THEVENIN’S AND NORTON’S THEOREMS
These are some of the most
powerful analysis results to be
discussed.
They permit to hide information that
is not relevant and concentrate in
what is important to the analysis
THEVENIN’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
RTH


vTH

i
a
vO
b
_

i
a
LINEAR CIRCUIT
vO
_
PART A
Thevenin Equivalent Circuit
for PART A
vTH
Thevenin Equivalent Source
RTH
Thevenin Equivalent Resistance
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
b
PART B
NORTON’S EQUIVALENCE THEOREM
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A

RN
i
Norton Equivalent Circuit
for PART A
Thevenin Equivalent Source
Thevenin Equivalent Resistance
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART B
a
LINEAR CIRCUIT
vO
PART A
RN
b
_
_
iN
a
vO

iN
i
b
PART B
ANOTHER VIEW OF THEVENIN’S AND NORTON’S THEOREMS
i a
RTH
vOC
+
_

+
i
vO
_
i SC
RTH

Norton
Thevenin
i SC
vO
b
vOC

RTH
This equivalence can be viewed as a source transformation problem
It shows how to convert a voltage source in series with a resistor
into an equivalent current source in parallel with the resistor
A General Procedure to Determine the Thevenin Equivalent
Open Circuit voltage
vTH
voltage at a - b if Part B is removed
i SC
Short Circuit Current
current through a - b if Part B is replaced
by a short circuit
RTH 
vTH
i SC
Thevenin Equivalent Resistance
1. Determine the
Thevenin equivalent
source
Remove part B and
compute the OPEN
CIRCUIT voltage Vab
One circuit problem
i 0 a
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A


vOC
Vab
_
b
_
Second circuit problem
2. Determine the
SHORT CIRCUIT
current
Remove part B and
compute the SHORT
CIRCUIT current I ab
vTH  vOC , RTH
vOC

i SC
LINEAR CIRCUIT
May contain
independent and
dependent sources
with their controlling
variables
PART A
i SC

a
v0
I ab
_
b
LEARNING EXAMPLE
USE THEVENIN TO COMPUTE Vo
You have the choice on the way to partition
the circuit.
Make “Part A” as simple as possible
Since there are only independent sources,
for the Thevenin resistance we set to zero all
sources and determine the equivalent resistance
RTH  2  (2 || 4)
“Part B”

10
k
3
For the open circuit voltage we analyze the
following circuit (“Part A”) ...
Loop Analysis
I 2  2mA
 6V  4kI1  2k ( I1  I 2 )  0
VOC
6  2I2
5
I1 
mA  mA
6
3
 4k * I1  2k * I 2  20 / 3  4V  32 / 3[V ]
The circuit becomes...
LEARNING EXAMPLE: COMPUTE Vo USING NORTON
4k
I
RN
IN
I SC
RN  RTH  3k
PART B
12V
I SC  I N 
 2mA  2mA
3k
COMPUTE Vo USING THEVENIN
2k
 RN

VO  2kI  2k 
I N 
 RN  6k

3
4
VO  2 (2)  [V ]
9
3
PART B
VTH
RTH

+
-
VTH
VTH  12
 2mA  0
3k
RTH  3k  4k
VO 
2k
VO

2
4
(6V )  [V ]
27
3
LEARNING EXAMPLE Find the Thevenin Equivalent circuit at A - B
Only dependent sources. Hence V_th = 0
To compute the equivalent resistance we
must apply an external probe
We choose to apply a current probe R  VP
TH

VP

IP
“Conventional” circuit with dependent
sources - use node analysis
IP
@V_1
@V_2
(IP )
Controlling variable
3(V1  2V1 )  6V1  2(V1  V2 )  0
2(V2  V1 )  3V2  6[V ]
RTH
A
B
Thevenin equivalent
5V1  2V2  0 * / 2
 2V1  5V2  6 * / 5
30 10
V2  
21 7
(VP  V2 )  ( I P  1mA)  RTH 
V2
 (10 / 7)k
1mA