Chapter 5, Solution 1. (a) (b) (c) Rin = MΩ Rout = 60 Ω A = 8x104 Therefore AdB = 20 log 8x104 = dB Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 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. | Chapter 5 Solution 1. a Rin MQ b Rout 60 Q c A 8x104 Therefore AdB 20 log 8x104 dB Chapter 5 Solution 2. vo Avd A v2 - vi 105 20-10 x 10-6 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 v0 - 4 v2 - 1 ------ A 2x105 -20pV If v1 and v2 are in mV then v2 - v1 -20 mV 1 - v1 Y1 mV Chapter 5 Solution 5. -Vi Avd Ri - Ro I 0 But 1 Vd RiI -Vi Ri Ro RiA I 0 ViRi Vd Ro 1 A Ri 2 -Avd - RoI Vo 0 Vo AVd RoI Ro RiA I Ro RiA Vi Ro 1 A Ri Vo _ Ro RiA _ loo io4xio5 4 Vi Ro 1 A Ri loo 1 io5 io9 1 1o5 1o4 1oo ooo ----_ 1oo oo1 Chapter 5 Solution 6. Ro Ri R Vi Avd 0 But vd RiI Vi Ro Ri RiA I 0 I Ro 1 A R 1 -Avd - RoI Vo 0 Vo Avd RoI Ro RiA I Substituting for I in 1 -f Ro RiA 1 v IRo 1 A Ri so 2xio6x2xio5 -io-3 5o 1 2xio5 x2xio6 - 2oo ooox2x1o6 mV 2oo oo1x2x1o6 vo mV Chapter 5 Solution .
