hw05solution

# hw05solution - P 5.2 = 0 therefore 1 = 9211 — 81[a...

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Unformatted text preview: P 5.2 = 0, therefore 1),, = 9211) — 81),, [a] var—1.5 V, vb=0 V, voz—12 V [b] 1),, 2 3.0 V, vb = 0 V, 110 = ~18 V (sat) [c] va=1.0 V, 711,22 V, 110210 V [d] 2),, z: 4.0 V, vb = 2 V, 11,, = —14 V [e] 7),, = 6.0 V, vb 2 8 V, 110 = 18 V (sat) [f] If vb = 4.5 V, v, = 40.5 — 8v, 2 :|:18 2.8125 _<_ 1),, 3 7.3125 V P 5.4 Since the current into the inverting input terminal of an ideal op—amp is zero, the voltage across the 3.3 M9 resistor is (3.3 x 106)(2.5 x 10—6) or 8.25 V. Therefore the voltmeter reads 8.25 V. P 5.9 [a] The gain of an inverting ampliﬁer is the ratio of the feedback resistor to the input resistor. If the gain of the inverting ampliﬁer is to be 4, the feedback resistor must be 4 times as large as the input resistor. There are many possible designs that use only 10 k9 resistors. We present two here. Use a single 10 k9 resistor as the input resistor, and use four 10 k9 resistors in series as the feedback resistor to give a total of 40 k9. IOKQ lﬁkﬂ Alternately, Use a single 10 k9 resistor as the feedback resistor and use four 10 k9 resistors in parallel as the input resistor to give a total of 2.5 k9. 10m 10m 10m IOkQ W00 101:0 E “8 ‘Vcc v0 \$ G [b] To amplify a 2.5 V signal without saturating the op amp, the power supply voltages must be greater than or equal to the product of the input voltage and the ampliﬁer gain. Thus, the power supplies should have a magnitude of (2.5)(4) = 10 V. P 5.12 [a] This circuit is an example of an inverting summing ampliﬁer. 180 180 180 [b] ’00 2 —Eva — 36—111, — 736416 2 ~45 — 9 + 7.5 = —6 V [c] '00 = ——l3.5 — 3'06 :2 i9 Dc 2 —7.5 V when '00 = 9 V; Dc: 1.5 V when '00 = ~9 V —7.5 ‘ngc g 1.5 V 95(100 + 5) + 100(5 + 95) _ (5)(95) — 5(100) __ _ [b] Acm _ Mﬁ—(QG + 95) _ 0.05 [c] CMRR = 1 19%| = 399.5 0.05 ...
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hw05solution - P 5.2 = 0 therefore 1 = 9211 — 81[a...

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