Ch16s - Chapter 16 Problem Solutions 16.1 (a) VTN = Cax =...

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Chapter 16 Problem Solutions 16.1 (a) () ( ) 14 8 8 1/2 19 14 15 8 2 22 (3.9)(8.85 10 ) 7.67 10 450 10 2 2 1.6 10 11.7 8.85 10 8 10 5.15 10 sa TN fp SB fp ax ax ax ax eN VV C C t φφ −− ⎡⎤ Δ= +− ⎣⎦ × == = × × ∈= × × × Then 8 8 5.15 10 2(0.343) 2(0.343) 7.67 10 TN SB × × For 1: SB = 0.671 1.686 0.686 0.316 TN TN V For SB = 0.671 2.686 0.686 0.544 TN TN V (b) For 2.5 V, 5 V, GS DS transistor biased in the saturation region. [] 2 2 2 2 For 0, 0.2(2.5 0.8) 0.578 mA For 1, 0.2 2.5 0.8 0.316 0.383 mA For 2, 0.2 2.5 0.8 0.544 0.267 mA Dn G S T N SB D SB D SB D IK V I V I V I =− = = = + = = + = 16.2 (a) ( ) 2 2 3 5 42 2( ) 5( 0 . 1 ) 25 0 .8 0 .1 0 40 10 81 0 or 1.476 10 / 2 DD O G S T N O O D n n Vv I KVV v v R K W KA V L × × ⎛⎞ = ⎜⎟ ⎝⎠ So that 3.69 W L = b. From Equation (16.10). 2 2 2 0 (0.1476)(40) 0.8 0.8 5 0 1 (1) 4(0.1476)(40)(5) or 0.8 2(0.1476)(40) or 0.8 0.839 n D It TN It TN DD It It It It K RVV VV V V V −+ = = −± + −=
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So that 1.64 V It V = (max) D DD PI V =⋅ and 5( 0 . 1 ) (max) 0.1225 mA 40 D I == or 0.6125 mW P = 16.3 a. From Equation (16.10), the transistor point is found from 2 2 2 0 () ()0 50 / , 20 0.8 ( 0 . 0 5 ) ( 2 0 ) 5 0 1 1 4(0.05)(20)(5) 1.79 V So 2.59 V 2(0.05)(20) 1.79 V n D It TN It TN DD nD T N It TN It TN It TN It t KR V V V V V KA V R k VV V V μ −+ = Ω = = −± + −= = = = Output voltage for 5V I v = is determined from Equation (16.12): 2 00 0 2 2 0 5 (0.05)(20) 2(5 0.8) 9.4 5 0 9.4 (9.4) 4(1)(5) So 0.566 V 2(1) vv v v ⎡⎤ =− ⎣⎦ = ±− b. For 200 k , D R () () ( ) 0 2 0 2 2 0 1 1 4(0.05)(200)(5) 0.659 V So 1.46 V 2(0.05)(200) 0.659 V 5 (0.05)(200) 2 5 0.8 or 10 85 5 0 85 85 4 10 5 0.0592 V 210 It TN It t V V v v + = = = = 16.4 (a) V =
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() 2 2 0.25 (33) 75.76 μ A 3.3 0.15 41.6 K 0.07576 2 80 75.76 3.3 0.8 0.303 2 n GS TN II RR k W IV V L WW LL = = = = ⎛⎞ =− ⎜⎟ ⎝⎠ = (b) 2 2 2 2 (sat) (sat) 3.3 0.8 0.08 0.303 1.6 0.64 24 1 . 6 0.504 1.6 0.64 4.1 0.504 0.1936 3.777 0 0.1936 0.03748 4(0.504)(3.777) 2(0.504) 2.55V DS GS TN DD DS Dn G S T N GS GS GS GS GS GS GS GS GS GS VV V IK R V V V V = −− −+ = = +− = −± + = = For 0.8 2.55V GS V ≤≤ Transistor biased in saturation region 16.5 (a) 0.25 (3.3) 75.76 μ A DD PIV =⋅ = = For Sat Load ( ) ( ) 2 2 2 80 75.76 3.3 0.15 0.8 0.343 2 80 80 0.343 3.3 0.15 0.8 75.76 2 2.5 0.8 0.15 0.15 22 3.89 D D I W I L W L == = = = = = Eq 16.21 3.89 3.3 0.8 0.8 1 0.343 5.994 4.3677 3.89 1 0.343 0.8 1.372 V It GS V V + + 16.6 (a) From Equation (16.23) ( ) ()( ) 2 3 0.5 0.25 0.25 3 0.25 0.5 4.26 D D L L KK ⎡⎤ = = ⎣⎦
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(b) ( ) () ()( ) 22 2 2.5 0.5 0.25 0.25 3 0.25 0.5 5.4 D D L L KK ⎡⎤ −− = = ⎣⎦ (c) 2 2 2 0.080 (1)(3 0.25 0.5) 0.203 2 (0.203)(3) 0.608 DL G S L T N L L D D O T N L D DD D iK V V K V v V im A PiV P m W =− = ⎛⎞ = ⎜⎟ ⎝⎠ =⋅ = = for both parts (a) and (b). 16.7 ( ) 2 2 0.4 (3) 0.1333 0.080 0.1333 3 0.1 0.5 0.2304 2 D D D D D O T N L L L Pm W i V i i m A V v V WW LL == = = = So 0.579 L W L = ( ) ( ) ( ) 2 2.5 0.5 0.1 0.1 3 0.1 0.5 D L K K = 14.8 D L K K = so that 8.55 D W L = 30 . 50 . 51 1 4 . 8 11 4 . 8 It V −+ + = + or 1.02 , 0.52 It Ot VV V V 16.8 We have ( ) ( ) ( ) ( ) [] 2 2 2 2 / 2 0.08 0.08 0.08 / / 2 2 0.2 0.08 0.0064 0.92 0.2 0.5184 / // 0.096 0.5184 5.4 D I TND O O DD O TNL L D DD TN TN DD DD DD DD TN L D D D D D D D D D D D L K vV v v V vV K WL VVV V V V V V V V = = = = = = 16.9 Logic 1 OH B TN V =− = So (a) 43 BO H V V = = (b) 54 H V V = = (c) 65 H V V = = (d) 75 , s i n c e 0 H D S V V V = For I OH vV =
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() [] 2 2 2 DI T O O L B O T K vVv v KV vV ⎡⎤ −− = ⎣⎦ Then (a) () ( ) ( ) [] 2 2 1231 0 .
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Ch16s - Chapter 16 Problem Solutions 16.1 (a) VTN = Cax =...

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