zen barcode ssrs in Fig. 5-40. not readily available. However, a resistor of such a large magnitude is expensive and in Software

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Design 2: The conversion gain of 108 V=A is also obtained in the circuit of Fig. 5-41. The rst op amp with R 106 converts i1 to v1 106 i1 . The second ampli er with a gain of 100 (e.g., R1 1 k
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and R2 100 k
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) ampli es v1 to v2 100v1 108 i1 . The circuit requires two op amps and three resistors (1 M
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Fig. 5-41 Design 3: See Fig. 5-42 and Problem 5.19.
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Determine the resistor values which would produce a current-to-voltage conversion gain of v2 =i1 108 V=A in the circuit of Fig. 5-42.
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CHAP. 5]
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AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
Fig. 5-42 Apply KCL at node C. Note that vB vA 0. Thus,
vC vC vC v2 0 R R1 R2 Substituting vC Ri1 and solving for v2 we get v2 Req i1 where Req   R2 R2 R 1 R1 R
For a conversion gain of v2 =i1 Req 108 V=A 100 M
, we need to nd resistor values to satisfy the following equation:   R R R 1 2 2 108
R1 R One solution is to choose R 1 M
, R1 1 k
, and R2 99 k
. The design of Fig. 5-42 uses a single op amp and three resistors which are not expensive and are readily available.
Find i2 as a function of v1 in the circuit of Fig. 5-43.
Fig. 5-43 We have vB vA 0 i1 v1 =R1 i2 i1 v1 =R1 The voltage-to-current conversion
The op amp converts the voltage source to a oating current source. ratio is R1 and is independent of R2 .
A practical current source (is in parallel with internal resistance Rs ) directly feeds a load Rl as in Fig. 5-44(a). (a) Find load current il . (b) Place an op amp between the source and the load as in Fig. 5-44(b). Find il and compare with part (a).
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
[CHAP. 5
Fig. 5-44 (a) In the direct connection, Fig. 5-44(a), il is Rs = Rs Rl , which varies with Rl . (b) In Fig. 5-44(b), the op amp forces vB to zero causing the current in Rs to become zero. Therefore, il is which is now independent of Rl . The op amp circuit converts the practical current source to an ideal current source. See Figure 5-44(c).
Find vo in the circuit of Fig. 5-45.
Fig. 5-45 The rst op amp is a unity gain inverter with v3 v2 . The second op amp is a summing circuit with a gain of R2 =R1 for both inputs v1 and v3 . The output is vo R2 R v v3 2 v2 v1 R1 1 R1
CHAP. 5]
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
The circuit is a di erence ampli er.
Find vo in the circuit of Fig. 5-46.
Fig. 5-46 Apply KCL at node B. Note that vB vA v2 . Thus,
v2 v1 v2 vo 0 R1 R2 Solving for vo , we get vo v2 R2 =R1 v2 v1 .
Find vo in the circuit of Fig. 5-47.
Fig. 5-47 The left part of the circuit has a gain of 1 R1 =R2 . Therefore, v3 1 R1 =R2 v1 . Using results of Problem 5.23 and substituting for v3 results in       R R R R R vo v2 2 v2 v3 1 2 v2 2 1 1 v1 1 2 v2 v1 R1 R1 R1 R2 R1
In Fig. 5-48 choose resistors for a di erential gain of 106 so that vo 106 v2 v1 .
The two frontal op amps are voltage followers. vA v1 From (16), Sec. 5.9, we have vo R2 R v vA 2 v2 v1 R1 B R1 and vB v2
To obtain the required di erential gain of R2 =R1 106 , choose R1 100
and R2 100 M
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
[CHAP. 5
Fig. 5-48 The circuit of Fig. 5-48 can have the same gain as that of Fig. 5-45, but its input resistance is in nite. However, it employs two small and large resistors which are rather out of ordinary range.
Resistors having high magnitude and accuracy are expensive. Show that in the circuit of Fig. 549 we can choose resistors of ordinary range so that vo 106 v2 v1 .
Fig. 5-49 The two frontal op amps convey the input voltages v1 and v2 to the terminals of RG , creating an upward current i v2 v1 =RG in the resistor. The current also goes through the two R3 resistors, creating voltage drops iR3 across them. Therefore, vA v1 R 3 i v1 R3 R v v1 vB v2 R3 i v2 3 v2 v1 RG 2 RG   2R vB vA 1 3 v2 v1 RG   R R 2R vo 2 vB vA 2 1 3 v2 v1 R1 R1 RG   vo R 2R 2 1 3 106 v2 v1 R1 RG Choose R1 RG 1 k
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