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Superposition. In the circuit of Fig. 4-50 nd the contribution of each source to I1 , I2 , I3 , and show that they add up to values found in Problem 4.43. From the source on the left: Ans. From the source on the right: From both sources: I1 36=51 I1 4=51 I1 32=51 I2 9=51 I2 18=51 I2 9=51 I3 27=51 I3 20=51 I3 7=51 (All in A)
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Node Voltage Method. In the circuit of Fig. 4-51 write three node equations for nodes A, B, and C, with node D as the reference, and nd the node voltages.
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ANALYSIS METHODS
Loop Current Method. In the circuit of Fig. 4-51 write two loop equations using I1 and I2 as loop currents, then nd the currents and node voltages.   Loop 1: 4I1 I2 2 I1 1 A; I2 2 A Ans: from which, Loop 2: I1 2I2 3 VA 9 V; VB 5 V; VC 2 V Superposition. In the circuit of Fig. 4-51 nd the contribution of each source to VA , VB , VC , and show that they add up to values found in Problem 4.45. From the current source: Ans. From the voltage source: From both sources: VA 7:429 VA 1:571 VA 9 VB 3:143 VB 1:857 VB 5 VC 1:429 VC 0:571 VC 2 (all in V)
Verify that the circuit of Fig. 4-52(a) is equivalent to the circuit of Fig. 4-51. Ans. Move node B in Fig. 4-51 to the outside of the loop.
Fig. 4-52
4.49 4.50
Find VA and VB in the circuit of Fig. 4-52(b).
Ans:
VA 9; VB 5, both in V
Show that the three terminal circuits enclosed in the dashed boundaries of Fig. 4-52(a) and (b) are equivalent (i.e., in terms of their relation to other circuits). Hint: Use the linearity and superposition properties, along with the results of Problems 4.48 and 4.49.
Ampli ers and Operational Ampli er Circuits
5.1 AMPLIFIER MODEL An ampli er is a device which magni es signals. The heart of an ampli er is a source controlled by an input signal. A simpli ed model of a voltage ampli er is shown in Fig. 5-1(a). The input and output reference terminals are often connected together and form a common reference node. When the output terminal is open we have v2 kv1 , where k, the multiplying factor, is called the open circuit gain. Resistors Ri and Ro are the input and output resistances of the ampli er, respectively. For a better operation it is desired that Ri be high and Ro be low. In an ideal ampli er, Ri 1 and Ro 0 as in Fig. 5-1(b). Deviations from the above conditions can reduce the overall gain.
Fig. 5-1
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CHAP. 5]
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
EXAMPLE 5.1 A practical voltage source vs with an internal resistance Rs is connected to the input of a voltage ampli er with input resistance Ri as in Fig. 5-2. Find v2 =vs .
Fig. 5-2 The ampli er s input voltage, v1 , is obtained by dividing vs between Ri and Rs . v1 The output voltage v2 is v2 kv1 from which v2 Ri k vs Ri Rs The ampli er loads the voltage source. The open-loop gain is reduced by the factor Ri = Ri Rs . 1 kRi v Ri Rs s Ri v Ri Rs s
EXAMPLE 5.2 In Fig. 5-3 a practical voltage source vs with internal resistance Rs feeds a load Rl through an ampli er with input and output resistances Ri and Ro , respectively. Find v2 =vs .
Fig. 5-3 By voltage division, v1 Similarly, the output voltage is v2 kv1 Rl Ri Rl k v Rl Ro Ri Rs Rl Ro s or V2 Ri Rl k vs Ri Rs Rl Ro 2 Ri v Ri Rs s
Note that the open-loop gain is further reduced by an additional factor of Rl = Rl Ro , which also makes the output voltage dependent on the load.
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