Tài liệu Bài giải mạch P5 doc

Chapter 5, Solution 1. (a) R in = 1.5 MΩ (b) R out = 60 Ω (c) A = 8x10 4 Therefore A dB = 20 log 8x10 4 = 98.0 dB Chapter 5, Solution 2. v 0 = Av d = A(v 2 - v 1 ) = 10 5 (20-10) x 10 -6 = 0.1V Chapter 5, Solution 3. v 0 = Av d = A(v 2 - v 1 ) = 2 x 10 5 (30 + 20) x 10 -6 = 10V Chapter 5, Solution 4. v 0 = Av d = A(v 2 - v 1 ) v 2 - v 1 = V20 10x2 4 A v 5 0 µ−= − = If v 1 and v 2 are in mV, then v 2 - v 1 = -20 mV = 0.02 1 - v 1 = -0.02 v 1 = 1.02 mV Chapter 5, Solution 5. + v 0 - - v d + R 0 R in I v i - + Av d + - -v i + Av d + (R i - R 0 ) I = 0 (1) But v d = R i I, -v i + (R i + R 0 + R i A) I = 0 v d = i0 ii R)A1(R Rv ++ (2) -Av d - R 0 I + v 0 = 0 v 0 = Av d + R 0 I = (R 0 + R i A)I = i0 ii0 R)A1(R v)ARR( ++ + 4 5 54 i0 i0 i 0 10 )101(100 10x10100 R)A1(R ARR v v ⋅ ++ + = ++ + = ≅ () =⋅ + 4 5 9 10 101 10 = 001,100 000,100 0.9999990 Chapter 5, Solution 6. - v d + + v o - R 0 R in I v i + - Av d + - (R 0 + R i )R + v i + Av d = 0 But v d = R i I, v i + (R 0 + R i + R i A)I = 0 I = i0 i R)A1(R v ++ − (1) -Av d - R 0 I + v o = 0 v o = Av d + R 0 I = (R 0 + R i A)I Substituting for I in (1), v 0 =       ++ + i0 i0 R)A1(R ARR   − v i = () () 65 356 10x2x10x2150 1010x2x10x250 ++ ⋅+ − − ≅ mV 10x2x001,200 10x2x000,200 6 6 − v 0 = -0.999995 mV Chapter 5, Solution 7. 100 k Ω 1 2 10 kΩ - + + V d - + V out - R out = 100 Ω R in AV d + - V S At node 1, (V S – V 1 )/10 k = [V 1 /100 k] + [(V 1 – V 0 )/100 k] 10 V S – 10 V 1 = V 1 + V 1 – V 0 which leads to V 1 = (10V S + V 0 )/12 At node 2, (V 1 – V 0 )/100 k = (V 0 – AV d )/100 But V d = V 1 and A = 100,000, V 1 – V 0 = 1000 (V 0 – 100,000V 1 ) 0= 1001V 0 – 100,000,001[(10V S + V 0 )/12] 0 = -83,333,334.17 V S - 8,332,333.42 V 0 which gives us (V 0 / V S ) = -10 (for all practical purposes) If V S = 1 mV, then V 0 = -10 mV Since V 0 = A V d = 100,000 V d , then V d = (V 0 /10 5 ) V = -100 nV Chapter 5, Solution 8. (a) If v a and v b are the voltages at the inverting and noninverting terminals of the op amp. v a = v b = 0 1mA = k2 v0 0 − v 0 = -2V (b) 10 kΩ 2V + - + v a - 10 kΩ i a + v o - + v o - v a v b i a 2 k Ω 2V - + 1V - + - + (b) (a) Since v a = v b = 1V and i a = 0, no current flows through the 10 kΩ resistor. From Fig. (b), -v a + 2 + v 0 = 0 v a = v a - 2 = 1 - 2 = -1V Chapter 5, Solution 9. (a) Let v a and v b be respectively the voltages at the inverting and noninverting terminals of the op amp v a = v b = 4V At the inverting terminal, 1mA = k2 0 v4 − v 0 = 2V Since v a = v b = 3V, -v b + 1 + v o = 0 v o = v b - 1 = 2V + v b - + v o - + - (b) 1V Chapter 5, Solution 10. Since no current enters the op amp, the voltage at the input of the op amp is v s . Hence v s = v o 2 v 1010 10 o =       + s o v v = 2 Chapter 5, Solution 11. 8 k Ω v b = V2)3( 510 10 = + i o b a + − 5 k Ω 2 k Ω 4 k Ω + v o − 10 k Ω − + 3 V At node a, 8 vv 2 v3 oaa − = − 12 = 5v a – v o But v a = v b = 2V, 12 = 10 – v o v o = -2V –i o = mA1 4 2 8 22 4 v0 8 vv ooa =+ + = − + − i o = -1mA Chapter 5, Solution 12. 4 k Ω b a + − 2 k Ω 1 k Ω + v o − 4 k Ω − + 1.2V At node b, v b = ooo v 3 2 v 3 2 v 24 4 == + At node a, 4 vv 1 v2.1 oaa − = − , but v a = v b = o v 3 2 4.8 - 4 x ooo vv 3 2 v 3 2 −= v o = V0570.2 7 8.4x3 = v a = v b = 7 6.9 v 3 2 o = i s = 7 2.1 1 v2. a − = 1 − p = v s i s = 1.2 =     − 7 2.1   -205.7 mW Chapter 5, Solution 13. By voltage division, i 1 i 2 90 k Ω 10 k Ω b a + − 100 k Ω 4 k Ω 50 k Ω + − i o + v o − 1 V v a = V9.0)1( 100 = 90 v b = 3 v v 150 o o = 50 But v a = v b 9.0 3 v 0 = v o = 2.7V i o = i 1 + i 2 = =+ k150 v k10 v oo 0.27mA + 0.018mA = 288 µA Chapter 5, Solution 14. Transform the current source as shown below. At node 1, 10 vv 20 vv 5 v10 o1 211 − + − = − But v 2 = 0. Hence 40 - 4v 1 = v 1 + 2v 1 - 2v o 40 = 7v 1 - 2v o (1) 20 k Ω v o 10 k Ω + − v 1 − + v 2 5 k Ω 10 k Ω + v o − 10V At node 2, 0v, 10 vv 20 vv 2 o2 21 = − = − or v 1 = -2v o (2) From (1) and (2), 40 = -14v o - 2v o v o = -2.5V Chapter 5, Solution 15 (a) Let v 1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives 332 1 3 1 2 1 11 R v RR v R vv R v i oo s −         += − += (1) At the inverting terminal, 11 1 1 0 Riv R v i ss −=→ − = (2) Combining (1) and (2) leads to         ++−=→−=         ++ 2 31 31 33 1 2 1 1 R RR RR i v R v R R R R i s oo s (b) For this case, Ω=Ω       ++−= k 92- k 25 4020 4020 x i v s o Chapter 5, Solution 16 10k Ω i x 5k Ω v a i y - v b + v o + 2k Ω 0.5V - 8k Ω Let currents be in mA and resistances be in k Ω . At node a, oa oaa vv vvv −=→ − = − 31 105 5.0 (1) But aooba vvvvv 8 10 28 8 =→ + == (2) Substituting (2) into (1) gives 14 8 8 10 31 =→−= aaa vvv Thus, A 28.14mA 70/1 5 5.0 µ −=−= − = a x v i A 85.71mA 14 8 4 6.0 ) 8 10 (6.0)(6.0 102 µ ==−=−= − + − = xvvvv vvvv i aaao aobo y Chapter 5, Solution 17. (a) G = =−=−= 5 12 R R v v 1 2 i o -2.4 (b) 5 80 v v i o −= = -16 (c) =−= 5 2000 v v i o -400 Chapter 5, Solution 18. Converting the voltage source to current source and back to a voltage source, we have the circuit shown below: 3 20 2010 = kΩ 1 M Ω 3 v2 3 20 50 1000 v i o ⋅ + −= =−= 17 200 v 1 o v -11.764 Chapter 5, Solution 19. We convert the current source and back to a voltage source. 3 4 42 = 5 k Ω v o 0V ( 4/3 ) k Ω 10 k Ω − + 4 k Ω + − (2/3)V ( 20/3 ) k Ω + v o − − + 50 k Ω + − 2v i /3