1

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.

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The Function Generator Model

Vs

FREQ = VAMPL = VOFF =

0

Rs

50

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

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Thevenin Model

Vs

FREQ = VAMPL = VOFF = RL

Rs

0

Load ResistorRth

Vth

FREQ = VAMPL = VOFF =

0

RL

4

Note:

We might also see a circuit with no load resistor, like this voltage divider.

R2

0

R1

Vs

FREQ = VAMPL = VOFF =

5

Thevenin Method

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

Vth

FREQ = VAMPL = VOFF =

0

RL

Rth

A

B

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Find Vth

Remove Load

RL

0

Vo

0Vdc

R4R2

R1 R3

0

Vo

0Vdc

R4R2

R1 R3

A AB B

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Find Vth

Let Vo=12, R1=2k, R2=4k, R3=3k, R4=1k

V Bk

k kV

1

1 31 2 3

V Ak

k kV

4

4 21 2 8

V th V A V B V 8 3 5

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Find Rth

Short out the voltage source (turn it off) & redraw the circuit for clarity.

0

R4R2

R1 R3

A BR3 R4

R2R1A

B

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Find Rth First find the parallel combinations of R1 & R2

and R3 & R4.

Then find the series combination of the results.

RR R

R R

k k

k k

kk1 2

1 2

1 2

4 2

4 2

8

61 3 3

.

RR R

R R

k k

k k

kk3 4

3 4

3 4

1 3

1 3

3

40 7 5

.

R th R R k k

1 2 3 44

3

3

42 1.

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Redraw Circuit as a Thevenin Source

Then add any load and treat it as a voltage divider.

0

Vth

5V

Rth

2.1k

thLth

LL V

RR

RV

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Thevenin Method Tricks

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.

R

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Thevenin Applet (see webpage)

Test your Thevenin skills using this applet from the links for Exp 3

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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.

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Thevenin Method Example Checking the answer with PSpice

Note the identical voltages across the load.• 7.4 - 3.3 = 4.1 (only two significant digits in Rth)

R3

3kVth

5Vdc

0

7.448V

0V

4.132V

3.310VVo

12VdcRload

10k

R1

2k

12.00V

RL

10kR2

4k

5.000V Rth

2.1k

R4

1k

0