Using the Thévenin Equivalent in the Amplifier Circuit
4.13 Superposition
A linear system obeys the principle of superposition, which states that whenever a linear system is excited, or driven, by more than one inde- pendent source of energy, the total response is the sum of the individual responses. An individual response is the result of an independent source acting alone. Because we are dealing with circuits made up of inter- connected linear-circuit elements, we can apply the principle of superposi- tion directly to the analysis of such circuits when they are driven by more than one independent energy source. At present, we restrict the discussion to simple resistive networks; however, the principle is applicable to any linear system.
Superposition is applied in both the analysis and the design of circuits.
In analyzing a complex circuit with multiple independent voltage and cur- rent sources, there are often fewer, simpler equations to solve when the effects of the independent sources are considered one at a time. Applying superposition can thus simplify circuit analysis. Be aware, though, that sometimes applying superposition actually complicates the analysis, produc- ing more equations to solve than with an alternative method.Superposition is required only if the independent sources in a circuit are fundamentally different. In these early chapters, all independent sources are dc sources, so superposition is not required. We introduce superposition here in anticipa- tion of later chapters in which circuits will require it.
Superposition is applied in design to synthesize a desired circuit response that could not be achieved in a circuit with a single source. If the desired circuit response can be written as a sum of two or more terms, the response can be realized by including one independent source for each term of the response. This approach to the design of circuits with complex responses allows a designer to consider several simple designs instead of one complex design.
A S S E S S M E N T P R O B L E M S
4.13 Superposition 123
4 3
6
120 V i1
2 i3
i4 i2
12 A
Figure 4.62 A circuit used to illustrate superposition.
4 3
120 V
6 i1
2 i3
i4
i2
v1
Figure 4.63 The circuit shown in Fig. 4.62 with the current source deactivated.
We demonstrate the superposition principle by using it to find the branch currents in the circuit shown in Fig. 4.62. We begin by finding the branch currents resulting from the 120 V voltage source. We denote those currents with a prime. Replacing the ideal current source with an open cir- cuit deactivates it; Fig. 4.63 shows this. The branch currents in this circuit are the result of only the voltage source.
We can easily find the branch currents in the circuit in Fig. 4.63 once we know the node voltage across the resistor. Denoting this voltage
we write
(4.78) from which
(4.79) Now we can write the expressions for the branch currents directly:
(4.80)
(4.81)
(4.82) To find the component of the branch currents resulting from the current source, we deactivate the ideal voltage source and solve the circuit shown in Fig. 4.64. The double-prime notation for the currents indicates they are the components of the total current resulting from the ideal current source.
We determine the branch currents in the circuit shown in Fig. 4.64 by first solving for the node voltages across the 3 and resistors, respec- tively. Figure 4.65 shows the two node voltages. The two node-voltage equations that describe the circuit are
(4.83)
(4.84)
Solving Eqs. 4.83 and 4.84 for and we get
(4.85)
(4.86) Now we can write the branch currents through directly in terms of the node voltages and :
(4.87) ifl1 =
-v3
6 =
12
6 = 2A, v4
v3
ifl4 ifl1
v4 = -24V.
v3 = -12V, v4, v3
v4 - v3
2 +
v4
4 + 12 = 0.
v3
3 + v3
6 +
v3 - v4
2 = 0, 4 Æ iœ3 = iœ4 =
30
6 = 5A.
iœ2 = 30
3 = 10A, iœ1 =
120 - 30
6 = 15A,
iœ1 -iœ4
v1 = 30 V.
v1 - 120
6 +
v1 3 +
v1 2 + 4 = 0, v1,
3 Æ
4 3
6 i1
2 i3
i4
i2 12 A
Figure 4.64 The circuit shown in Fig. 4.62 with the voltage source deactivated.
4 3
6 2
v4 12 A
v3
Figure 4.65 The circuit shown in Fig. 4.64 showing the node voltages v3and v4.
(4.88)
(4.89)
(4.90)
To find the branch currents in the original circuit, that is, the currents and in Fig. 4.62, we simply add the currents given by Eqs. 4.87–4.90 to the currents given by Eqs. 4.80–4.82:
(4.91)
(4.92)
(4.93)
(4.94)
You should verify that the currents given by Eqs. 4.91–4.94 are the correct values for the branch currents in the circuit shown in Fig. 4.62.
When applying superposition to linear circuits containing both independ- ent and dependent sources, you must recognize that the dependent sources are never deactivated. Example 4.13 illustrates the application of superposi- tion when a circuit contains both dependent and independent sources.
i4 = i4œ + i4fl = 5 - 6 = -1 A.
i3 = i3œ + i3fl = 5 + 6 = 11A, i2 = i2œ + i2fl = 10 - 4 = 6A, i1 = i1œ + i1fl = 15 + 2 = 17 A, i4
i3, i2, i1,
ifl4 = v4
4 = -24
4 = -6A.
ifl3 =
v3 - v4
2 =
-12 + 24
2 = 6A,
ifl2 = v3
3 = -12
3 = -4A,
Example 4.13 Using Superposition to Solve a Circuit
Use the principle of superposition to find in the circuit shown in Fig. 4.66.
Figure 4.66The circuit for Example 4.13.
Solution
We begin by finding the component of resulting from the 10 V source. Figure 4.67 shows the circuit.
With the 5 A source deactivated, vϢ must equal vo
10 20
5
10 V
i
5 A 0.4 v
2 i vo
v
vo Hence, must be zero, the branch
containing the two dependent sources is open, and
Figure 4.67The circuit shown in Fig. 4.66 with the 5 A source deactivated.
10 20
5
10 V
i
0.4 v
2 i vo
v
vœo = 20
25(10) = 8V.
vϢ
(-0.4vϢ)(10).
Practical Perspective 125
When the 10 V source is deactivated, the circuit reduces to the one shown in Fig. 4.68. We have added a reference node and the node designations a, b, and c to aid the discussion.Summing the cur- rents away from node a yields
Summing the currents away from node b gives
We now use
to find the value for Thus,
5vߢ = 50, or vߢ = 10 V.
vߢ.
vb = 2iߢ + vߢ 4vߢ + vb - 2iߢ = 50.
0.4vߢ +
vb - 2iߢ
10 - 5 = 0, or vflo
20 + vflo
5 - 0.4vfl¢ = 0, or 5vflo - 8vfl¢ = 0.
From the node a equation,
The value of is the sum of and or 24 V.
Figure 4.68The circuit shown in Fig. 4.66 with the 10 V source deactivated.
NOTE: Assess your understanding of this material by trying Chapter Problems 4.93 and 4.98.
10 20
5 a b
c i
5 A 0.4 v
2 i vo
v
vflo, vœo
vo
5vfl0 = 80, or vfl0 = 16 V.