Ch13s - Chapter 13 Problem Solutions 13.1 13.2 13.3 (a) g...

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Chapter 13 Problem Solutions 13.1 Computer Simulation 13.2 Computer Simulation 13.3 (a) () 124 6 dm o o i A grrR = 1 1 2 2 2 4 4 2 66 1 7 7 6 1 6 20 0.769 / 0.026 80 4 20 80 20 (1 ) (120)(0.026) 15.6 0.2 0.6 0.030 20 (120)(0.026) 104 0.030 C m T A o C A o C in BE C I g mA V V V rM I V I Rr R r rk Vo n I mA R ππ π β == =+ + ⎡⎤ ⎣⎦ Ω ≅= = Ω Then 6 104 (121) 20 15.6 1.16 i RM Ω Then 769 4 4 1.16 565 dd AA = = Now 1 77 7 7 7 6 17 1 1 76 6 6 ) and oc o n b o n o c o nn o b b i R VI r I r r I V R rI I R ββ ⎛⎞ =− ⎜⎟ + ⎝⎠ + = + Then 7 1 2 16 1 7 7 7 ) 80 400 0.2 on n o v oi A o C Vr R A VR R r V I −+ + === Ω So 22 (120)(121)(400) 20 2813 1160 20 15.6 vv = + 6 2 Overall gain (565)( 2813) 1.59 10 dv A =⋅ = × (b) 1 2 id = and 1 (80)(0.026) 104 0.020 Ω 208 id Rk (c) 1 and (10)(1 2813) 28,140 2 PD M eq M f Cp F RC + = 24 6 61 2 4 4 1.16 0.734 1 7.71 2 (0.734 10 )(28,140 10 ) eq o o i PD r R M f Hz = Ω ××
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6 Gain-Bandwidth Product (7.71)(1.59 10 ) 12.3 MHz 13.4 a. 3 Q acts as the protection device. b. Same as part (a). 13.5 If we assume (on) 0.7 V, BE V = then in 0.7 0.7 50 5 V =++ + So breakdown voltage 56.4 V. 13.6 (a) 5 5 15 0.6 0.6 ( 15) 0.50 57.6 REF I Rk R −−− == 10 4 10 44 ln 0.026 0.50 ln 2.44 0.030 0.030 REF CT C I IR V I RR k ⎛⎞ = ⎜⎟ ⎝⎠ = (b) 50 . 60 . 6(5 ) 0.153 57.6 REF REF I Im A = = 10 10 10 0.153 (2.44) (0.026)ln By trial and error, 21.1 C C C I I I A μ = 13.7 (a) 0.50 REF I mA 3 11 12 14 0.50 10 ln (0.026)ln 0.641 10 REF BE T BE EB S I VV V V V I × Then 55 3 10 10 14 15 0.641 0.641 ( 15) 57.4 0.50 0.026 0.50 ln 2.44 0.030 0.030 0.030 10 0.026ln 0.567 10 BE BE k k V = = × = = (b) From Problem 13.6, 0.15 REF I mA 3 11 12 14 0.15 10 0.026ln 0.609 10 5 0.609 0.609 ( 5) Then 0.153 57.4 BE EB REF REF V I A × = = = Then 10 21.1 C I A from Problem 13.6 13.8 a. . . ) 0.22 mA 40 REF REF II = = 10 4 10 10 10 ln 0.22 (5) (0.026)ln REF C C C I I I I = =
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By trial and error; 10 10 66 17 17 13 13 14.2 A 7.10 A 2 0.75 0.165 mA 0.25 0.055 mA C C CC C REF C CA R E F I I II I I μ = = = = = (b) Using Example 13.4 17 16 16 16 16 31.5 k 50 [31.5 (201)(0.1)] 50 51.6 25.4 k and 0.165 (0.165)(0.1) 0.6 0.0132 mA 200 50 394 k E nT C C r R V r I I r π β =+ == Ω = + = Then 2 6 6 06 394 (201)(25.4) 5.5 M 732 k 0.00710 0.273 mA / V 0.026 50 7.04 M 0.0071 i m R r g r Ω ==Ω Then act1 04 7.04[1 (0.273)(1 732)] 8.96 M 50 7.04 M 0.0071 R r = Ω Then 7.1 (7.04 8.96 5.5) 0.026 d A ⎛⎞ =− ⎜⎟ ⎝⎠ or 627 d A Gain of differential amp stage Using Example 13.5, and neglecting the input resistance to the output stage: act2 13 2 50 303 k 0.165 (200)(201)(50)(303) (303) (5500)[50 31.5 (201)(0.1)] A CB v V R I A = ++ or 2 545 v A Gain of second stage 13.9 10 19 A C I = From Equation (13.6) 2 2 10 22 (10) 2(10) 2 3 2 (10) 3(10) 2 122 2 132 PP C I I ββ ⎡⎤ ⎢⎥ + + ⎣⎦ = So 132 2 (19) 20.56 A 122 I 2 10.28 A C
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99 9 44 2 20.56 17.13 A 2 2 1 1 10 17.13 1.713 A 10 10.28 0.9345 A (1 ) 11 10 (10.28) 9.345 A 1 CC P C BB P P P P I II I I I μ β == = ⎛⎞ + + ⎜⎟ ⎝⎠ = = + = + 13.10 55 (on) (1) 0.6 (0.0095)(1) 0.6095 E C VVV I −= + =+ = 77 89 13 14 0.6095 12.2 A 50 19 A 0.72 mA 0.72 mA 138 A REF E REF C I I = = = = () [ ] 789 1 31 4 Power 30[0.0122 0.019 0.019 0.72 0.72 0.138] Power 48.8 mW C C C REF E C VVI I I I I I +− =− + + +++ = + + = 1 4 Current supplied by and 1.63 mA CCCR E F E C VV I I I III =+++ + + = 13.11 (a) (min) 15 0.6 0.6 0.6 0.6 12.6 cm vV + + + + (max) 15 .6 14.4 So 12.6 14.4 cm cm = −≤ (b) (min) 5 4(0.6) 2.6 cm =− + (max) 5 0.6 4.4 So 2.6 4.4 cm cm = −≤≤ 13.12 If 0 15 , V the base voltage of 14 Q is pulled low, and 18 Q and 19 Q are effectively cut off. As a first approximation 14 14 0.6 22.2 mA 0.027 22.2 0.111 mA 200 C B I I Then 15 13 14 0.18 0.111 0.069 mA A B I = = Now 15 15 3 14 ln 15 0.069 10 (0.026)ln 10 0.589 V C BE T I = × = =
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As a second approximation 14 14 14 0.589 21.8 mA 0.027 21.8 0.109 mA 200 CC B II I = = == and 15 15 0.18 0.109
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This homework help was uploaded on 02/10/2008 for the course ECE 138 taught by Professor N/a during the Spring '08 term at CSU Fresno.

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Ch13s - Chapter 13 Problem Solutions 13.1 13.2 13.3 (a) g...

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