Chapter17

Microelectronic Circuit Design

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CHAPTER 17 17.1 a () T = A β =∞ | A v = 1 = 5 FGE = 0 b A = 10 80 20 = 10000 T = 10000 0.2 = 2000 A v = A 1 + A = 10000 1 + 2000 = 5.00 FGE = 100% 1 + A = 100% 2001 = 0.05% c T = 10 0.2 = 2 A v = A 1 + A = 10 1 + 2 = 3.33 FGE = 100% 1 + 2 = 33.3% 17.2 a = R 1 R 1 + R 2 = 1 k 101 k = 1 101 b T = A = 10 86 20 1 101 = 198.6 A v = A 1 + A = 2 x 10 4 200 = 100 17.3 a s = R 1 R 1 + R 2 = 1 k 101 k = 1 101 Ts = A = 10 80 20 1 101 ⎟ = 99.0 A v =− R 2 R 1 A 1 + A 100 k 1 k 99 100 ⎟ =− 99 17.4 s = Z 1 s Z 1 s + Z 2 s = R R + 1 sC = s s + 1 RC = s s + 5000 A = 10 80 20 = 10 4 = A = 10 4 s s + 5000 A v Z 2 Z 1 A 1 + A 1 sRC 10 4 s s + 5000 1 + 10 4 s s + 5000 1 RC 1 s + 0.5 Instead of a pole at the origin, the integrator has a low- pass response with a pole a ω = 0.5 rad/s. ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-1
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17.5 A V = A 1 + A β = A 1 + A | From Chapter 12, GE = 1 1 + A = 1 1 + A 1 1 + A 10 4 A 9999 A 80 dB 17.6 -2.0V -1.0V 0V 1.0V 2.0V VS 2.0V 0V -2.0V Output Voltage Input Voltage Feedback No feedback *Problem 17.6 – Figure P17.6- Class-B Amplifiers VCC 3 0 DC 10 VEE 4 0 DC -10 VI 1 0 DC 0 Q1 3 1 2 NBJT Q2 4 1 2 PBJT RL1 2 0 2K RID 1 7 100K E1 5 0 1 7 5000 RO 5 6 100 Q3 3 6 7 NBJT Q4 4 6 7 PBJT RL2 7 0 2K .MODEL NBJT NPN .MODEL PBJT PNP .OP .DC VS -10 10 .01 .PROBE V(1) V(2) V(7) .END 17.7 A v = A 1 + A From Chapter 12, GE = 1 1 + A 1 A 1 = 200 GE 200 A 0.002 A 200 0.002 = 10 5 A 100 dB 17.8 17-2 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07
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S A A v = A A v A v A A v = A 1 + A β A v A = 1 + A () 1 A 1 + A 2 = 1 1 + A 2 S A A v = A A 1 + A 1 1 + A 2 = 1 1 + A 1 A S A A v = 1 1 + 10 5 0.01 = 1 1001 A v A v = S A A v A A = 1 1001 10% = 9.99 x 10 3 % 17.9 (a) Series-series feedback (b) Shunt-series feedback (c) Shunt-shunt feedback (d) Series-shunt feedback 17.10 (a) Series-shunt feedback (b) Shunt-series feedback (c) Series-series feedback (d) Shunt-shunt feedback 17.11 (a) Shunt-series and series-series feedback (b) Shunt-shunt and series-shunt feedback 17.12 (a) Series-shunt and series-series feedback (b) Shunt-series and shunt-shunt and feedback 17.13 a A v = 10 86 20 = 20000 | A i = i o i i i o = i i 40 k 20000 1 k A i = 8.00 x 10 5 With resistive feedback, the closed-loop gain cannot exceed the open-loop gain. Therefore, A i 8.00 x 10 5 . b A tr = i o v i = i o i i 40 k = A i 40 k A tr 8 x 10 5 4 x 10 4 = 20 S ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 17-3
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17.14 A = 10 90 20 = 31600 a () R in = R id 1 + A β | For =1, R in = 40 k 1 + 31600 = 1.26 G b R in = R id 1 + A For R in = 40 k 1 + 31600 = 1.27 c R out = R o 1 + A For R out = 1 k 1 + 31600 = 31.6 M d R out = R o 1 + A For R out = 1 k 1 + 31600 = 31.6 m 17.15 The circuit topology is identical to Fig. 17.8. h 11 F = 5 k 45 k Ω= 4.50 k h 22 F = 45 k Ω+ 5 k -1 =5 0 . 0 k -1 = h 12 F = v 1 v 2 i 2 = 0 = 5 k 5 k 45 k = 1 10 R L 1 h 22 F = 5 k 50 k = 4.55 k A = 20 k 1 k +20 k 4.5 k 4000 4.55 k 1 k 4.55 k = 2570 A v = A 1+ A = 2570 1+ 2570 1 10 = 2570 258 = 9.96 R in = R in A A = 1 k k 4.5 k 258 = 6.58 M R out = R out A A = 5 k 50 k 1 k 258 = 3.18 17-4 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07
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17.16 (a) v o + R 1 v I R 2 + A - R L R I Feedback Network b () h 11 A = v 1 i 1 v 2 = 0 = 15 k | h 11 F = 4.3 k 39 k Ω= 3.87 k h 11 T = 18.9 k h 22 A = i 2 v 2 i 1 = 0 = 1 k -1 =1 k -1 h 22 F = 39 k Ω+ 4.3 k -1 =4 3 . 3 k -1 h 22 T =+ 1.02 mS h 21
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Chapter17 - CHAPTER 17 17.1 (a) T = A = (b) A = 10 Av = 80...

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