zen barcode ssrs Fig. 5-37 At nodes B and A, vB vA 0. Applying KCL at node C, we get from which vC 3 V in Software

Maker QR Code in Software Fig. 5-37 At nodes B and A, vB vA 0. Applying KCL at node C, we get from which vC 3 V

Fig. 5-37 At nodes B and A, vB vA 0. Applying KCL at node C, we get from which vC 3 V
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vC 9 =4 vC =6 vC =3 0 Then i1 9 vC =4 1:5 A and
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Rin v1 =i1 9=1:5 6
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From the inverting ampli er circuit we have v2 5=3 vC 5 V and i2 5=10 0:5 A
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Find v2 in Problem 5.11 by replacing the circuit to the left of nodes A-B in Fig. 5-37 by its Thevenin equivalent.
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RTh 3 Then v2 5=5:4 5:4 5 V. 6 4 5:4
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6 4 and vTh 6 9 5:4 V 4 6
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Find vC , i1 , v2 , and Rin , the input resistance seen by the 21-V source in Fig. 5-38.
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From the inverting ampli er we get v2 5=3 vC Note that vB vA 0 and so KCL at node C results in 39
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AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
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[CHAP. 5
Fig. 5-38 vC 21 vC vC vC v2 0 3 6 3 8 Substituting vC 3=5 v2 from (39) into (40) we get v2 10 V. Then vC 6 V i1 21 vC =3000 0:005 A 5 mA Rin 21=i1 21=0:005 4200
4:2 k
40
In the circuit of Fig. 5-38 change the 21-V source by a factor of k. Show that vC , i1 , v2 in Problem 5.13 are changed by the same factor but Rin remains unchanged.
Let vs 21k (V) represent the new voltage source. From the inverting ampli er we have [see (39)] v2 5=3 vC Apply KCL at node C to obtain [see (40)] vC vs vC vC vC v2 0 3 6 3 8 Solving for vC and v2 , we have vC 6=21 vs 6k V and v2 10=21 vs 10k V i1 vs vC =3000 21 6 k=3000 0:005k A Rin vs =i1 21k=0:005k 4200
These results are expected since the circuit is linear.
Find v2 and vC in Problem 5.13 by replacing the circuit to the left of node C in Fig. 5-38 (including the 21-V battery and the 3-k
and 6-k
resistors) by its Thevenin equivalent.
We rst compute the Thevenin equivalent: RTh 6 3 2 k
6 3 and vTh 6 21 14 V 3 6
Replace the circuit to the left of node C by the above vTh and RTh and then apply KCL at C: vC 14 vC vC v2 0 2 3 8 41
For the inverting ampli er we have v2 5=3 vC or vC 0:6 v2 , which results, after substitution in (41), in v2 10 V and vC 6 V.
CHAP. 5]
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
(a) Find the Thevenin equivalent of the circuit to the left of nodes A-B in Fig. 5-39(a) and then nd v2 for Rl 1 k
, 10 k
, and 1. (b) Repeat for Fig. 5-39(c) and compare with part (a).
(a) The Thevenin equivalent of the circuit in Fig. 5-39(a) is shown in Fig. 5-39(b).
Fig. 5-39 vTh 6 15 10 V 6 3 and RTh 3 6 2 k
By dividing vTh between RTh and Rl we get v2 For Rl For Rl For Rl The output 1 k
, 10 k
, 1 v2 depends on Rl 10 Rl 2
v2 3:33 V v2 8:33 V v2 10 V Rl . The operation of the voltage divider is also a ected by Rl . Here we have vTh 10 V and RTh 0
(b) The Thevenin equivalent of the circuit in Fig. 5-39(c) is shown in Fig. 5-12(d).
and v2 vTh 10 V for all values of Rl , that is, the output v2 depends on R1 , R2 , and vs only and is independent of Rl .
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
[CHAP. 5
Find v2 as a function of i1 in the circuit of Fig. 5-40(a).
Fig. 5-40 Current i1 goes through resistor R producing a voltage Ri1 across it from right to left. Since the inverting terminal B is zero potential, the preceding voltage appears at the output as v2 Ri1 [see Fig. 540(b)]. Therefore, the op amp converts the current i1 to a voltage v2 with a gain of jv2 =i1 j R. The current source i1 delivers no power as the voltage vAB across it is zero.
A transducer generates a weak current i1 which feeds a load Rl and produces a voltage v1 across it. It is desired that v1 follow the signal with a constant gain of 108 regardless of the value of Rl . Design a current-to-voltage converter to accomplish this task.
The transducer should feed Rl indirectly through an op amp. The following designs produce v1 108 i1 independently of Rl . Design 1: Choose R 100 M
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