Chapter 5: Additional Analysis
Techniques
(Part 3)
Minkyu Je
Additional Analysis Techniques
EE201 Circuit Theory
1
Outline
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Introduction
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Superposition
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Thévenin’s and Norton’s Theorems
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Maximum Power Transfer
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Design Example
Additional Analysis Techniques
EE201 Circuit Theory
2
Source Transformation
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There is yet one other aspect of Thévenin’s and Norton’s theorems that we find
useful in circuit analysis and design.
¢
The relationship between Thévenin and Norton equivalent circuits represents
what is called a source transformation or source exchange.
voc
v oc = i sc RTh
i=
v o = v oc - iRTh
Additional Analysis Techniques
EE201 Circuit Theory
- vo
+ i sc
RTh
3
Source Transformation
¢
Source Transformation
£
If we have embedded within a network a current source i in parallel with a
resistor R, we can replace this combination with a voltage source of value v
= iR in series with the resistor R.
£
The reverse is also true; that is a voltage source v in series with a resistor R
can be replaced with a current source of value i = v/R in parallel with R.
Additional Analysis Techniques
EE201 Circuit Theory
4
Source Transformation
¢
Find Vo using the repeated application of source transformation.
£
The series combination of the 12-V source and 3-kW resistor is converted to
a 4-mA current source in parallel with the 3-kW resistor.
£
If we combine the 3-kW resistor with the 6-kW resistor, we obtain the circuit
shown below.
Additional Analysis Techniques
EE201 Circuit Theory
5
Source Transformation
¢
Find Vo using the repeated application of source transformation.
£
Continuing the reduction, we convert the 4-mA source and 2-kW resistor into
an 8-V source in series with this same 2-kW resistor.
£
The two 2-kW resistors that are in series are now combined to produce the
network shown below.
Additional Analysis Techniques
EE201 Circuit Theory
6
Source Transformation
¢
Find Vo using the repeated application of source transformation.
£
If we now convert the combination of the 8-V source and 4-kW resistor into a
2-mA source in parallel with the 4-kW resistor.
£
Then, we combine the resulting current source with the other 2-mA source,
arriving at the circuit shown below.
Additional Analysis Techniques
EE201 Circuit Theory
7
Source Transformation
¢
Find Vo using the repeated application of source transformation.
£
Now, we can apply current division to the two parallel resistance paths.
4k
æ
ö
Io = (4m)ç
÷ = 1 mA
è 4k + 4k + 8k ø
Vo = (1m)(8k ) = 8 V
Additional Analysis Techniques
EE201 Circuit Theory
8
Application of Theorems: Benefits
¢
Find Vo using nodal analysis.
£
KCL equation at the supernode:
-
£
2 V x V x - V3 3V x - V3 3V x - 4
+
+
+
+
=0
k 1k
1k
1k
1k
KCL equation at the node labeled V3:
-
£
£
Simplified equations:
8V x - 2V3 = 6
Constraint equations:
- 4V x + 2V3 = 2
V4 = 4
V1 - V2 = 2 Vx where Vx = V2
à V1 = 3Vx
2 V3 - 3V x V3 - V x
+
+
=0
k
1k
1k
£
Solving these equations, we obtain
Vx = 2 V, V3 = 5 V, and
Vo = 3Vx - V3 = 1 V.
Additional Analysis Techniques
EE201 Circuit Theory
9
Application of Theorems: Benefits
¢
Find Vo using loop analysis.
£
KVL equations for loops 2 and 4:
-2Vx + 1kI2 + 1k(I2 - I3) = 0
1k(I4 + I3 - I1) - 2Vx + 1kI4 + 4 = 0
£
Controlling equation:
Vx = 1k(I1 - I3 - I4)
£
£
£
The network has 4 loops, and thus
4 linearly independent equations
are required.
Constraint equations:
I1 = 2/k
Substituting the equations for I1 and
I3 into the 2 KVL equations yields
2kI2 + 2kI4 = 6
4kI4 = 8.
£
Solving these equations for I2 and I4,
we obtain I4 = 2 mA and I2 = 1 mA,
leading to Vo = 1 V.
I3 = -2/k
Additional Analysis Techniques
EE201 Circuit Theory
10
Application of Theorems: Benefits
¢
Find Vo using Thévenin’s theorem.
£
The circuit can be simplified by
forming a Thévenin equivalent for the
leftmost and rightmost branches.
£
Voc1 =
£
2
(1k ) + 4 = 6 V, RTh1 = 1 kW
k
The resultant Thévenin equivalent
circuit is connected to the remaining
portion of the circuit.
£
Additional Analysis Techniques
EE201 Circuit Theory
Now we break the
network at the
output terminals to
find Voc2 .
11
Application of Theorems: Benefits
¢
Find Vo using Thévenin’s theorem.
£
£
£
V2
Additional Analysis Techniques
Node equations:
V - 6 V1 - 2V x¢ 2
V1 = 3V x¢ and 1
+
=
1k
1k
k
Thus, V x¢ = 2 V and V1 = 6 V.
Voc2 is obtained using the KVL equation:
2
- 2V x¢ + Voc 2 + (1k ) = 0
k
resulting in Voc2 = 2 V.
£ Isc2
is now derived.
V - 6 V2 - 2V x¢¢ 2
+
=
V2 = 3V x¢¢ and 2
1k
1k
k
£
Thus, V x¢¢ = 2 V and V2 = 6 V.
£
From KCL, isc = 2 mA.
£
Then, RTh2 = Voc2/Isc2 = 1kW.
EE201 Circuit Theory
12
Application of Theorems: Benefits
¢
Find Vo using Thévenin’s theorem.
Additional Analysis Techniques
£
The Thévenin equivalent circuit now
consists of a 2-V source in series with a 1kW resistor.
£
Connecting this Thévenin equivalent circuit
to the load resistor yields the final network
to solve.
£
By voltage division, Vo = 1 V.
EE201 Circuit Theory
13
Application of Theorems: Benefits
¢
¢
Let us compare the use of node or loop analysis with that of Thévenin’s or
Norton’s theorem.
£
We have found that in some cases, the theorems do not necessarily simplify
the problem, and a straightforward approach using node or loop analysis is
as good an approach as any.
£
This is a valid point, provided that we are simply looking for some particular
voltage or current.
However, the real value of the theorems is the insight and understanding that
they provide about the physical nature of the network.
£
The theorems help us to understand the concept of loading a network, the
idea of matching a network for maximum power transfer, or the effect of
using a transducer at the input of an amplifier with a given input resistance.
£
We derive none of this information from a node or loop analysis.
Additional Analysis Techniques
EE201 Circuit Theory
14
Application of Theorems: Benefits
¢
As a simple example, suppose that a network at a specific pair of terminals has
a Thévenin equivalent circuit consisting of a voltage source in series with a 2kW resistor.
£
If we connect a 2-W resistor to the network at these terminals, the voltage
across the 2-W resistor will be essentially nothing.
£
This result is obvious using the Thévenin theorem approach.
£
However, a node or loop analysis gives no clue as to why we have obtained
this result.
Additional Analysis Techniques
EE201 Circuit Theory
15
Maximum Power Transfer
¢
There are situations in circuit design when we want to select a load so that the
maximum power can be transferred to it.
¢
By employing Thévenin’s theorem, we can determine the maximum power that
a circuit can supply and the manner in which the load is adjusted to achieve
maximum power transfer.
£
The power that is delivered to the load is given by
Pload
£
2
æ v ö
÷÷ RL .
= i RL = çç
è R + RL ø
2
To determine the value of RL that maximizes this
quantity, we differentiate this expression with
respect to RL and equate the derivative to zero:
dPload (R + RL )2 v 2 - 2v 2RL (R + RL )
=
=0
4
dRL
(R + RL )
which yields RL = R.
Additional Analysis Techniques
EE201 Circuit Theory
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Maximum Power Transfer
¢
In conclusion,
£
the maximum power transfer takes place when RL = R, and
£
the maximum power delivered under such condition is
Pload ,max
2
v2
æ v ö
=i R=ç
.
÷ R=
4R
èR +Rø
2
=R
Additional Analysis Techniques
EE201 Circuit Theory
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Example
¢
Find the value of RL for maximum power transfer and the maximum power that
can be transferred to this load.
£
We derive the Thévenin equivalent circuit for
the network exclusive of the load.
£
Calculation of Voc:
I1 = 2 mA and 3k(I2 - I1) + 6kI2 + 3 = 0
Solving these equations yields I2 = 1/3 mA.
Therefore, Voc = 4kI1 + 6kI2 = 10 V.
£
RTh = 4k + (6k//3k)
= 6 kW
Additional Analysis Techniques
EE201 Circuit Theory
18
Example
¢
Find the value of RL for maximum power transfer and the maximum power that
can be transferred to this load.
£
Therefore, RL = RTh = 6 kW for maximum power transfer.
£
The maximum power transferred to the load is
2
25
æ 10 ö
PL = ç
mW.
÷ (6k ) =
6
è 12k ø
Additional Analysis Techniques
EE201 Circuit Theory
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Example
¢
For the network shown below, let us examine a variety aspects of maximum
power transfer by plotting the parameters Vout, I, Pout, Pin, and the efficiency =
Pout/Pin as a function of the resistor ratio R2/R1.
Additional Analysis Techniques
£
æ R2 ö
5R2
÷÷Vin =
Vout = çç
2 + R2
è R1 + R2 ø
£
I=
Vin
5
=
R1 + R2 2 + R2
V in2
25
=
=
R1 + R2 2 + R2
£
Pin = IVin
£
æ Vin öæ VinR2 ö
25R2
÷÷çç
÷÷ =
Pout = IVout = çç
2
è R1 + R2 øè R1 + R2 ø (2 + R2 )
£
efficiency =
EE201 Circuit Theory
Pout
R2
R2
=
=
Pin
R1 + R2 2 + R2
20
Example
¢
For the network shown below, let us examine a variety aspects of maximum
power transfer by plotting the parameters Vout, I, Pout, Pin, and the efficiency =
Pout/Pin as a function of the resistor ratio R2/R1.
5R2
Vout =
2 + R2
5
I=
2 + R2
Pin =
25
2 + R2
Pout =
Pout
Pin
25R2
(2 + R2 )2
R2
=
2 + R2
£
Max. power doesn’t correspond to max. output voltage, current, or efficiency.
£
At max. power transfer, the efficiency is always 0.5, or 50%.
Additional Analysis Techniques
EE201 Circuit Theory
21
Design Example
¢
The terminal voltage of a voltage source is 12 V when connected to a 2-W load.
When the load is disconnected, the terminal voltage rises to 12.4 V. Calculate
the source voltage vs and internal resistance Rs.
Additional Analysis Techniques
£
The terminal voltage when the load is
disconnected is the open-circuit voltage,
vs = voc = 12.4 V.
£
When the load is connected, vL = 12 V
and PL = 2 W.
£
Hence, PL = vL2/RL = 2 à RL = 72 W.
£
The load current, iL = vL/RL = 1/6 A
£
RsiL = 12.4 - 12 = 0.4 à Rs = 2.4 W
EE201 Circuit Theory
22
Design Example
¢
An attenuator is an interface circuit that reduces the voltage level without
changing the output resistance.
(a) By specifying Rs and Rp, design an attenuator that will meet the following
requirements: Vo/Vg = 0.125, Req = RTh = Rg = 100 W.
Rp
Vo
=
º a = 0.125
Vg Rg + Rs + Rp
®
Rp
a
= Rg + Rs + Rp
® Rg + Rs = Rp
Req = Rp //(Rg + Rs ) =
Rp (Rg + Rs )
Rp + Rg + Rs
= Rg
® RpRs = Rg (Rg + Rs )
Additional Analysis Techniques
EE201 Circuit Theory
1- a
a
1- a
Req = 700 W
a
Rg (Rg + Rs )
Rp =
= 114.29 W
Rs
Rs =
23
Design Example
¢
An attenuator is an interface circuit that reduces the voltage level without
changing the output resistance.
(b) Using the interface designed, calculate the current through a load of RL =
50 W when Vg = 12 V.
VTh = 0.125 Vg = 1.5 V
RTh = Rg = 100 W
I=
Additional Analysis Techniques
EE201 Circuit Theory
VTh
1 .5
=
= 10 mA
RTh + RL 150
24
Summary
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Thévenin’s and Norton’s Theorems
£
Source transformation
£
Benefits of using Thévenin’s or Norton’s theorem
¢
Maximum power transfer
¢
Design example
Additional Analysis Techniques
EE201 Circuit Theory
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End of Slides
Additional Analysis Techniques
EE201 Circuit Theory
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