Chapter 5: Additional Analysis
Techniques
(Part 1)
Minkyu Je
Additional Analysis Techniques
EE201 Circuit Theory
1
Outline
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Introduction
¢
Superposition
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Thévenin’s and Norton’s Theorems
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Maximum Power Transfer
¢
Design Example
Additional Analysis Techniques
EE201 Circuit Theory
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Introduction: Equivalence
¢
Quick review of equivalent circuits that have been employed in our analyses
Additional Analysis Techniques
EE201 Circuit Theory
3
Introduction: Linearity
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Linearity requires both additivity and homogeneity (scalability).
¢
Determine Vout .
£
Straight forward approach: calculate Io, then I1, then I2, and so on.
£
Here, we use linearity and make an initial assumption that Vout = 1 V.
£
This assumption will yield a value for the source voltage.
£
We will then use the actual value of the source voltage and linearity to
compute the actual value of Vout.
Additional Analysis Techniques
EE201 Circuit Theory
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Introduction: Linearity
¢
Determine Vout .
Vout = V2 = 1 V
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The assumption that Vout = 1 V produced a
source voltage of 6 V.
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Since the actual source voltage is 12 V, the
actual output voltage is 1 V ´ (12/6) = 2 V.
I2 = V2/2k = 0.5 mA
V1 = 4kI2 + V2 = 3 V
I1 = V1/3k = 1 mA
Io = I1 + I2 = 1.5 mA
Vo = 2kIo + V1 = 6 V
Additional Analysis Techniques
EE201 Circuit Theory
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Example
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Change only the value of the sources to obtain a factor of 10 increase in the
current i1. The value of both sources must be changed and neither must be set
to zero.
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The current i1 is composed of two terms: one that depends solely on the 4-V
source, and the other that depends solely on the 10-V source.
£
This may be written as i1 = 4k1 + 10k2, where k1 and k2 are constants that
depend on the circuit topology and resistor values.
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We may not change k1 or k2, as only the source voltages may be changed.
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Thus, 4 V à 40 V and 10 V à 100 V.
Additional Analysis Techniques
EE201 Circuit Theory
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Superposition
¢
Concept of superposition
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The mesh equations for this network are
6ki1(t) - 3ki2(t) = v1(t) and
-3ki1(t) + 9ki2(t) = -v2(t).
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Solving these equations for i1(t) yields
v (t ) v (t )
i1(t ) = 1 - 2 .
5k
15k
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The current i1(t) has a component due to v1(t) and a component due to v2(t).
£
In view of the fact that i1(t) has two components, one due to each
independent source, it would be interesting to examine what each source
acting alone would contribute to i1(t).
Additional Analysis Techniques
EE201 Circuit Theory
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Superposition
¢
Concept of superposition
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For v1(t) to act alone, v2(t) must be zero.
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Determining the component of i1(t) due to
v1(t) only and referring to it as i1’(t),
i1' (t ) =
=
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Let us now determine the component due
to v2(t) acting alone and refer to it as i1’’(t).
i2'' (t ) =
- v 2 (t )
- 2v 2 (t )
=
(3k )(3k )
15k
6k +
3k + 3k
i1'' (t ) =
- 2v 2 (t ) æ 3k ö - v 2 (t )
.
ç
÷=
15k è 3k + 3k ø
15k
+
Additional Analysis Techniques
v 1(t )
v (t )
= 1 .
(3k )(6k )
5k
3k +
3k + 6k
EE201 Circuit Theory
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Superposition
¢
Concept of superposition
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Now, by adding i1’(t) and i1’’(t), we obtain
i1(t ) = i1' (t ) + i1'' (t ) =
£
v 1(t ) v 2 (t )
.
5k
15k
=
We have superposed the value of i1’(t) on
i1’’(t), or vice versa, to determine the
unknown current.
+
Additional Analysis Techniques
EE201 Circuit Theory
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Superposition
¢
Principle of superposition
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In any linear circuit containing multiple independent sources, the current or
voltage at any point in the network may be calculated as the algebraic sum
of the individual contributions of each source acting alone.
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When determining the contribution due to an independent source,
£
˜
any remaining voltage sources are made zero by replacing them with
short circuits, and
˜
any remaining current sources are made to zero by replacing them with
open circuits.
Superposition provides some insight in determining the contribution of each
source to the variable under investigation.
Additional Analysis Techniques
EE201 Circuit Theory
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Example
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Use superposition to find Vo.
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Calculating the contribution of the 2-mA
source to Vo with the 3-V source set to
zero,
æ 1k + 2k ö 2
Io = 2m ´ ç
÷ = mA
1
k
+
2
k
+
6
k
è
ø 3
=
Vo' = Io (6k ) = 4 V.
£
6k
æ
ö
Vo' = 3 ´ ç
÷ = 2 V.
1
k
+
2
k
+
6
k
è
ø
+
Additional Analysis Techniques
Determining contribution of the 3-V
source with the 2-mA source set to zero,
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Therefore, Vo = Vo’ + Vo’’ = 6 V.
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Although we solved two separate
circuits, both were very simple.
EE201 Circuit Theory
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Example
¢
For comparison, use nodal and loop analysis to find Vo.
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To use nodal analysis, we form a
supernode containing the 3-V source.
Vo - 3
V
- 2m + o = 0,
1k + 2k
6k
resulting in Vo = 6 V.
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By using loop analysis, we obtain
I1 = -2 mA and 3k(I1 + I2) - 3 + 6kI2 = 0,
which yield I2 = 1 mA and hence Vo = 6 V.
Additional Analysis Techniques
EE201 Circuit Theory
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Example
¢
Use superposition to find Vo.
£
£
Contribution of V1: inverting
amplifier configuration
R
Vo1 = - 2 V1
R1
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Therefore,
Vo = Vo1 + Vo2 = -
æ
R2
R ö
V1 + çç 1 + 2 ÷÷V2
R1
R1 ø
è
Contribution of V2: noninverting
amplifier configuration
æ
R ö
Vo2 = çç 1 + 2 ÷÷V2
R1 ø
è
Additional Analysis Techniques
EE201 Circuit Theory
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Summary
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¢
Introduction
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Equivalence
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Linearity
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Example
Superposition
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Principle of superposition
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Examples
Additional Analysis Techniques
EE201 Circuit Theory
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End of Slides
Additional Analysis Techniques
EE201 Circuit Theory
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