Thevenin Voltage Equivalents

Thevenin invented a model called a Thevenin
Source for representing a complex circuit using
• A single “pseudo” source
• A single “pseudo” resistance

The Thevenin source, Vth, with its associated
resistance, Rth, “looks” to the load on the circuit
like the actual complex combination of resistances
and sources.
 This model can be used interchangeably with the
more complex circuit when doing analysis.
1
The Function Generator Model
Rs
50
Vs
VOFF =
VAMPL =
FREQ =
0

Recall that the function generator
has an internal impedance of 50
Ohms.
 Could the internal circuitry of the
function generator contain only a
single source and one resistor?
 This is actually the Thevenin
equivalent model for the circuit
inside the function generator
2
Thevenin Model
Vs
VO FF =
VA MP L =
FREQ =
RL
0
Rs
Load Resistor
Rth
V th
V O FF =
V AMP L =
F RE Q =
RL
0
3
Note:

We might also see a circuit with no load
resistor, like this voltage divider.
R1
Vs
V O FF =
V AMP L =
F RE Q =
R2
0
4
Rth
Thevenin Method
A
V th
V O FF =
V AMP L =
F RE Q =
RL
B
0

Find Vth (open circuit voltage)
• Remove load if there is one so that load is open
• Find voltage across the open load

Find Rth (Thevenin resistance)
• Set voltage sources to zero (current sources to open) –
in effect, shut off the sources
• Find equivalent resistance from A to B
5
Find Vth

Remove Load
R1
R1
R3
Vo
RL
Vo
A
0Vdc
R2
R3
B
A
0Vdc
R2
R4
0
0
6
B
R4
Find Vth

Let Vo=12, R1=2k, R2=4k, R3=3k, R4=1k
 1k 
VB  
12  3V
 1k  3k 
 4k 
VA  
12  8V
 4k  2k 
Vth  VA  VB  8  3  5V
7
Find Rth

Short out the voltage source (turn it off) &
redraw the circuit for clarity.
R1
A
R2
R3
A
R1
R2
B
R3
R4
R4
B
0
8
Find Rth

First find the parallel combinations of R1 & R2
and R3 & R4.
R1R2
4k 2k
8k
R12 


 133
. k
R1  R2 4k  2k
6
R3R4
1k 3k
3k
R34 


 0.75k
R3  R4 1k  3k
4

Then find the series combination of the results.
 4 3
Rth  R12  R34     k  21
.k
 3 4
9
Redraw Circuit as a Thevenin
Source
Rth
Vth
2.1k
5V
0

Then add any load and treat it as a voltage divider.
RL
VL 
Vth
Rth  RL
10
Thevenin Method Tricks
R

Note
• When a short goes across a resistor, that resistor
is replaced by a short.
• When a resistor connects to nothing, there will
be no current through it and, thus, no voltage
across it.
11
Thevenin Applet (see webpage)

12
Test your
Thevenin
skills
using this
applet
from the
links for
Exp 3
Does this really work?


To confirm that the Thevenin method
works, add a load and check the voltage
across and current through the load to see
that the answers agree whether the original
circuit is used or its Thevenin equivalent.
If you know the Thevenin equivalent, the
circuit analysis becomes much simpler.
13
Thevenin Method Example

Checking the answer with PSpice
12.00V
5.000V
R1
2k
2.1k
3k
RL
Vo
Rt h
R3
Vt h
3.310V
12Vdc
Rload
7.448V
5Vdc
10k
10k
R2
R4
4k
1k
0
0V
0

4.132V
Note the identical voltages across the load.
• 7.4 - 3.3 = 4.1 (only two significant digits in Rth)
14