10-Source Transformations
Text: 4.4 – 4.6
ECEGR 210
Electric Circuits I
Overview
•
•
•
•
Introduction
Source Transformation
Thevenin’s Theorem
Norton’s Theorem
Dr. Louie
2
Introduction
• Consider the two circuits (A and B) shown below
• Compute the voltage across the 2W resistor in
each circuit
Circuit A
Circuit B
4V
0.5W
4V
2W
0.5V
Dr. Louie
+
-
+
-
0.5W
+
-
1A
2W
3
Introduction
• Circuit A (superposition):
 VR1 = 2IR1 = 2 x 1(0.5/2.5)= 0.4V (current source)
 VR2 = 4(2/2.5) = 3.2V (voltage source)
 VR = VR1 + VR2 = 0.4 + 3.2 = 3.6V
• Circuit B: VR = 4.5(2/2.5)=3.6V (voltage divider)
Circuit A
Circuit B
4V
0.5W
4V
2W
0.5V
Dr. Louie
+
-
+
-
0.5W
+
-
1A
2W
4
Introduction
• Solving Circuit B was much easier
• Same voltage across (current through) the
resistor
• Circuits are equivalent looking into the terminals
Circuit A
Circuit B
4V
0.5W
4V
2W
0.5V
Dr. Louie
+
-
+
-
0.5W
+
-
1A
2W
5
Introduction
• Clearly there can be advantageous of
transforming sources
• Source transformations and equivalence are the
focus of this lecture
source
source
R
a
Vs
a
+
Is
-
b
R
b
Dr. Louie
6
Introduction
• Two ways of modeling real (non-ideal) voltage and
current sources are shown
 Rs: small value, prevents infinite current from
flowing if terminals (a,b) are shorted
 R||: large value, prevents infinite voltage at the
terminals (a,b) under open circuit conditions
• Generically: can be any voltage source in series with
resistance, or any current source in parallel with
resistance
Rs
a
Vs
a
+
Is
-
b
R||
b
Dr. Louie
7
Source Transformation
• Not possible to transform current (voltage)
sources to voltage (current) sources directly
Vs
+
Is
-
• But we can transform sources with series or
parallel resistances as seen by terminals
Rs
a
Vs
a
+
Is
-
b
R||
b
Dr. Louie
8
Source Transformations
• For the two circuits to be equivalent, they must
have the same i-v characteristics at their
terminals under all external circuit connections
 Due to linearity, only need to verify i-v
characteristics under two different external
connections (short, open circuit)
• How are Vs, Is, Rs and R|| related?
Rs
a
Vs
a
+
Is
-
b
R||
b
Dr. Louie
9
Source Transformation
• Consider when the external circuit is a short
• Both circuits must have same short current out of
their terminals
• Isc = Vs/Rs
• Isc = Is
• Therefore: Is = Vs/Rs
Rs
a
Vs
+
-
a
I
Is
b
R||
I
b
Dr. Louie
10
Source Transformations
•
•
•
•
•
Consider an open circuit
Both circuits must same open circuit voltage Voc
Voc = Vs
Voc =R||Is
Therefore: Is = Vs/R||
Rs
Vs
+
-
a
+
Voc
-b
Is
Dr. Louie
R||
a
+
Voc
b
11
Source Transformations
• Relationships:
 Is = Vs/R||
 Is = Vs/Rs
• Therefore
• Rs = R|| = R
• Is = Vs/R
• Vs = IsR
Source transformation equations
R
a
Vs
a
+
-
Is
b
R
b
Dr. Louie
12
Source Transformations
• Verify the results hold for Circuit A and Circuit B
Circuit A
Circuit B
4V
0.5W
4V
2W
Dr. Louie
0.5V
+
-
+
-
0.5W
+
-
1A
2W
13
Example
• Use source transformations to find V0
 Consider the current source first. Which resistor
can we associate it with?
• 4W (they are in parallel)
 Should we transform them?
• Yes, the resistor will be in series with the 2W resistor,
and we can combine the two
2W
4W
3A
8W
Dr. Louie
3W
+
V0
-
+
-
12V
14
Example
• Transform the source:
 R = 4W
 Vs = IsR = 3 x 4 = 12V
• Pay careful attention to the polarity
3A
8W
+
V0
-
4W
+
-
12V
12V
Dr. Louie
+
4W
3W
-
2W
2W
8W
3W
+
V0
-
+
-
15
12V
Example
• Transform the 12V source (on the right), if it is
beneficial
6W
+
12V
3W
8W
+
V0
-
+
-
12V
After combining series resistance
Dr. Louie
16
Example
• Yes, beneficial (results in parallel combination
with V0). Combine with 3W resistor to get:
 R = 3W
 Is = Vs/R = 12/3 = 4A
8W
+
V0
-
6W
+
-
12V
Dr. Louie
12V
+
+
12V
3W
-
6W
8W
+
V0
-
4A
3W
17
Example
• Now transform voltage source with 6W resistor
6W
+
12V
8W
+
V0
-
3W
4A
Dr. Louie
18
Example
• Result:
 R = 6W
 Is = Vs/R = 12/6 = 2A
6W
+
12V
8W
+
V0
-
3W
4A
2A
Dr. Louie
6W
8W
+
V0
-
4A
3W
19
Example
• The rest is easy.
• Current division: I0 = (4-2)x(2/10) = 0.4A
• V0 = 3.2V
2A
6W
8W
+
V0
-
3W
8W
4A
Dr. Louie
+
V0
-
2W
2A
20
Source Transformations
• Dependent sources are handled using the same
procedure
• Be careful
Dr. Louie
21
Thevenin’s Theorem
• Often, most elements of a circuit are fixed and
only one element (the load) changes
• Do not want to re-solve the entire circuit every
time the load changes
• Better approach: represent unchanging part of
circuit with voltage source with series resistance
RTH
+
-
a
+
Vab
b
VTH
a
+
-
b
(Variable
Resistor)
Dr. Louie
22
Thevenin’s Theorem
• Thevenin’s Theorem: a linear two-terminal circuit
can be replaced by an equivalent circuit
consisting of a voltage source in series with a
resistor
• RTH: Thevenin Resistance
• VTH: Thevenin Voltage
Linear
TwoTerminal
Circuit
RTH
a
a
=
VTH
b
+
-
b
Dr. Louie
23
Thevenin’s Theorem
• How do we find VTH and RTH?
• One way: keep applying resistance and source
transformations until there is a voltage source in
series with a resistance between the terminals
RTH
+
-
a
VTH
b
a
+
-
b
Dr. Louie
24
Thevenin’s Theorem
• Better way: recognize that
 VTH = VOC and
 RTH = input resistance (looking into terminals a,
and b), or RTH =VOC/ISC
+
-
a
RTH
+
VOC
-
VTH
b
a
+
-
b
Dr. Louie
25
Finding Thevenin Resistance
• No Dependent Sources:
 short all voltage sources
 open all current source
 then find equivalent resistance RTH = Req
• Dependent Sources:




short all voltage sources
open all current source
Apply test voltage V0, compute current I0
RTH = V0/I0
Dr. Louie
26
Example
• Find the current through the load resistor if RL is
6, 16 and 36W
• Perfect situation for Thevenin Equivalent
 Find Thevenin Equivalent, then solve equivalent
circuit for various values of RL
4W
32V
+
-
12W
1W
a
2A
RL
b
Dr. Louie
27
Example
• Start with finding the Thevenin voltage
 VTH = VOC
• By superposition (or mesh analysis)
 VOC1 = 32(12/16) = 24V
 VOC2 = 4x2x(12/16) = 6V
 VTH = VOC1 + VOC2 = 30V
4W
32V
+
-
12W
1W
a
2A
b
Dr. Louie
28
Example
• Now find the Thevenin resistance
 RTH = Req
 deactivate all sources
 Req = 1 + (4x12)/16 = 4W
4W
1W
a
12W
Req
b
Dr. Louie
29
Example
• Thevenin equivalent circuit is shown below
• Current through various load resistances can be
easily computed
4W
32V
+
-
12W
1W
4W
a
2A
30V
b
Dr. Louie
a
+
-
b
30
Example
• Find the Thevenin Equivalent of the circuit
between the terminals a, b
60W
2A
30W
a
b
Dr. Louie
+
-
30V
31
Example
• Via superposition
 VOC1 = 30x(30/90) = 10V
 VOC2 = 30x2(60/90) = 40V
• VTH = VOC1 + VOC2 = 50V
60W
2A
30W
a
b
Dr. Louie
+
-
30V
32
Example
• Now find RTH
 RTH = (30x60)/(30+60) = 20W
60W
30W
a
b
Dr. Louie
33
Thevenin’s Theorem
• Find the Thevenin equivalent
 Note the dependent source
• Finding VTH is the same procedure as before
5W
Ix
3W
a
6V
+
-
1.5Ix
4W
b
Dr. Louie
34
Thevenin’s Theorem
• Mesh Analysis
 6 = 5I1 + 7Ix (Supermesh)
 1.5Ix + I1 = Ix (current source constraint equation)
Solving…
Ix = 1.33A
Therefore
VOC = 1.33x4 = 5.33V = VTH
5W
3W
a
6V
+
-
I1
Ix
4W
b
Dr. Louie
35
Thevenin’s Theorem
• To find RTH




apply either test voltage or current to the terminals
deactivate independent sources
compute either terminal current or voltage
RTH = V0/I0
5W
Ix
3W
a
6V
+
-
1.5Ix
4W
b
Dr. Louie
36
Thevenin’s Theorem
• Apply a test voltage V0
• Let V0 = 1V
• Now find I0 (note polarity if I0)
5W
Ix 3W
I0
a
1.5Ix
4W
+
V0
-
b
Dr. Louie
37
Thevenin’s Theorem
• I1 + 1.5Ix = Ix (Nodal Analysis)







I1 + 0.5Ix = 0
-0.2V1 + 0.5(V1 - 1)/3=0
-0.0333V1 – 0.16667 =0
V1=-5V
Ix = -2A (Ohm’s Law)
I2 = 0.25A (Ohm’s Law)
I0 = 2.25A (KCL)
5W I1
Ix 3W
I0
a
1.5Ix
I2
4W
+
V0
-
b
Dr. Louie
38
Thevenin’s Theorem
• Therefore RTH = V0/I0 = 1/2.25 = 0.444W
0.444W
5.33V
a
+
-
b
Dr. Louie
39
Norton’s Theorem
• Thevenin equivalent circuit can be transformed
into current source in parallel with a resistor
• From discussion on source transformation:
 RN = RTH
Thevenin Equivalent
RTH
Norton Equivalent
a
VTH
a
+
IN
-
b
RN
b
Dr. Louie
40
Norton’s Theorem
• IN is found by shorting the terminals of the circuit
 IN = ISC
Linear
TwoTerminal
Circuit
a
ISC
a
IN
b
IN
b
Dr. Louie
41
Norton’s Theorem
• Process for finding RTH is identical to that for
Thevenin’s theorem
• No Dependent Sources:
 short all voltage sources
 open all current source
 then find equivalent resistance RTH = Req
• Dependent Sources:
 short all voltage sources
 open all current source
 Apply test voltage V0 (or current), compute current
I0 (or voltage)
 RTH = V0/I0
Dr. Louie
42
Practical Sources
14
Ideal source
12
RS
L
-
+
VL
-
V (V)
12V
+
10
Rs = 1W
8
Rs = 0.25W
6
4
2
0
0
2
Dr. Louie
4
RL (W)
6
8
10
43
Practical Sources
6
Ideal source
5
R||
IL
R|| = 100W
3
L
5A
I (A)
4
R|| = 50W
2
1
0
0
20
Dr. Louie
40
60
RL (W)
80
100
44