Thevenin equivalent circuits
We have seen the idea of equivalency
IS1
used in several instances already.
VS1 +
–
VS2 +
–
IS2
same
as
IS1 +IS2
R1
same
as
+
– VS1 +VS2
R2
R3 same
as
=
+
R1
VS
0V
same
as
IS
0A
same
as
+ V
S
–
same
as
IS
R1
=
EE 201
Thevenin – 1
The behavior of any circuit, with respect to a pair of terminals (port)
can be represented with a Thevenin equivalent, which consists of a
voltage source in series with a resistor.
load
some
circuit
two terminals
(two nodes) =
port
some
circuit
RTh
VTh +
–
RTh
Thevenin
equivalent
VTh +
–
+
vRL
–
RL
load
RL
+
vRL
–
Need to determine VTh and RTh so that the model behaves just like the
original.
EE 201
Thevenin – 2
Norton equivalent
load
load
+
vRL
–
some
circuit
IN
RN
+
vRL
–
Ideas developed independently (Thevenin in 1880’s and Norton in
1920’s). But we recognize the two forms as identical because they are
source transformations of each other. In EE 201, we won’t make a
distinction between the methods for finding Thevenin and Norton.
Find one and we have the other.
RTh
=
VTh +
–
EE 201
IN
RN
=
Thevenin – 3
Example
R1 1.5 k!
VS +
–
9V
R2
3 k!
RTh 1 k!
IS
6 mA
RL
VTh
12 V
R
RL
+
–
v
i
P
Attach various load resistors to the
original circuit. Do the same for
the equivalent circuit. For each
load resistance, calculate the load
voltage (and current and power)
for each of the circuits. The results
are identical. In terms of the load
that is attached at the port, the two
circuits are indistinguishable.
1 k!
6.0 V
6.0 mA
36 mW
10 k!
10.91 V
1.09 mA
11.9 mW
Check it yourself.
100 k!
11.88 V
0.119 mA 1.41 mW
EE 201
1!
11.99 mV 11.99 mA 0.144 mW
10 !
118.8 mV 11.88 mA 1.41 mW
100 !
1.091 V 10.91 mA 11.90 mW
Thevenin – 4
Determining the Thevenin (or Norton) components
How to find VTh and RTh?
Need two components, so two measurements or calculations should
suffice. Use two different load resistors.
RTh
load
unknown
circuit
R1
+
v1
–
VTh +
–
unknown
circuit
R2
2 equations, 2 unknowns:
EE 201
+
v2
–
VTh +
–
=
+
v1
–
R1
RTh
load
load
+
load
+
v2
–
R2
(
=
)
=
=
+
(
)
Thevenin – 5
More directly: open-circuit voltage, short-circuit current
1. Leave port open-circuited. (RL →∞, iL = 0) Measure open-circuit voltage.
RTh
+
unknown
circuit
voc = VTh
+
VTh +
–
voc
–
voc
–
open-circuit voltage is a
direct measure of VTh.
2. Short the output port. (RL = 0, vL = 0) Measure short-circuit current.
RTh
unknown
circuit
isc
VTh +
–
isc
=
=
=
Note, that isc can also be interpreted as a direct measurement of IN: isc = IN.
EE 201
Thevenin – 6
Calculating Thevenin equivalent
The open-circuit voltage / short-circuit current approach can be used to
calculate the Thevenin equivalent for a known circuit.
R1 1.5 k!
R2
3 k!
VS +
9V –
Consider the circuit from slide 4:
IS
6 mA
Open-circuit voltage – Use whatever method you prefer. We’ll use
node voltage in this case.
R1
VS +
–
va
R2
IS
+
voc
=
+
–
=
+
+
=
=
+( .
+
)(
.
.
)
= 12 V
VTh = voc = 12 V.
EE 201
Thevenin – 7
Short-circuit current – Use whatever method you prefer. We’ll use
node voltage in this case. But proceed carefully – the short circuit
introduces some unusual wrinkles into the circuit analysis.
R1
va
a: Because of the short circuit, va = 0!
R2
VS +
–
isc
IS
R1
b: Because of the short, vR2 = 0 and iR2
= 0. So R2 plays no role and can be
removed.
va
VS +
–
isc = iR1 + IS
isc
IS
=
+
=
=
EE 201
=
=
=
+
.
+
=
Thevenin – 8
Thevenin
Norton
RTh 1 k!
VTh +
–
12 V
IN
12 mA
RN
1 k!
Alternate method for RTh
If the circuit consists of independent sources and resistors only, then
the Thevenin resistance can also be found by de-activating the
independent sources and finding the equivalent resistance as seen from
the port.
De-activate the sources
R1
R2
EE 201
RTh =
=( .
) (
)=
Thevenin – 9
Summary
To measure VTh and RTh
1. Use a voltmeter to measure the open-circuit voltage at the port of the
circuit: voc = VTh.
2. Connect a short circuit across the output and use an ammeter to
measure the short-circuit current: isc = IN.
3. Calculate RTh = VTh / IN.
Note that shorting the output may not always be practical. For
example, some devices may have over-current protection circuitry that
prevents large short-circuit currents from flowing. Or the device might
not be able to handle the large current that might flow when the output
is shorted without being damaged. In those cases:
1. Use a voltmeter to measure the open-circuit at the port of the circuit: voc
= VTh.
2. Attach a load resistance, RL that is small enough so that an appreciable
current is flowing. Measure the resulting load voltage, vL.
3. Calculate
EE 201
=
Thevenin – 10
Summary
To calculate VTh and RTh
1. Using whatever techniques are appropriate, calculate the opencircuit voltage at the port of the circuit: voc = VTh.
2. Connect a short circuit across the output. Using whatever techniques
are appropriate, calculate the short-circuit current: isc = IN.
3. Calculate RTh = VTh / IN.
Alternate method (for circuits that consist only of independent sources
and resistors).
1. Using whatever techniques are appropriate, calculate the opencircuit voltage at the port of the circuit: voc = VTh.
2. De-activate all independent sources. Calculate the equivalent
resistance as seen from the port. (If dependent sources are present in
the circuit, the test generator method can be used to find equivalent
resistance. See the equivalent resistance notes to review the test
generator technique.)
EE 201
Thevenin – 11
Example 1
R2 1 k!
Find the Thevenin and Norton
equivalents of the circuit at
IS
right, with the port as shown. 24 mA
R1
6 k!
+
voc
R3
3 k!
–
Find voc. Start with a current divider.
=
+
=
+
+
=(
=
.
+
+
)(
(
)=
.
+
)=
VTh = voc = 43.2 V
.
R2
Find isc. Note that R3 is shorted out.
Use a current divider again.
=
=
EE 201
IS
R1
isc
+
+
(
)=
.
IN = isc = 20.57 mA
Thevenin – 12
=
.
=
= .
.
RTh 2.1 k!
VTh +
–
43.2 V
IN
RN
2.1 k!
20.57 mA
Alternatively, we could use the short-cut method to find RTh.
De-activating the current source:
R2
R1
R3
Req
=
=(
EE 201
(
+
) (
)
+
)= .
Thevenin – 13
Example 2
R1 1.5 k!
VS +
20 V –
R4 3.3 k!
R2
R5
4.7 k!
4.7 k!
R3 1.5 k!
R1
Find the Thevenin and Norton
equivalents of the circuit at
left, with the port as shown.
R6 3.3 k!
R4
Find voc. Use mesh current method.
+
VS +
–
ia
ib
R2
R5
voc
(
–
R3
R6
Insert values and solve: ib = 0.930 mA.
EE 201
voc = ib R5 = 4.37 V.
=
=
(
)
=
)
(
+
(
=
+
+
)
+
=
+
)
=
Thevenin – 14
R1
Find isc. Note that R5 is shorted out
by the short circuit.
R4
iS
VS +
–
isc
R2
Use equivalent resistance to find iS.
=
=
R3
+
+
(
+
)= .
= .
R6
RTh 3.27 k!
Use current divider to find isc.
VTh +
–
4.37 V
+
=
+
+
isc = 1.45 mA
=
EE 201
=
.
.
= .
IN
1.45 mA
RN
3.27 k!
Thevenin – 15
Alternatively, we can use the short-cut method to find RTh.
R1
R4
R2
R3
R5
RTh
R6
=
[
+
+
(
+
)]
= 3.02 k!
EE 201
Thevenin – 16
R5 15 !
Example 3
IS
Find the Thevenin and Norton
equivalents of the circuit at
right, with the port as shown.
4A
R1 30 ! R3 30 !
VS +
30 V –
Find voc. Use node voltage.
R5
=
+
R2
20 !
+
+
R4
15 !
=
=
+
IS
R1
VS +
–
+
a R3
b
+
+
+
=
=
+
R2
R4
voc
+
+
+
=
+
–
EE 201
solve: va = 20 V, vb = 40 V. "
voc = vb = 40 V
Thevenin – 17
Find isc. Use node voltage, again. Note that R4 is shorted out (so ignore it)
and node b is shorted to ground, vb = 0.
=
+
+
R5
IS
=
R1
VS
a R3
+
–
=
b
isc
R2
+
=
+
.
=
Need va.
=
+
EE 201
+
+
= .
= .
RTh 6.36 !
=
=
+
+
.
+
+
+
40 V
VTh +
–
= 8.57 V
IN
6.28 A
RN
6.36 !
Thevenin – 18
Maximum power transfer
Now that we have the ability to model any circuit using a simple Thevenin
(or Norton) equivalent, we can answer another important question: How
much power can a given circuit supply to an attached load?
Start with the Thevenin equivalent and determine the load resistance
that would lead to the maximum amount of power being dissipated in
the load.
RTh
VTh
+
–
load
RL
=
+
vRL
–
=
(
+
)
In the usual way, find the max by setting the derivative to zero and solving.
=
(
+
)
+
EE 201
(
+
=
)
=
=
For maximum
power to the load
See slide 4 for an example – look at power column in the table.
Thevenin – 19