Transcript ECE 3144 Lecture 4

ECE 3144 Lecture 21
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
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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
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(e)
Vo
The equivalent Norton equivalent
circuit is shown in figure (e).
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Case 2 example
Find Vo in the circuit shown using
Thevenin’s Theorem.
4K
6V
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Vx
Vo
-
2K
4K
2K
-
12V
Vx/1000
Find Voc
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-
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 = -0.1875mA
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
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vS
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We can see that i=-1A. Apply KVL nodal
analysis at node a:
v S  1.5(1) v S

1
3
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=>
vS = 0.6V =>
RTH = vs/i= 0.6
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Problem solving strategy by
using Thevenin theorem/Norton theorem
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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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Operational Amplifier
•An operational amplifier is a linear circuit network.
•Based on Thevenin theorem or Norton theorem, any linear circuit can be considered
as equivalent to a Thevenin circuit or Norton circuit at a specified terminal pair . The
following is the detailed modeling for an opamp. Ri is the input Thevenin resistance
and Ro is the output Thevenin resistance.
i+ ->
->iout
i- ->
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Ideal Operational Amplifier
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Now we want to derive the ideal conditions for the op-amp.
If an external network A is connected to the op-amp at the input terminals, then op-amp is
considered as the load of the external network A.
A
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In order to gain maximum output voltage (v+-v-) for op-amp with any external network A => the
input Thevenin resistance for the ideal op-amp is infinite. Ri-> => i+ = i- = 0
If an external network B is connected to the op-amp at the output terminals, the external
network B is considered as the load of the op-amp.
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B
vo
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In order for the load to gain the maxim output vo for any external network B=> the output
Thevenin resistance for the ideal op-amp is zero, Ro = 0.
Finally, the gain of an ideal operational amplifier is infinite => A-> and v+ = v- = 0.
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Ideal Operational Amplifier
Ideal model for an operational amplifier
Modeling parameters
i+ = i- = 0 and v+ = v-
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Example : Differential operational amplifier
Apply KCL in the “-” terminal(inverting
terminal) of the opamp:
(1)
Apply KCL in the “+” terminal
(noninverting terminal) of the opamp:
(2)
v+ = v -
(3)
Using equations (1), (2) and (3) to solve vout
The output of this amplifier clearly is proportional to the difference of the voltages of the
driving circuits. For this reason, this amplifier is called a differential amplifier.
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Homework for Lecture 21
• Problems 4.61, 4.67, 4.69.
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