Ch17s - Chapter 17 Problem Solutions 17.1 a vI =-1.5 Vi Q1 off Q2 on-0.7-3.5 iE = 0.56 mA iE = 5 iC1 = 0 v01 = 3.5 V iC 2 = iE v02 = 3.5 iE RC 2 =

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Chapter 17 Problem Solutions 17.1 a. 12 1.5 V off, on Ii vQ Q =− () ( ) 10 1 20 2 2 0.7 ( 3.5) 0.56 mA 5 03 . 5 V 3.5 3.5 0.56 2 EE C CE E C ii iv iiv i R −− = = = = = or 02 2.38 V v = b. 1.0 V on, off Q = ( ) 2 . 7 3 . 5 0.76 mA 5 . 5 V C = = = = c. logic 0 at 02 v (low level) 2.38 V = Then 01 1 2.38 3.5 (0.76) C vR == or 1 1.47 k C R 17.2 (a) 22 2 30 0.5 6 K CQ C C iI R R == = = (b) 11 1 31 0.5 4 K C C R R = (c) 1 1 21 exp exp exp 1 1e x p BE S T C Q BE BE S TT BE BE T IB EB E V I V i I VV I V vV V ⎛⎞ ⎜⎟ ⎝⎠ = ⎡⎤ + ⎢⎥ ⎣⎦ = + So 1 1 x p 0.1 1 0.2 0.5 x p C Q I T I T i I v V v V = + +
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() ( ) 1 exp 1 4 0.2 0.026 ln (4) 0.0360 V I T I I v V v v ⎛⎞ =− = ⎜⎟ ⎝⎠ −= 17.3 (a) 0.5 V, I v = 1 Q on, 2 Q off 02 3 V v == 01 3 (1)(0.5) 2.5 V v = (b) 0.5 V I v 1 Q off, 2 Q on 01 3V v = 02 3 (1)(0.5) 2.5 V v = 17.4 (a) 2 on, 1.2 0.7 1.9 E Qv V 2 22 2 2 2 1.9 5.2 1.32 2.5 11 . 3 2 0.758 EC CC C C ii m A vV i R R Rk −− = (b) 1 on, 0.7 0.7 1.40 E V ( ) 1 1 1 1 1.4 5.2 1.52 2.5 . 5 2 0.658 C C m A i R R = (c) 12 For 0.7 , on, off in Q Q 1 0.70 O 10 . 7 1 . 7 OO vv V =− − For 1.7 , off, on in Q Q 2 0.7 O . 7 1 . 7 V (d) (i) For 0.7 , 1.52 in E im A = ( ) 4 3 43 1.7 5.2 1.17 3 0.7 5.2 1.5 3 5.2 1.52 1.17 1.5 5.2 C C EC C A A Pii i =++ = + + or 21.8 Pm W = (ii) For 1.7 , 1.32 in E A = ( ) 4 3 0.7 5.2 1.5 3 1.7 5.2 1.17 3 1.32 1.5 1.17 5.2 C C A A P =+ + or 20.7 W =
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17.5 a. 3 3.7 0.7 1.5 mA 0.67 1.33 I == + () ( ) 34 1.5 1.33 0.7 R VI R V γ =+ = + or 2.70 V R V = b. logic 1 level 3.7 0.7 3.0 V =− For logic 1. XY vv ( ) 1 3 01 30 . 7 2.875 mA 0.8 3.7 2.875 0.21 3.10 V logic 0 2.4 V ER C B ii v v = = = For logic 0, on R Q 2 4 02 2.7 0.7 2.5 mA 0.8 3.7 2.5 0.24 3.1 V logic 0 2.4 V C B v v = = = 17.6 (a) 1.1 0.7 1.9 0.115 mA 20 0.115 mA REF Q REF I II −− (b) 1 1 5 5 56 (max) 1.1 0.7 0.4 V (max) 1.9 (max) 0.4 1.9 11.5 K 0.2 o o v v i R RR =− = = + = (c) 15 1 0.4 V 0.3 V 1.1 0.3 6.96 K 0.115 oB C R (d) 26 2 0.4V 0.3V 1.1 0.3 6.96 K 0.115 C R 17.7 logic 1 logic 0 1 0 0.5 V 22 R V ++ = For 2 1mA i = 55 0.5 2.3 2.8 k 1 = For R Q on, 2.3 RB E E E VV i R = or
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() 2 1 2 0.5 0.7 2.3 2.1 k 1 0.5 0.7 1.2 V 1.2 1.4 2.3 EE BR B E RR VV V i R −+ = =+ = + = −− = or 22 11 33 3 44 4 1.2 1.4 2.3 2.1 k 1 1.7 1.2 0.5 k 1 1(2 . 3 ) 31 . 1 k 0(2 . 3 ) 30 . 7 6 7 k iR R R = = == For R Q on, 03 logic 0 = 0 V 0.7 V RB vv = = 1mA EC R ii So 1.7 0.7 1k 1 CC = For logic1 1 V, 10 . 7 2 . 3 1.238 mA 2.1 I E v i For NOR 4 logic 0 0 V, 0.7 V B v v = Then 1.7 0.7 0.808 k 1.238 = 17.8 Maximum E i for I v = logic 13 . 3 V = Then 3.3 0.7 5mA 0.52k E R For 02 3 3 logic1 3.3 V 3.3 0 5 mA E v i R or 3 0.66 k R By symmetry, 2 0.66 k R For 1 Q on, 1 1 42 . 70 . 7 ER C C R = So 1 0.12 k C R
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For R Q on, 2 30 . 7 4.423 mA 0.52 E RC ii == = and () 2 2 2 42 . 70 . 7 4.423 0.136 k RC C C i R R −+ 17.9 Neglecting base currents: (a) 13 0, 0 EE II 55 50 . 7 1.72 2.5 0.7 I Im A YV = = = (b) 11 3 . 7 0.239 18 0 . 7 1.72 2.5 0.7 E I A I I A = = = = = = (c) . 7 0.239 18 I I m A 5 0, 5 E I (d) Same as (c). 17.10 (a) (1)(1) 0.7 1.7 RR VV V =− (b) 1 off, then 1 0.7 RO Qv L o g i c V 1 1 on, then (1)(2) 0.7 02 . 7 O vL o g i c V 2 2 2 / off, then Logic 1 0.7 / on, then (1)(2) 0.7 . 7 AB O O O QQ v V v o g i c V (c) . 7 ,o n , 1.7 0.7 2.4 R ABL o g i c VQ V = 10 . 7 ,/o n , 0.7 0.7 1.4 A B Logic V Q Q V = (d) . 7 n , A B Logic V Q Q =
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3 2 32 2.7 ( 5.2) 1.67 1.5 0.7 ( 5.2) 3 1.5 (1.67 1 1 1 3)(5.2) 39.9 02 . 7 3 , 1.67 39.9 C C CC im A A PP m W A B Logic V A i m A Pm W −− == =+ + + + = = = 17.11 a. AND logic function b. logic 0 0 V = 3 2 5( 1 .
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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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Ch17s - Chapter 17 Problem Solutions 17.1 a vI =-1.5 Vi Q1 off Q2 on-0.7-3.5 iE = 0.56 mA iE = 5 iC1 = 0 v01 = 3.5 V iC 2 = iE v02 = 3.5 iE RC 2 =

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