# Ch16s - Chapter 16 Problem Solutions 16.1(a VTN = Cax = 2e...

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Chapter 16 Problem Solutions 16.1 (a) ( ) ( ) ( )( ) 14 8 8 1/ 2 19 14 15 8 2 2 2 (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 s a TN fp SB fp ax ax ax ax s a e N V V C C t e N φ φ Δ = + × = = = × × = × × × = × Then 8 8 5.15 10 2(0.343) 2(0.343) 7.67 10 TN SB V V × Δ = + × For 1 : SB V V = 0.671 1.686 0.686 0.316 TN TN V V V Δ = Δ = For 1 : SB V V = 0.671 2.686 0.686 0.544 TN TN V V V Δ = Δ = (b) For 2.5 V, 5 V, GS DS V V = = 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 D n GS TN SB D SB D SB D I K V V V I V I V I = = = = = = + = = = + = 16.2 (a) ( )( ) ( ) 2 2 3 5 4 2 2( ) 5 (0.1) 2 5 0.8 0.1 0.1 40 10 8 10 or 1.476 10 / 2 DD O D n GS TN O O D n n V v I K V V v v R K W K A 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 R V V V V V V V V V + = + = − ± + = =

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So that 1.64 V It V = (max) D DD P I 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 (0.05)(20)( ) ( ) 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 n D TN It TN It TN It TN It t K R V V V V V K A V R k V V V V V V V V V V μ + = = = Ω = + = − ± + = = = = Output voltage for 5 V I v = is determined from Equation (16.12): 2 0 0 0 2 0 0 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) v v v v v v =