Ch06s - Chapter 6 Problem Solutions 6.1 a gm = r = r0 = I...

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Chapter 6 Problem Solutions 6.1 a. () ( ) 00 2 76.9 mA/V 0.026 180 0.026 2.34 k 2 150 75 k 2 CQ mm T T CQ A CQ I gg V V rr I V I ππ β == = b. ( ) 0.5 19.2 mA/V 0.026 180 0.026 9.36 k 0.5 150 300 k 0.5 = = = = 6.2 (a) ( ) 0.8 30.8 mA/V 0.026 120 0.026 3.9 K 0.8 120 150 K 0.8 CQ m T T CQ A o CQ I g V V r I V r I π = = === (b) ( ) 0.08 3.08 mA/V 0.026 120 0.026 39 K 0.08 120 1500 K 0.08 m o g r r 6.3 ( ) 200 5.2 mA 0.026 125 0.026 0.625 k 5.2 200 38.5 k 5.2 CQ CQ mC Q T T CQ A CQ II gI V V I V I = = = 6.4
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() 80 2.08 mA 0.026 0.026 1.20 96 2.08 CQ CQ mC Q T T CQ II gI V V r I π β = = = = = = 6.5 (a) ( ) ( ) 20 . 7 0.0052 250 120 0.0052 0.624 0.624 24 / 0.026 120 0.026 5 0.624 BQ C mm o Im A I mA g gm A V rr k r ππ == = = = =∞ (b) ( ) 5 24 4 1.88 5 250 vm C v B r Ag R A rR ⎛⎞ =− ⎜⎟ ++ ⎝⎠ (c) 0.426sin100 1.88 OO SS v vv t V A 6.6 , 1.08 1.32 mA CQ Q T I V =≤ 1.08 1.32 41.5 50.8 mA/V 0.026 0.026 gg ≤≤ ( ) ( ) 120 0.026 ; max 2.89 k 1.08 80 0.026 min 1.58 k 1.32 1.58 2.89 k T CQ V I r r = Ω Ω Ω 6.7 a. ( ) ( ) 120 0.026 5.4 0.578 mA 11 52 . 5 V 22 2.5 5.0 0.578 4.33 k 0.578 0.00482 mA 120 on 0.00482 25 0.70 0.820 V T CQ CQ CQ CEQ CC CEQ CC CQ C C C CQ BQ BB BQ B BE BB V rI VV I R R R I I VI R V V = = === = =+ = b.
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() ( ) ( ) 0 00 0 0 120 0.026 5.40 k 0.578 0.578 22.2 mA/V 0.026 100 173 k 0.578 , 120 173 4.33 120 4.22 16.7 5.40 25 30.4 T CQ CQ m T A CQ mC S B C vm C BB vv V r I I g V V r I r Vg r R V V V rR r Ag r R AA π ππ β == = Ω = = Ω ⎛⎞ =− = ⎜⎟ + ⎝⎠ ++ ⎡⎤ ⎣⎦ + 6.8 a. () ( )( ) 1 5 V 2 10 5 10 0.5 10 k 0.5 0.005 100 on 0.70 0.005 50 0.95 V ECQ CC ECQ CQ C C C CQ BQ EB BQ B BB BB VV VI R R R I I R V V === += =+ = b. ( ) 0.5 19.2 mA/V 0.026 100 0.026 5.2 k 0.5 0.5 CQ mm T T CQ A CQ I gg V V rr I V I = =∞ c. ( ) 100 10 18.1 5.2 50 C B R 6.9 ( ) 10 4 1.5 mA 4 1.5 0.015 mA 100 100 0.026 1.73 K 1.5 5sin mV 2.89sin A 1.73 k CQ BQ be b I I r t v it r ω ωμ = Ω So
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() ( ) () () ( ) () () [] () 11 1 1 15 2.89sin A 1.5 0.289sin mA 10 10 1.5 0.289sin 41 . 1 5 6 s i n 1.156 231 0.005 BB Q b CB C CC C C C vv be it I i E t it i it t vt i tR t t v AA ωμ βω ω γ =+= + = =+ =− =− + == 6.10 1.2sin V 1.2sin 0 2 0.60sin mA 6sin A 100 2 K 50 12sin mV o o C C C b be b m be t itR v t t vt itr g r r t ππ π β = += = =⋅ = 6.11 a. CQ EQ I I 51 0 10 1.2 0.2 CEQ CQ C E CQ VI R R I == − + + 3.57 mA 3.57 0.0238 mA 150 CQ BQ I I = ( ) 12 0.1 1 0.1 151 0.2 3.02 k TH E RR R R + Ω & ()( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 2 2 2 1 10 5 on 1 5 1 3.02 10 5 0.0238 3.02 0.7 151 0.0238 0.2 5 1 30.2 1.50 20.1 20.1 3.02 3.55 k 20.1 TH TH BQ BQ E TH TH BE VR R R V I R R Rk R R R R + =⋅ ⋅ − + −= + + = = + b. ( ) 150 0 026 109±k 357 137 mA/V 0 026 m . r. . . g . Ω p
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R E r ± V ± V 0 g m V ± R C V S ± ² ± ² R 1 ±± R 2 () ( ) 150 1 2 575 11 0 9 1 5 1 0 2 C vv E . R A A. rR . . β == = ++ + p 2 22 6.12 a. ( ) ( ) ( ) 2 12 50 12 10 V 50 10 50 10 8 33 k 12 0 7 10 0 0119 mA 8 33 101 1 119±mA 120±mA 12 120 1 119 2 842±V TH CC TH BQ CQ EQ ECQ ECQ R VV RR R . . I. . V. . ⎛⎞ == = ⎜⎟ ⎝⎠ ===Ω −− + =− = 4 1.19 8.42 12 i C ² EC b. R E r ± V ± V 0 g m V ± R C V S ± ² ² ± R 1 ±± R 2
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() ( ) ( ) ( ) 0 100 0 026 218±k 119 1 100 2 194 12 1 8 1 0 1 1 mC Sm E E C vv E . r. . Vg V R V VV g V R r rR V r R A A. . π β == Ω = ⎛⎞ =−+ ⎜⎟ ⎝⎠ ++ ⎡⎤ =− ⎢⎥ ⎣⎦ = + p p p pp p p p p 2 2 2 2 c. Approximation: Assume r p does not vary significantly. ( ) ( ) ( ) 2 k 5% 2.1 k or 1.9 k 1 k 5% 1.05 k or 0.95 k For max 2.1 k and min 100 2.1 2.14 2.18 101 0.95 For min 1.9 k and max 1.05 k 100 1.9 1.76 2.18 101 1.05 C E CE v v R R RR A A = Ω = Ω = + Ω + So 1 76 2 14 v .A .
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Ch06s - Chapter 6 Problem Solutions 6.1 a gm = r = r0 = I...

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