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ECE 3144 Lecture 19
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Sources Transformations
Practical voltage source
Practical current source
If vS = iS RS
+
Ideal source
Practical source
Ideal source
vo
Practical source
Ideal source
vo=vS
vo
If vS = iS RS
vo = vS - ioRS
Practical
source
io= vS/RS
io
vo= iSRS
vo = RS (iS – io)
Ideal source
Practical
source
io
2
Source transformations
• Recall voltage source formula vo = vS – ioRS and current source formula vo =
RS (iS – io). The voltage source circuit is equivalent to the current source
circuit if they have the same i-v characteristics to the load resistor, i.e., vS =
RSiS
• After all, one circuit can't see another's structure, count the number of its
components, notice differences in color, smell or weight. All one circuit can
"see" about another is its i-v characteristic.
• To another circuit, two linear circuits are equivalent (even if they are entirely
different inside) provided only that the two circuits have the same relationship
between voltage and current at corresponding load terminals (that is, have the
same i-v characteristic).
3
Source Transformation example 1
Find I in the circuit given.
The equivalent voltage source: 9 mA*5k = 45V in series with a 5k resistor.
Apply KVL to the only loop:
-45 +5000I+4700I+3000I+3 = 0
Which can be easily solved to find that the current
I =3.3mA
4
Source Transformation example 2
12K
2K
6V
Find Io in the network shown
using source transformation.
18K
4K
2mA
2mA
8K
Io
18K
1/2mA
12V
8K
2mA
12K
6K
Io
Io
1/2mA
12K
24K
1/2mA
8K
Io
2mA
12K || 24K
8K
8K
2mA
Io = 2m [ 8K / (8K + 8K) ]
Io = 1 mA
5
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 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
6
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 vTH(t) and 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 )
 ishortcircuit (t )  isc (t )
RTH
•If vTH(t) ≠ 0, then ishortcircuit(t) ≠ 0 and we find
RTH 
voc (t )
isc (t )
7
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 to a 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 )
8
Norton theorem
io
+
vo
(c)
•Based on source transformation we have learned, we
can determine iN(t) and RN
RN  RTH
voc (t )

isc (t )
vTH
i N (t ) 
 ishortcircuit (t )  isc (t )
RTH
9
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.
10
Applications of Thevenin and Norton’s Theorems
•
•
•
•
•
•
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)
11
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
12
+
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
13
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
14
Homework for Lecture 19
• Problems 4.19, 4.22, 4.24, 4.28
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