Bài giải phần giải mạch P5

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Bài giải phần giải mạch P5

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Chapter 5, Solution 1. (a) (b) (c) Rin = 1.5 MΩ Rout = 60 Ω A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 0.1V Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V Chapter 5, Solution 4. v0 = Avd = A(v2 - v1) v −4 v2 - v1 = 0 = = −20µV A 2 x10 5 If v1 and v2 are in mV, then v2...

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  1. Chapter 5, Solution 1. (a) Rin = 1.5 MΩ (b) Rout = 60 Ω (c) A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 0.1V Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V Chapter 5, Solution 4. v0 = Avd = A(v2 - v1) v −4 v2 - v1 = 0 = = −20µV A 2 x10 5 If v1 and v2 are in mV, then v2 - v1 = -20 mV = 0.02 1 - v1 = -0.02 v1 = 1.02 mV Chapter 5, Solution 5. I R0 - vd Rin + + + Avd v0 - vi + - -
  2. -vi + Avd + (Ri - R0) I = 0 (1) But vd = RiI, -vi + (Ri + R0 + RiA) I = 0 vi R i vd = (2) R 0 + (1 + A)R i -Avd - R0I + v0 = 0 (R 0 + R i A) v i v0 = Avd + R0I = (R0 + RiA)I = R 0 + (1 + A)R i v0 R 0 + RiA 100 + 10 4 x10 5 = = ⋅ 10 4 v i R 0 + (1 + A)R i 100 + (1 + 10 ) 5 10 9 100,000 ≅ ⋅ 10 4 = = 0.9999990 ( 1 + 10 5 ) 100,001 Chapter 5, Solution 6. vi + - R0 - I vd Rin + + + Avd - vo -
  3. (R0 + Ri)R + vi + Avd = 0 But vd = RiI, vi + (R0 + Ri + RiA)I = 0 − vi I= (1) R 0 + (1 + A)R i -Avd - R0I + vo = 0 vo = Avd + R0I = (R0 + RiA)I Substituting for I in (1),  R 0 + R iA  v0 = −   R + (1 + A)R  vi   0 i  = − ( ) 50 + 2 x10 6 x 2 x10 5 ⋅ 10 −3 ( ) 50 + 1 + 2x10 5 x 2 x10 6 − 200,000 x 2 x10 6 ≅ mV 200,001x 2 x10 6 v0 = -0.999995 mV Chapter 5, Solution 7. 100 kΩ Rout = 100 Ω 10 kΩ 1 2 + VS + + - Vd Rin + AVd - Vout - -
  4. At node 1, (VS – V1)/10 k = [V1/100 k] + [(V1 – V0)/100 k] 10 VS – 10 V1 = V1 + V1 – V0 which leads to V1 = (10VS + V0)/12 At node 2, (V1 – V0)/100 k = (V0 – AVd)/100 But Vd = V1 and A = 100,000, V1 – V0 = 1000 (V0 – 100,000V1) 0= 1001V0 – 100,000,001[(10VS + V0)/12] 0 = -83,333,334.17 VS - 8,332,333.42 V0 which gives us (V0/ VS) = -10 (for all practical purposes) If VS = 1 mV, then V0 = -10 mV Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = -100 nV Chapter 5, Solution 8. (a) If va and vb are the voltages at the inverting and noninverting terminals of the op amp. va = v b = 0 0 − v0 1mA = v0 = -2V 2k (b) 10 kΩ - 2V + ia va - 2V 10 kΩ + +- vb + + + 2 kΩ vo va 1V + ia vo - - - - (a) (b)
  5. Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b), -va + 2 + v0 = 0 va = va - 2 = 1 - 2 = -1V Chapter 5, Solution 9. (a) Let va and vb be respectively the voltages at the inverting and noninverting terminals of the op amp va = vb = 4V At the inverting terminal, 4 − v0 1mA = v0 = 2V 2k (b) 1V +- + + vb vo - - Since va = vb = 3V, -vb + 1 + vo = 0 vo = vb - 1 = 2V Chapter 5, Solution 10. Since no current enters the op amp, the voltage at the input of the op amp is vs. Hence  10  v o vo vs = v o  = =2  10 + 10  2 vs
  6. Chapter 5, Solution 11. 8 kΩ 2 kΩ a io − + + 5 kΩ b + 3V 10 kΩ 4 kΩ vo − − 10 vb = (3) = 2V 10 + 5 At node a, 3 − va va − vo = 12 = 5va – vo 2 8 But va = vb = 2V, 12 = 10 – vo vo = -2V va − vo 0 − vo 2 + 2 2 –io = + = + = 1mA 8 4 8 4 i o = -1mA Chapter 5, Solution 12. 4 kΩ 1 kΩ a − + 2 kΩ b + + 1.2V − vo 4 kΩ −
  7. 4 2 2 At node b, vb = vo = vo = vo 4+2 3 3 1 .2 − v a v a − v o 2 At node a, = , but va = vb = v o 1 4 3 2 2 3x 4.8 4.8 - 4 x vo = vo − vo vo = = 2.0570V 3 3 7 2 9.6 va = vb = vo = 3 7 1 .2 − v a − 1 .2 is = = 1 7  − 1.2  p = vsis = 1.2   = -205.7 mW  7  Chapter 5, Solution 13. 10 kΩ a + io − 100 kΩ i2 i1 + b 4 kΩ vo + 90 kΩ 1V − 50 kΩ − By voltage division, 90 va = (1) = 0.9V 100 50 v vb = vo = o 150 3 v0 But va = vb = 0 .9 vo = 2.7V 3 v v io = i1 + i2 = o + o = 0.27mA + 0.018mA = 288 µA 10k 150k
  8. Chapter 5, Solution 14. Transform the current source as shown below. At node 1, 10 − v1 v1 − v 2 v1 − v o = + 5 20 10 10 kΩ vo 10 kΩ 5 kΩ 20 kΩ v1 v2 − + + + 10V − vo − But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo 40 = 7v1 - 2vo (1) v1 − v 2 v 2 − v o At node 2, = , v 2 = 0 or v1 = -2vo (2) 20 10 From (1) and (2), 40 = -14vo - 2vo vo = -2.5V Chapter 5, Solution 15 (a) Let v1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives v v −v  1 1  vo i s = 1 + 1 o = v1  R + − (1) R2 R3  2 R3  R3  At the inverting terminal, 0 − v1 is = → v1 = −i s R1 (2) R1 Combining (1) and (2) leads to  R R  v vo  RR  i s 1 + 1 + 1  = − o → = − R1 + R3 + 1 3    R  2  R3  R3 is  R2  (b) For this case, vo  20 x 40  = − 20 + 40 +  kΩ = - 92 kΩ is  25 
  9. Chapter 5, Solution 16 10k Ω ix 5k Ω va iy - vb + vo + 2k Ω 0.5V - 8k Ω Let currents be in mA and resistances be in k Ω . At node a, 0 .5 − v a v a − v o = → 1 = 3v a − vo (1) 5 10 But 8 10 v a = vb = vo  vo = v a → (2) 8+2 8 Substituting (2) into (1) gives 10 8 1 = 3v a − v a  → v a = 8 14 Thus, 0 .5 − v a ix = = −1 / 70 mA = − 14.28 µA 5 v − vb v o − v a 10 0 .6 8 iy = o + = 0 .6 ( v o − v a ) = 0 .6 ( v a − v a ) = x mA = 85.71 µA 2 10 8 4 14 Chapter 5, Solution 17. vo R 12 (a) G= = − 2 = − = -2.4 vi R1 5 vo 80 (b) =− = -16 vi 5 vo 2000 (c) =− = -400 vi 5
  10. Chapter 5, Solution 18. Converting the voltage source to current source and back to a voltage source, we have the circuit shown below: 20 10 20 = kΩ 3 1 MΩ (20/3) kΩ 50 kΩ − + + + 2vi/3 − vo − 1000 2v i vo 200 vo = − ⋅ =− = -11.764 20 3 v1 17 50 + 3 Chapter 5, Solution 19. We convert the current source and back to a voltage source. 4 24= 3 (4/3) kΩ 4 kΩ 10 kΩ 0V − vo + + (2/3)V − 5 kΩ
  11. 10k  2  vo = −   = -1.25V  4 3  4x k  3 v v −0 io = o + o = -0.375mA 5k 10k Chapter 5, Solution 20. 8 kΩ 2 kΩ 4 kΩ 4 kΩ a b − + + + 9V + vo − vs − − At node a, 9 − va va − vo va − vb = + 18 = 5va – vo - 2vb (1) 4 8 4 At node b, va − vb vb − vo = va = 3vb - 2vo (2) 4 2 But vb = vs = 0; (2) becomes va = –2vo and (1) becomes -18 = -10vo – vo vo = -18/(11) = -1.6364V
  12. Chapter 5, Solution 21. Eqs. (1) and (2) remain the same. When vb = vs = 3V, eq. (2) becomes va = 3 x 3 - 2v0 = 9 - 2vo Substituting this into (1), 18 = 5 (9-2vo) – vo – 6 leads to vo = 21/(11) = 1.909V Chapter 5, Solution 22. Av = -Rf/Ri = -15. If Ri = 10kΩ, then Rf = 150 kΩ. Chapter 5, Solution 23 At the inverting terminal, v=0 so that KCL gives vs − 0 0 0 − vo vo Rf = + → =− R1 R2 Rf vs R1 Chapter 5, Solution 24 v1 Rf R1 R2 - vs + - + + R4 R3 vo v2 - We notice that v1 = v2. Applying KCL at node 1 gives v1 (v1 − v s ) v1 − vo  1  + + =0  →  + 1 + 1 v1 − v s = vo (1) R1 R2 Rf R R Rf  R2 R f  1 2 
  13. Applying KCL at node 2 gives v1 v1 − v s R3 + =0 → v1 = vs (2) R3 R4 R3 + R4 Substituting (2) into (1) yields  R R R  R3  1  vo = R f  3 + 3 − 4   − vs   R1 R f R2  R3 + R4  R2     i.e.  R R R  R3  1  k = R f  3 + 3 − 4   −    R1 R f R2  R3 + R4  R2     Chapter 5, Solution 25. vo = 2 V + − + + va vo -va + 3 + vo = 0 which leads to va = vo + 3 = 5 V. Chapter 5, Solution 26 + vb - io + + 0.4V 5k Ω - 2k Ω vo 8k Ω - 8 vb = 0.4 = vo = 0.8vo  → vo = 0.4 / 0.8 = 0.5 V 8+ 2 Hence, v o 0 .5 io = = = 0.1 mA 5k 5k
  14. Chapter 5, Solution 27. (a) Let va be the voltage at the noninverting terminal. va = 2/(8+2) vi = 0.2vi  1000  v 0 = 1 +  v a = 10.2v i  20  G = v0/(vi) = 10.2 (b) vi = v0/(G) = 15/(10.2) cos 120πt = 1.471 cos 120πt V Chapter 5, Solution 28. − + + − 0 − v1 v1 − v o At node 1, = 10k 50k But v1 = 0.4V, -5v1 = v1 – vo, leads to vo = 6v1 = 2.4V Alternatively, viewed as a noninverting amplifier, vo = (1 + (50/10)) (0.4V) = 2.4V io = vo/(20k) = 2.4/(20k) = 120 µA
  15. Chapter 5, Solution 29 R1 va + vb - + + vi R2 R2 vo - R1 - R2 R1 va = vi , vb = vo R1 + R2 R1 + R2 R2 R1 But v a = vb  → vi = vo R1 + R2 R1 + R2 Or v o R2 = vi R1 Chapter 5, Solution 30. The output of the voltage becomes vo = vi = 12 30 20 = 12kΩ By voltage division, 12 vx = (1.2) = 0.2V 12 + 60 vx 0 .2 ix = = = 10µA 20k 20k v 2 0.04 p= x = = 2µW R 20k
  16. Chapter 5, Solution 31. After converting the current source to a voltage source, the circuit is as shown below: 12 kΩ 3 kΩ 6 kΩ v 1 o v1 + vo 2 − + 12 V − 6 kΩ At node 1, 12 − v1 v1 − v o v1 − v o = + 48 = 7v1 - 3vo (1) 3 6 12 At node 2, v1 − v o v o − 0 = = ix v1 = 2vo (2) 6 6 From (1) and (2), 48 vo = 11 vo ix = = 0.7272mA 6k Chapter 5, Solution 32. Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting amplifier.  50  v x = 1 +  (4 mV) = 24 mV  10  60 30 = 20kΩ
  17. By voltage division, 20 v vo = v o = o = 12mV 20 + 20 2 vx 24mV ix = = = 600nA (20 + 20)k 40k v o 144x10 −6 2 p= = = 204nW R 60x10 3 Chapter 5, Solution 33. After transforming the current source, the current is as shown below: 1 kΩ 4 kΩ vi + vo va − + 4V − 2 kΩ 3 kΩ This is a noninverting amplifier.  1 3 v o = 1 +  v i = v i  2 2 Since the current entering the op amp is 0, the source resistor has a OV potential drop. Hence vi = 4V. 3 vo = (4) = 6V 2 Power dissipated by the 3kΩ resistor is 2 v o 36 = = 12mW R 3k va − vo 4 − 6 ix = = = -2mA R 1k
  18. Chapter 5, Solution 34 v1 − vin v1 − vin + =0 (1) R1 R2 but R3 va = vo (2) R3 + R 4 Combining (1) and (2), R1 R v1 − va + v 2 − 1 va = 0 R2 R2  R  R v a 1 + 1  = v1 + 1 v 2  R   2 R2 R 3v o  R  R 1 + 1  = v1 + 1 v 2  R  R3 + R 4  2 R2 R3 + R 4  R  vo =  v1 + 1 v 2   R  R2  R 3 1 + 1    R    2 R3 + R 4 vO = ( v1R 2 + v 2 ) R 3 ( R1 + R 2 ) Chapter 5, Solution 35. vo R Av = = 1 + f = 10 Rf = 9Ri vi Ri If Ri = 10kΩ, Rf = 90kΩ
  19. Chapter 5, Solution 36 VTh = Vab R1 But vs = Vab . Thus, R1 + R2 R + R2 R VTh = Vab = 1 v s = (1 + 2 )v s R1 R1 To get RTh, apply a current source Io at terminals a-b as shown below. v1 + v2 - a + R2 vo io R1 - b Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes through R1 and consequently R2 . Thus, vo=0 and v RTh = o = 0 io Chapter 5, Solution 37. R R R  v o = −  f v1 + f v 2 + f v 3   R1 R2 R3   30 30 30  = −  (1) + (2) + (−3)  10 20 30  vo = -3V
  20. Chapter 5, Solution 38. R R R R  v o = −  f v1 + f v 2 + f v 3 + f v 4   R1 R2 R3 R4   50 50 50 50  = −  (10) + (−20) + (50) + (−100)  25 20 10 50  = -120mV Chapter 5, Solution 39 This is a summing amplifier.  Rf Rf Rf   50 50 50   R v1 + R v 2 + R v3  = − 10 (2) + 20 v 2 + 50 (−1)  = −9 − 2.5v 2 vo = −   1 2 3    Thus, vo = −16.5 = −9 − 2.5v 2  → v 2 = 3 V Chapter 5, Solution 40 R1 R2 va + R3 vb - + + v1 + - v2 Rf vo - + v3 R - - Applying KCL at node a, v1 − v a v 2 − v a v3 − v a v1 v 2 v3 1 1 1 + + =0  → + + = va ( + + ) (1) R1 R2 R3 R1 R2 R3 R1 R2 R3
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