0 06 16 V For a negative input there is no path for current through R so v O 0

0 06 16 v for a negative input there is no path for

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1.0 + 0.6 = 1.6 V For a negative input, there is no path for current through R, so v O = 0 V . The op - amp sees a -1V input so the output will limit at the negative power supply : v O = 10 V . The diode has a 10 - V reverse bias across it, so V Z > 10 V . Page 587 i ( ) v S = 2 V : Diode D 1 conducts, and D 2 is off. The negative input is a virtual ground. v 1 = v D 2 = 0.6 V . The current in R is 0, so v O = 0 V . v S = 2 V : Diode D 2 conducts, and D 1 is off. The negative input is a virtual ground. v O = R 2 R 1 v S = 68 k Ω 22 k Ω 2 V ( ) = + 6.18 V v 1 = v O + v D 1 = 6.78 V . v S = 15 V 3.09 = 4.85 V v 1 = v O + v D 1 = 6.78 V . When v O = 15 V , v D 2 = -15.6 V , so V Z = 15.6 V . ii ( ) v O = 20 k Ω 20 k Ω 10.2 k Ω 3.24 k Ω 2 V π = 2.00 V Page 589 V = R 1 R 1 + R 2 V EE = 1 k Ω 1 k Ω + 9.1 k Ω 10 V = 0.990 V V + = 1 k Ω 1 k Ω + 9.1 k Ω 10 V = + 0.990 V V n = 0.990 V − − 0.990 V ( ) = 1.98 V Page 591 T = 2 10 k Ω ( ) 0.001 μ F ( ) ln 1 + 0.5 1 0.5 = 21.97 μ s f = 1 T = 45.5 kHz
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11 Page 594 β = 22 k Ω 22 k Ω + 18 k Ω = 0.550 T = 11 k Ω ( ) 0.002 μ F ( ) ln 1 + 0.7 5 1 0.550 = 20.4 μ s T r = 11 k Ω ( ) 0.002 μ F ( ) ln 1 + 0.55 5 V 5 V 1 0.7 5 = 13.0 μ s
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12 CHAPTER 12 Page 612 i ( ) A ideal = 1 β = 100 A v = A 1 + A β = 10 5 1 + 10 5 0.01 ( ) = 99.90 v o = 99.9 0.1 V ( ) = 9.99 V v id = v o A = 9.99 V 10 5 = 99.9 μ V ii ( ) Values taken from OP - 27 specification sheet ( or ) ( iii ) Values taken from OP - 27 specification sheet Page 613 A v = R 2 R 1 A β 1 + A β FGE = R 2 R 1 R 2 R 1 A β 1 + A β R 2 R 1 = 1 A β 1 + A β = 1 1 + A β Page 614 FGE = 1 1 + A β = 1 1 + 10 4 1 k Ω 1 k Ω + 39 k Ω = 3.98 x 10 3 or 0.398 % FGE 1 A β = 0.40 % Page 615 Values taken from OP - 77 specification sheet ( or ) Page 616 A R o β R out = 50 Ω 0.025 ( ) 0.1 Ω ( ) = 20,000 Page 617 i ( ) A v = A 1 + A β = 10 4 1 + 10 4 0.025 ( ) = + 39.8 ii ( ) A v max = 39 k Ω + 1 k Ω ( ) 1.05 ( ) 1 k Ω 0.95 ( ) = 44.2 A v max = 39 k Ω + 1 k Ω ( ) 0.95 ( ) 1 k Ω 1.05 ( ) = 36.2 GE = 44.2 40.0 = 4.20 FGE = 4.20 40 = 10.5 % GE = 36.2 40.0 = 3.80 F GE = 3.80 40 = 9.5 %
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13 Page 618 A R o β R out = 200 Ω 0.01 ( ) 0.1 Ω ( ) = 200,000 A dB = 20log 2 x 10 5 ( ) = 106 dB Page 619 Values taken from op- amp specification sheets ( or ) Page 620 R in = R id 1 + A β ( ) = 1 M Ω 1 + 10 4 10 k Ω 10 k Ω + 390 k Ω = 251 M Ω i = v s R in = 1 V 251 M Ω = 3.98 nA i 1 = β v o R 1 = A β 1 + A β v s R 1 1 V 10 k Ω = 100 μ A i 1 >> i More exactly, i 1 = β v o R 1 = A β 1 + A β v s R 1 = 10 4 0.025 ( ) 1 + 10 4 0.025 ( ) 1 V 10 k Ω = 99.6 μ A Page 621 R in = R 1 + R id R 2 1 + A = 1 k Ω + 1 M Ω 100 k Ω 1 + 10 5 = 1001 Ω R in ideal = R 1 = 1000 Ω 1 Ω or 0.1 % Page 622 Values taken from op- amp specification sheets ( or ) Page 626 v o = A v id + v ic CMRR v o min = A v id + v ic CMRR = 2500 0.002 5.000 10 4 = 3.750 V v o max = A v id + v ic CMRR = 2500 0.002 + 5.000 10 4 = 6.250 V Page 627 A v = A 1 + 1 2 CMRR 1 + A 1 1 2 CMRR
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