Equivalent Circuits
Aims:
• Introduce Source Transformations.
• Be able to determine Thévenin and Norton equivalents.
• Use KCL.
• Know condition for maximum power transfer.
Lecture 5
1
How much detail do we need?
There are different ways of looking at one device
(an operational amplifier integrated circuit)
Package outline
V+
inv input
non-inv
input
-
output
+
gnd
V-
Detailed schematic
Circuit symbol
Lecture 5
2
1
Equivalent circuit
Describes the FUNCTION of the
circuit – the details of what is
inside the box are not important
Equivalent circuit must have
corresponding terminals
Electrical properties between
corresponding terminals must be
identical
Rout
inv input
-
+
output
V = A(V+ - V-)
Rin
non-inv input
+
- (controlled voltage
source)
Lecture 5
3
Source Transformations
A
We can simplify networks containing sources
(batteries and current sources) and resistors:
B
6Ω
A
3Ω
6V
8Ω
18Ω
B
9Ω
Lecture 5
4
2
Simplify the Circuit #1
A
We can simplify networks containing sources
(batteries and current sources) and resistors:
B
6Ω
A
3Ω
6V
6V
8Ω
18Ω
A
2Ω
8Ω
6Ω
B
B
9Ω
Lecture 5
5
Simplify the Circuit #2
A
We can simplify networks containing sources
(batteries and current sources) and resistors:
A
2Ω
6V
A
8Ω
6V
8Ω
6Ω
B
8Ω
B
B
Lecture 5
6
3
Output Characteristics
A
We can simplify networks containing sources
(batteries and current sources) and resistors:
1.0
A
8Ω
B
6V
Current (A)
0.8
8Ω
0.6
0.4
0.2
B
0.0
0
1
2
3
4
Voltage (V)
Lecture 5
7
Simplify the Circuit #3
A
We can simplify networks containing sources
(batteries and current sources) and resistors:
4Ω
B
1.0
A
Current (A)
0.8
3V
B
0.6
0.4
0.2
0.0
0
1
2
3
4
Voltage (V)
Lecture 5
8
4
Simplify the Circuit #4
A
We can simplify networks containing sources
(batteries and current sources) and resistors:
B
1.0
A
Current (A)
0.8
4Ω
0.75A
0.6
0.4
0.2
B
0.0
0
1
2
3
4
Voltage (V)
Lecture 5
9
Source Transformations
We can simplify networks
containing sources (batteries
and current sources) and
resistors:
Thévenin’s Theorem:
A
B
Replace any source
network with a voltage
source and a resistor in
series
A
Norton’s Theorem:
RT
VT
A
B
IN
RN
B
Lecture 5
Replace any source
network with a current
source and a resistor in
parallel
10
5
A confused history …
Helmholtz was one of the most eminent physicists of the 19th C. He published the
voltage source equivalent in 1853.
Thévenin was an engineer working for the French PTT. He published the voltage
source equivalent in 1883 - apparently unaware of Helmholtz’s work.
Mayer was a senior engineer with Siemens in Germany. (He leaked several
documents on German military electronics to the Allies in WW2).
Norton was a brilliant engineer with the Bell Telephone Labs in the USA.
Norton and Mayer published the current source transformation independently in the
same MONTH in 1926!
Lecture 5
11
Thévenin Equivalent Circuit
A
RT
VT
B
RT (the Thévenin resistance) is given by the value of the resistance
between the terminals A and B when all the sources in the real
circuit are set to zero (i.e. voltage sources short circuit and current
sources open circuit)
VT (the Thévenin voltage) is the voltage that appears across the
terminals A and B when there is nothing connected to them (the open
Lecture 5
12
circuit voltage)
6
Output Characteristics of Thévenin Sources
RT
VT
A
RL
Connect a LOAD resistance RL to
the terminals of a Thévenin source
B
By varying RL we trace out a line
on the I-V plot. This is called the
load line.
VL =
VT RL
RT + RL
IL =
VT
RT + RL
I
Low RL
ISC
Operating point
The load line tells us that as we
draw more current from a source
the voltage falls.
High RL
Load line:
Slope = -1/RT
The actual I and V (the operating
point) is where the load line and
the I-V curve for the load intersect
VOC V
Lecture 5
13
X
Example #1
Find the Thévenin equivalent circuit
between A and B
10
10
v
4
40
2A
Lecture 5
A
B
14
7
Real batteries are Thévenin sources
Batteries have internal resistance which is equivalent to
the Thévenin resistance
As the battery fades, VT remains constant, but RT increases
• (This is why you can not check for a dead battery just
using a voltmeter – you have to connect a load resistor)
RT
VT
Typical values:
AA cell: VOC = 1.5 V; ISC ~ 1 A: RT = 1.5 Ω
PP3
: VOC = 9 V; ISC ~ 0.1 A: RT = 90 Ω
Car battery: VOC = 12 V; ISC ~ 100 A: RT = 0.12 Ω
Lecture 5
17
Norton Equivalent Circuit
A
IN
RN
B
RN (the Norton resistance) is given by the value of the resistance
between the terminals A and B when all the sources in the real circuit
are set to zero (i.e. voltage sources short circuit and current sources
open circuit)
IN (the Norton current) is the
current that flows between A and B
when the terminals are shorted
together (the short circuit current).
Lecture 5
SOURCE TRANSFORMATIONS
For any real network
RN = R T
VT = INRN
18
8
Example #1
Find the Norton equivalent circuit
between A and B.
X
10
4
10
v
A
40
2A
B
X
10
4
10
v
Is/c
40
2A
B
Lecture 5
19
Example #2
Find the Norton equivalent circuit
between A and B
A
X
A
4
24
V
12
3A
B
X
4
24
V
Lecture 5
Isc
12
3A
A
B
21
9
Power Transfer
RT
VT
A
The power dissipated in the load is P = VI
RL
B
VL =
VT RL
VT
VT2 RL
; IL =
; so P =
2
RT + RL
RT + RL
( RT + RL )
P
W
For maximum power transfer
from a source to a load the
load resistance must be equal
to the source equivalent
resistance
Maximum
power in
load when
RL=RT
This is the
MATCHING THEOREM
RT
RL
Lecture 5
23
Power Efficiency
RT
VT
A
RL
B
The power in the load at the condition for maximum
power transfer (when RL=RT) is given by
Pmax =
VT2
4 RT
Note that the same amount of power is dissipated in the
Thévenin resistance.
This tells us that under the optimum conditions the maximum
power that can be transferred from a source to a load is 50% of
the power generated by the source.
(the maximum power efficiency in an electrical network is 50%)
Lecture 5
24
10
Load matching is common:
TV set is
designed to
present a 75 Ω
load
TV antenna has
75 Ω source
resistance
TV
Audio amplifier has
8Ω source resistance
Loudspeaker has
8Ω equivalent load
resistance
An amplifier with 20W output
will dissipate at least 20 W in
the output stages
Lecture 5
25
Controlled or Dependent Sources
These are current or voltage sources which have a value that depends on
another voltage or current in the equivalent circuit
These allow us to model devices like amplifiers using simple equivalent
circuits
+
-
V = -3V 2
Dependent
voltage source
I = K V1
The controlling
parameter must be
clearly labelled (and
visible in the circuit)
Dependent
current source
Lecture 5
26
11
Example – Operational Amplifier
Rout
inv input
-
+
output
V = A(V+ - V-)
Rin
non-inv input
-
+
gnd
The output is a Thévenin source with a voltage that depends on
the difference in voltage between the input terminals.
This is called a TWO PORT equivalent circuit and is often used to
model amplifiers (more in a few lectures time…)
Lecture 5
27
12