Thevenin and Norton Theorems
1.
Thevenin theorem:
Any electric circuit, no matter how complex it is, can be represented by a single voltage source and a single series resistor. This is called the Thevenin equivalent circuit of the original circuit.
2.
Thevenin equivalent circuit: a) Open circuit the load. b) Compute/measure the output voltage with the load open. This is the Thevenin equivalent voltage. c) Short all voltage sources in the circuit. d) Open all current sources in the circuit. e) Compute/measure the resistance. This is the Thevenin equivalent resistance.
Example: Find the Thevenin equivalent circuit for the following circuit. a) Open circuit R
L
. b) V th
= 12 (3000/4000) = 9 V c) Short circuit the voltage source. d) No current source. e) R th
= 2000 + 1000//3000 = 2.75 K


Now we can use the equivalent circuit to calculate the load voltage and current.
V
L
= V th
[ R
L
/ ( R th
+ R
L
)]
The advantage of using the Thevenin theorem is that we only need to compute the equivalent circuit one time and we can use it for any R
L
. On the other hand, if the circuit is analyzed using other techniques such as series/parallel, KCL, KVL …, you will need to repeat the calculations for a different load resistor.
3.
Norton theorem:
Any electric circuit, no matter how complex it is, can be represented by a single current source and a single parallel resistor. This is called the Norton equivalent circuit of the original circuit.
4.
Norton equivalent circuit: a) Find the Thevenin equivalent circuit. b) The Norton resistance is the same as the Thevenin resistance. c) The Norton current equals to V th
/ R th
.
Example: Find the Norton equivalent circuit for the above circuit. a) V th
= 9 V, R th
= 2.75 K
 b) R
N
= R th
= 2.75 K
 c) I
N
= V th
/ R th
= 9 V/2.75 K

= 3.273 mA

Maximizing circuit output
5.
Maximizing output (load) voltage:
To maximize the output voltage we need to make the load resistance much larger than the Thevenin resistance of the circuit. As a result, small loads are difficult for voltage amplifiers to drive.
6.
Maximize output (load) current:
To maximize the output current we need to make the load resistance much smaller than the Thevenin resistance of the circuit. As a result, large loads are difficult for current amplifiers to drive.
7.
Maximize output (load) power:
To maximize the output power we need to make the load resistance equal to the
Thevenin resistance of the circuit.
Example: Determine the optimum load for power transfer for the circuit in 2).
Since the Thevenin resistance is 2.75 K

, a load resistance of 2.75 K

will absorb the maximum power that this circuit can provide.
P = V
L
2
/ R
L
= ( V th
/ 2)
2
/ R th
= (9/2)
2
/2750 = 7.364 mW