zen barcode ssrs NONINVERTING CIRCUIT in Software

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NONINVERTING CIRCUIT
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In a noninverting circuit the input signal arrives at the noninverting terminal of the op amp. The inverting terminal is connected to the output through R2 and also to the ground through R1 (see Fig. 5-15).
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Fig. 5-15
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To nd the gain v2 =v1 , apply KCL at node B. op amp draws no current. v1 v1 v2 0 R1 R2
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Note that terminals A and B are both at v1 and the v2 R 1 2 v1 R1
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CHAP. 5]
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AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
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The gain v2 =v1 is positive and greater than or equal to one. The input resistance of the circuit is in nite as the op amp draws no current.
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EXAMPLE 5.11 Find v2 =v1 in the circuit shown in Fig. 5-16.
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Fig. 5-16 First nd vA by dividing v1 between the 10-k
and 5-k
resistors. vA From (12) we get     7 9 9 1 v1 1:5v1 v2 1 vA vA 2 2 2 3 and v2 1:5 v1 5 1 v v 5 10 1 3 1
Another Method
Find vB by dividing v2 between the 2-k
and 7-k
resistors and set vB vA . vB 2 2 1 v v v 2 7 2 9 2 3 1 and v2 1:5 v1
EXAMPLE 5.12
Determine vo in Fig. 5-17 in terms of v1 ; v2 ; v3 ; and the circuit elements.
Fig. 5-17 First, vA is found by applying KCL at node A. v1 vA v2 vA v3 vA 0 R R R From (12) and (13) we get     R 1 R 1 2 v1 v2 v3 vo 1 2 vA 3 R1 R1 14 or 1 vA v1 v2 v3 3 13
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
[CHAP. 5
VOLTAGE FOLLOWER
The op amp in the circuit of Fig. 5-18(a) provides a unity gain ampli er in which v2 v1 since v1 v , v2 v and v v . The output v2 follows the input v1 . By supplying il to Rl , the op amp eliminates the loading e ect of Rl on the voltage source. It therefore functions as a bu er.
EXAMPLE 5.13 (a) Find is ; vl ; v2 ; and il in Fig. 5-18(a). (b) Compare these results with those obtained when source and load are connected directly as in Fig. 5-18(b). (a) With the op amp present [Fig. 5-18(a)], we have is 0 v1 vs v2 v1 vs il vs =Rl
The voltage follower op amp does not draw any current from the signal source vs . Therefore, vs reaches the load with no reduction caused by the load current. The current in Rl is supplied by the op amp. (b) With the op amp removed [Fig. 5-18(b)], we have is il vs Rl Rs and v1 v2 Rl v Rl Rs s The load voltage v2
The current drawn by Rl goes through Rs and produces a drop in the voltage reaching it. depends on Rl .
Fig. 5-18
CHAP. 5]
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
DIFFERENTIAL AND DIFFERENCE AMPLIFIERS Such a signal may be
A signal source vf with no connection to ground is called a oating source. ampli ed by the circuit of Fig. 5-19.
Fig. 5-19
Here the two input terminals A and B of the op amp are at the same voltage. KVL around the input loop we get vf 2R1 i or i vf =2R1
Therefore, by writing
The op amp inputs do not draw any current and so current i also ows through the R2 resistors. Applying KVL around the op amp, we have vo R2 i R2 i 0 vo 2R2 i 2R2 vf =2R1 R2 =R1 vf 15
In the special case when two voltage sources v1 and v2 with a common ground are connected to the inverting and noninverting inputs of the circuit, respectively (see Fig. 5-20), we have vf v1 v2 and vo R2 =R1 v2 v1
EXAMPLE 5.14 Find vo as a function of v1 and v2 in the circuit of Fig. 5-20. Applying KCL at nodes A and B,
16
Fig. 5-20 vA v2 vA 0 R3 R4 vB v1 vB vo 0 R1 R2
Node A: Node B:
Set vA vB and eliminate them from the preceding KCL equations to get
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
[CHAP. 5
vo
R4 R1 R2 R v 2v R1 R3 R4 2 R1 1
17
When R3 R1 and R2 R4 , (17) is reduced to (16).
CIRCUITS CONTAINING SEVERAL OP AMPS
The analysis and results developed for single op amp circuits can be applied to circuits containing several ideal op amps in cascade or nested loops because there is no loading e ect.
EXAMPLE 5.15 Find v1 and v2 in Fig. 5-21.
Fig. 5-21 The rst op amp is an inverting circuit. v1 3=1 0:6 1:8 V The second op amp is a summing circuit. v2 2=1 0:5 2=2 1:8 2:8 V EXAMPLE 5.16 Let Rs 1 k
in the circuit of Fig. 5-22, nd v1 ; v2 ; vo ; is ; i1 ; and if as functions of vs for (a) Rf 1 and (b) Rf 40 k
Fig. 5-22 (a) Rf 1. The two inverting op amps are cascaded, with v 0. v1 From the inverting ampli ers we get 5 5 v v 5 1 s 6 s By voltage division in the input loop we have 18
CHAP. 5]
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