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ECE 3144 Lecture 20
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Thevenin and Norton theorems
•
•
•
Let us suppose that we need to only make a partial analysis of the a circuit. For
example, perhaps we need to determine the current, voltage and power delivered to
a single “load” resistor by the remainder of the circuit., which may consist of a
sizable number of sources and resistors.
Thevenin theorem tells us that we can replace the entire linear network, exclusive
of the load resistor, by an equivalent circuit that contains only an independent
voltage source in series with a resistor in such a way that the current-voltage (i-v)
relationship at the load resistor is unchanged.
Norton theorem tells us that we can replace the entire network, exclusive of the
load resistor, by an equivalent circuit that contains only an independent current
source in parallel with a resistor in such a way that the current-voltage (i-v)
relationship at the load resistor is unchanged.
io
io
io
+
vo
-
+
+
vo
vo
-
(a)
A complex network
including a load resistor RL.
(b)
A Thevenin equivalent
network
(c)
A Norton equivalent
network
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Thevenin Theorem
io
+
vo
-
• This result for Thevenin theorem is not too surprising
if we note that the Thevenin equivalent circuit can
produce any linear (straight line) i-v characteristic for
some choice of Thevenin voltage vTH(t) and Thevenin
resistance RTH.
•A natural question is how to determine vTH(t) and RTH
for a particular circuit. Sometimes they are measured
experimentally. Sometimes they are calculated.
•In either case, it is helpful to note that if we connect no
load and therefore io(t) = 0, then we can determine vTH(t)
from
vTH (t )  vopencircuit (t )  voc (t )
where voc(t) is called the open circuit voltage
•If we short circuit the two terminals to force vo(t) = 0,
then we get vTH (t )
RTH
 ishortcircuit (t )  isc (t )
•If vTH(t) ≠ 0, then ishortcircuit(t) ≠ 0 and we find
RTH 
voc (t )
isc (t )
3
Thevenin Theorem: cont’d
•During the measurement, one difficulty is that if we short the terminals of some circuits
(a wall plug, for example), the circuit can be damaged by the resulting large current. In
that case, we can simply place a resistor, R*, across the terminals that causes the terminal
voltage to drop below the open-circuit voltage by a measurable amount voltage, v*(t). By
the voltage divider rule, we see, in this case that
R*
R*
v (t ) 
v (t ) 
v (t )
* TH
* oc
RTH  R
RTH  R
*
RTH  R* (
=>
voc (t )
 1)
v* (t )
•Consider, now, the case in which vTH(t) = 0. Then, the Thevenin equivalent circuit is
simply a resistor. To measure the Thevenin resistance in this case, we can apply a voltage
vS(t) (usually a constant) and measure the resulting current iS(t):
R
v S (t )
iS (t )
4
Norton theorem
io
+
vo
(c)
•Based on source transformation we have learned, we can
determine Norton current iN(t) and Norton resistance RN
RN  RTH
voc (t )

isc (t )
vTH
i N (t ) 
 ishortcircuit (t )  isc (t )
RTH
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Summary for Thevenin and Norton theorems
• Thevenin’s theorem requires that, for any linear circuit consisting of
resistance elements and energy sources with an identified terminal
pair, the circuit can be replaced by a series combination of an ideal
voltage source vTH and a resistance RTH, where vTH is the open-circuit
voltage at the two terminals and RTH is the ratio of the open-circuit
voltage to the short-circuit current at the terminal pair.
• Norton’s theorem requires that, for any linear circuit consisting of
resistance elements and energy sources with an identified terminal
pair, the circuit can be replaced by a parallel combination of an ideal
current source iN and a resistance RN, where iN is the short-circuit
current at the two terminals and RN is the ratio of the open-circuit
voltage to the short-circuit current at the terminal pair.
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Applications of Thevenin and Norton’s Theorems
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•
•
•
•
•
Note that this systematic transformation allows us to reduce the network to a simpler
equivalent form with respect to some other circuit elements.
Although this technique is applicable to networks containing dependent sources, it
may not be as useful as other techniques and care must be taken not transform the
part of the circuit which contains the control variables.
How to apply these theorems depends on the structure of the circuits.
Case 1: if only independent sources are present, we can calculate the open-circuit
voltage and short circuit current and then the Thevenin equivalent resistance.
Case 2: if both independent sources and dependent sources are present, we will
calculate the open-circuit and short circuit current first. Then determine the Thevenin
equivalent resistance.
Case 3: For circuit only contains dependent sources, since both open-circuit and
short-circuit current are zero (no independent sources here to provide the controlling
variables), we cannot determine RTH in this case by using voc/isc. Remember during
the discussion of Thevenin theorem, we can apply an external voltage source vS(t)
(usually a constant) and measure the resulting current iS(t) => RTH = vS(t)/iS(t)
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Case 1 example 1
Find the Thevenin equivalent circuit at
terminal pair a and b for the circuit
shown.
+
This specific problem can be solved by
using different approaches. We solve the
problem by using source transformation
technique.
-
Thus we have
RTH = 12
and
vTH = -8V
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Case 1 example 2
Find Vo in the circuit given in figure (a).
(a)
Vo
Vo is the voltage across the 6k resistor. Thus 6k resistor can
be considered as the load resistor and the rest network can
be replaced by the equivalent Thevenin circuit.
The open circuit voltage Voc is found from figure (b).
(b)
I1
Voc
I2
(c)
RTH
Apply mesh analysis technique:
Mesh 1: -6 +4kI1+2k(I1-I2) = 0
=> I1 = 5/3 mA
Mesh 2: I2 = 2 mA
Applying KVL Voc = 4kI1+2kI2
=4*5/3 + 2*2 = 32/3 V
RTH can be derived by calculating the short circuit current
Isc. Then RTH = Voc/Isc. For the circuits containing only
independent sources, RTH can also be derived by zeroing
out all sources as shown in (c).
RTH = (4k//2k)+2k = 10/3 k
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Case 1 example 2: cont’d
Thus the original circuit is equivalent to
the network shown in figure (d).
(d)
Thus using voltage divider =>
Vo
Vo 
32
6k
48
(
) V
3 6k  10 / 3k
7
(e)
Vo
The equivalent Norton equivalent
circuit is shown in figure (e).
10
+
Case 2 example
Find Vo in the circuit shown using
Thevenin’s Theorem.
4K
6V
+
Vx
Vo
-
2K
4K
2K
-
12V
Vx/1000
Find Voc
+
-
Vx = 6*2k/(4k+2k)
- = 2V
Vy = 12-4k*Vx/1000 = 12-8= 4V
Voc = Vx-Vy = -2V
Find Isc
Vx
(6-Vx)/4k=Vx/2k+Isc
Isc = Vx/1K+(Vx-12)/4k
=> Isc = -0.19mA
Rth = Voc/Isc=10.67k
Vo = Voc*2k/(2k+10.67k) = -0.32V
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Case 3 example
a
Find the Thevenin equivalent of the
circuit given
b
Since the rightmost terminals are already open-circuit, i=0. Consequently, the dependent
source is dead. For this network, since there is no independent source in the network,
both voc and isc are zero. Apply a 1-A source iS externally, measure the voltage vS across the
terminal pairs. Then we have RTH = vS/iS = vS.
a
+
vS
-
We can see that i=-1A. Apply KVL nodal
analysis at node a:
v S  1.5(1) v S

1
3
2
vS = 0.6V =>
=>
RTH = 0.6
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Problem solving strategy by
using Thevenin theorem/Norton theorem
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•
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•
Remove the load and the find the voltage across the open-circuit terminals, Voc. All the circuit
analysis techniques presented (KVL/KCL, current/voltage divider, nodal/loop analysis,
superposition, source transformation)
Determine the Thevenin equivalent resistance of the network at the open terminals with the
load removed. Three different types of circuits may encountered in determining the resistance
RTH.
– If the circuits contains only independent sources, they are made zero by replacing voltage
sources with short circuits and current sources with open circuits. RTH is then found by
computing the resistance of the purely resistive network at the open terminals.
– If the circuit contains only dependent sources, an independent voltage or current source is
applied at the open terminals and the corresponding current or voltage at these terminals
is calculated. The voltage/current ratio at the terminals is the Thevenin equivalent
resistance. Since there is no energy source, the open circuit voltage is zero here.
– If the circuits contains both independent and dependent sources, the open circuit terminals
are shorted and the short-circuit current between these terminals are decided. The ratio of
the open-circuit voltage to the short-circuit current is the Thevenin resistance RTH.
The load is now connected to the Thevenin equivalent circuit, consisting of Voc in series with
RTH, the desired output is obtained.
The problem solving strategy for Norton theorem is essentially the same as that for the
Thevenin theorem with the exception that we are dealing with the short-circuit current instead
of open-circuit voltage.
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Homework for Lecture 20
• Problems 4.32, 4.33, 4.34, 4.35, 4.39, 4.47
• Exam 2 (covering Chapter 3 and 4) will be
on March 20.
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