
------ + -----R1 R2
Voltage across resistor is then given by:
i
----------------------- .
1 N 1
--+
----R ∑
1 Ri
e1 e2
------ + -----R1 R2
V = ----------------------------1- ----1- -1-----+
+
R1 R 2 R
~
-41-
Circuit matching
Power matching
What value of load resistor should we choose to maximise the power in the load in
this circuit? The current
through the load is
Rs
V
i = -----------------R s + RL
Then the power delivered to
the load R L is given by
+
load
V
RL
–
RL
2
2
P L = i R L = -------------------------⋅V
2
( Rs + R L )
and this is a maximum when
dP L
dRL
Maximum power in load when
= 0.
It follows from this that the
condition for maximum
power to be dissipated in the
load is R L = R s , a result you
will have met previously at
A-level; we refer to this as
power matching.
RL = Rs
Voltage matching
Voltage across load is given by:
RL
V L = ------------------ ⋅ V
Rs + RL
For maximum voltage we need:
RL » Rs
In many situations we
want to maximise the voltage delivered to a load. For
example, in the above circuit
the load R L might represent
the input resistance of an
amplifier, being fed by a
transducer or an audio pickup, represented by a voltage
source of internal resistance
R s . In this case, to maximise
the voltage delivered to R L
we need to make R L much
larger than the source resistance R s . Although many
books emphasise power
matching, it is of less practical importance than maximising input voltage, and most
circuits are designed so that
their input resistance is much
larger than the source driving them.
-42-
Current matching
This situation is met less
often than the previous two.
Sometimes we may need to
maximise the current delivered to a load from a current
source. In this case, it is easy
to see from the expression for
the current through the load
that we need to make the load
resistance much smaller than
the resistance of the current
source.
I
Rs
RL
Current through load is
Rs
I L = ------------------ ⋅ I
Rs + RL
For maximum current we need
RL « Rs
-43-
A more difficult example
Design Example 4.4
Use Thévenin’s theorem to find the
charging current I B in the following cirThis is a more difficult
cuit of a battery charger.
example. It is difficult not
LM317
Vi n
Vo u t
Adj
V ref
Rs
R1
R2
=
=
=
=
1.25 V
0.2
240
2.4 k
Rs
A
IB
V ref
R1
I≈0
X
VB
R2
A'
because of harder mathematics, but because you need to
think more carefully about
the circuit.
The circuit is of a 12V
battery charger and is taken
from a National Semiconductors data sheet, who also
make the LM317 used in the
circuit. The LM317 is a
device known as a voltage
regulator, and is typical of the
wide range of integrated circuits available today. It is
essentially a three terminal
device which takes a voltage
at the input Vi n of between
4–40V and converts it to a
very stable reference voltage
Vr e f of 1.25V between the
terminal Vo u t and the
adjuster terminal Adj. The
data sheet suggests the component values shown, and
tells us that the battery
charger will gradually reduce
the charging current as the
battery gets close to its fully
charged 12V state. No other
explanation of the circuit is
given. Our task, which is typical of real life electronic
design, is to analyse the circuit and discover how it
works.
-44We will find Thévenin equivalent of the
circuit to the left of AA'
We begin by making AA'
an open circuit, that is we
completely remove the battery V B . The voltage across
AA' is then the same as the
voltage drop across R 1 and
R 2 since the current I flowing from the Adj terminal is
negligible. This means that
we need the current i flowing
through these resistors. This
is also the same as the current
through R s and R 1 . The voltage drop across R s and R 1 is
V ref , so that the current i
through these resistors is
(i) Find the open circuit voltage V oc
Current through R s and R 1 :
V ref
i = -----------------Rs + R1
Voc = voltage drop across R 1 and R 2
V ref
i = ------------------ .
R s + R1
V oc = i ( R 1 + R2 )
Therefore, the voltage drop
across R 1 and R 2 is
i ( R 1 + R 2 ) , giving us the
expression for the open circuit voltage opposite.
R 1 + R2
= ------------------ ⋅ V ref
Rs + R1
(ii) Find short circuit current i sc
This is conceptually more
difficult. When AA' is short
circuited all the current will
flow through R s , and there
will be no current through R 1
and R 2 . In particular, the
voltage drop across R 2 will
be zero. This means that the
node marked X is at earth
potential (zero volts) and that
V ref is effectively dropped
across R s , giving us a short
circuit current of V ref ⁄ R s .
With AA' short circuited no current
flows through R 1 , R 2 .
Hence
i sc
V ref
= --------Rs
-45(iii) Thévenin parameters:
V Th = V oc
R1 + R2
= ------------------ ⋅ V ref
Rs + R1
RTh
V oc
= -------i sc
Rs
= ------------------ ⋅ ( R 1 + R 2 )
Rs + R1
(iv) Draw the Thévenin equivalent
circuit:
R Th
IB
+
V Th
VB
–
V Th – V B
I B = --------------------R Th
We are now in a position
to construct the Thévenin
equivalent
circuit.
The
Thévenin voltage is the open
circuit voltage derived above,
and the Thévenin resistance
is defined as V oc ⁄ i sc . This is
an example of a situation
where we have to use the formal
definition
of
the
Thévenin resistance. Without knowing the details of the
internal construction of the
voltage regulator we cannot
easily use the method of calculating the resistance seen
looking to the left of AA'
with voltage sources short
circuited and current sources
open circuited as we did with
previous examples.
And this is our Thévenin
equivalent circuit. The complexities of the original circuit have been swept away to
leave a very simple equivalent circuit from which it is
now easy to write down an
expression for the battery
charging current I B . Note
that this circuit contains everything needed to predict the
behaviour of the original circuit as far as connecting anything to the right of AA' is
concerned. This is the real
power of the Thévenin equivalent circuit approach.
-46Inserting values for VTh and R Th :
V ref V B R 1 + R s
I B = --------- – ------ ⋅  ------------------
R s R s R1 + R 2
Inserting component values, we find
We can now calculate I B
as a function of V B using the
component values suggested
by National Semiconductors.
In fact, the 240Ω value for
R 1 is suggested in the data
sheet as the best choice
between V o u t and Adj for a
wide variety of applications
for this device. We see that,
as expected, the charging current I B gradually reduces as
the battery voltage builds up.
The table opposite shows
how the current starts at
about 2.6A for a battery voltage of 8V, and falls to zero at
around 13.9V. Most batteries will charge to a little
above their nominal working
voltage, so this is probably an
acceptable design. Modifying
the component values would
allow the charging rate to be
adjusted, but bear in mind
that the values of resistors
chosen must fall within the
range of those manufactured,
which will place some constraints on the design.
I B = 6.25 – 0.45V B amps .
Consider how I B varies with V B
V B (volts)
8
10
12
I B (amps) 2.65 1.75 0.85
13.9
0
Charging current decreases as battery
voltage increases.
~