Ch05p - Chapter 5 Exercise Problems EX5.1 = 1- 0.980 = 49 1...

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Chapter 5 Exercise Problems EX5.1 1 α β = For 0.980 0.980, 49 1 0.980 αβ == = For 0.995 0.995, 199 10 . 9 9 5 So 49 199 ≤≤ EX5.2 3 200 120 CBO CEO n BV BV == or 40.5 CEO BV V = EX5.3 () ( ) ( ) () ( ) ( )( ) ( ) 20 . 7 6.5 200 120 6.5 0.78 5 0.78 4 1.88 0.0065 0.7 0.78 1.88 1.47 BB BE B B CC CE CC C C CE BB E E VV o n IA R II A I m A VVI R o r V V PI V o n I V m W μ βμ = =− = = =+ = + EX5.4 50 . 72 . 8 or 4.62 325 80 4.615 0.369 2 5 0.369 which yields 8.13 EB BB B CB C EC C C VVo nV I R I m A VR R k + −− = = ==− = Ω EX5.5 (a) ( ) ( ) . 7 5.91 220 100 5.91 0.591 . 5 9 7 10 0.591 4 or 7.64 BB BE B B EB CE CC C C CE o n R A m A m A R V V = = = (b) 6.5 0.7 26.4 220 B I A = Transistor is biased in saturation mode, so 0.2 10 0.2 or 2.45 4 2.45 0.0264 2.48 CE CE CC CE C ECB s a t V s a t I Im A R III m A = =+= + EX5.6 For 00 . 7 , I ≤< Q n is cutoff, 9 O = When Q n is biased in saturation, we have ( ) ( ) 100 0.7 4 0.2 9 5.1 200 I I V = So, for 5.1 , I V O = 0.2 V EX5.7
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() ( ) 80 . 7 13 0 7 6 1 . 2 BB BE B BE VV o n I RR β == ++ + or 60.2 B I A μ = ( ) ( ) 75 60.23 4.52 14 . 5 8 12 4.517 0.4 4.577 1.2 CB EB CE CC C C E E I IA m A II m A VVI R I R βμ =+ = =− or 4.70 CE = EX5.8 For V C = 4 V and I CQ = 1.5 mA , 10 4 4 1.5 11 0 1 1.5 1.515 100 C CC CQ EC k I I Im A + ⎛⎞ + = ⎜⎟ ⎜⎟ ⎝⎠ also 0.7 10 Then 6.14 1.515 BE E E EE Vo nV I R k −− = = EX5.9 30 . 7 0.25 9.2 k 75 0.25 0.2467 mA 76 0.7 3 0 . 72 6.89 k 0.2467 EQ E E CQ CEQ CQ C I R R I VI R ⎜⎟ −+ + = +− = EX5.10 5o n 2 E B BB IR V IR (a) 180 5 2 0.7 2 0.9859 mA 41 0.962 mA C I = + = = (b) 180 6.3 2 1.2725 mA 61 1.25 mA C I = = (c) 180 6.3 2 1.6657 mA 101 1.64 mA C I = = (d) 180 6.3 2 1.97365 mA 151 1.94 mA C I = = EX5.11
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() 40 . 7 1.0 BB EB EE E VV o n IR R = = or 3.3 E Rk ( ) ( ) 0.992 1 0.992 1.0 0.992 or 8 0.992 1 5 CE BEC B CB C C CC II m A I I A VI R V α μ == = =−= − = =− = or 4.01 CB EX5.12 51 . 50 . 2 15 CE C V s a t R I γ + −+ or 220 R 15 1 15 . 7 4.3 1 B IB E B B Im A vVo n I = Ω ( ) ( ) ( ) 1 0.7 15 0.2 3.7 BB E CC E PI Vo n I V mW =+ = EX5.13 (a) For V 1 = V 2 = 0, All currents are zero and V O = 5 V . (b) For V 1 = 5 V , V 2 = 0; I B 2 = I C 2 = 0, 1 11 . 7 4.53 0.95 . 2 8 0.6 0.2 B R O A I m A = = (c) For V 1 = V 2 = 5 V , I B 1 = I B 2 = 4.53 mA ; I R = 8 mA , I C 1 = I C 2 = I R /2 = 4 mA , V O = 0.2 V EX5.14 55 OC C B C OB C vi R i R R β Δ= −Δ BB I BE B B I B B Vv V o n i R v i R = Δ Then vR Δ− = Δ Let 100, 5 , 100 CB R k Ω = Ω So ( ) 100 5 5 100 o I v v Δ Δ Want Q-point to be ( ) ( ) 2.5 5 100 5 oB Q vQp t I −== Then 0.7 0.005 0.005 100 BB BQ BQ V A I = so 1.2 BB = ( ) Also, 100 0.005 0.5 CQ BQ I A = EX5.15
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2.5 5 CEQ CQ C VI R == 52 . 5 or 10 0.25 C Rk Ω 0.25 0.002083 120 CQ BQ I I mA β = 50 . 7 Then 2.06 0.002083 BB RR M = EX5.16 (a) () 12 2 9 2.25 1.8
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Ch05p - Chapter 5 Exercise Problems EX5.1 = 1- 0.980 = 49 1...

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