# Ch03s - Chapter 3 Problem Solutions 3.1 W k 10 0.08 2 Kn =...

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Chapter 3 Problem Solutions 3.1 2 10 0.08 0.333 mA/V 2 1.2 2 n n k W K L ⎞⎛ = = = ⎟⎜ ⎠⎝ For 0.1 V DS V = Non Sat Bias Region (a) 0 0 GS D V I = = (b) ( )( ) ( ) 2 1 V 0.333 2 1 0.8 0.1 0.1 0.01 mA GS D V I = = = (c) ( )( ) ( ) 2 2 V 0.333 2 2 0.8 0.1 0.1 0.0767 mA GS D V I = = = (d) ( )( ) ( ) 2 3 V 0.333 2 3 0.8 0.1 0.1 0.143 mA GS D V I = = = 3.2 All in Sat region 2 10 0.08 0.333 mA/V 1.2 2 n K ⎞⎛ = = ⎟⎜ ⎠⎝ (a) 0 D I = (b) [ ] 2 0.333 1 0.8 0.0133 mA D I = = (c) [ ] 2 0.333 2 0.8 0.480 mA D I = = (d) [ ] 2 0.333 3 0.8 1.61 mA D I = = 3.3 (a) Enhancement-mode (b) From Graph V T = 1.5 V Now ( ) ( ) ( ) ( ) 2 2 2 2 0.03 2 1.5 0.25 0.12 0.15 3 1.5 2.25 0.0666 0.39 4 1.5 6.25 0.0624 0.77 5 1.5 12.25 0.0629 n n n n n n n n n n n n K K K K K K K K K K K K = = = = = = = = = = = = From last three, 2 (Avg) 0.0640 mA/V n K = (c) 2 2 (sat) 0.0640(3.5 1.5) (sat) 0.256 mA for 3.5 V (sat) 0.0640(4.5 1.5) (sat) 0.576 mA for 4.5 V D D GS D D GS i i V i i V = = = = = = 3.4 a. ( ) ( ) 0 0 2.5 2.5 GS DS GS TN V V sat V V V = = = − − = i. ( ) ( )( ) ( ) 2 0.5 V Biased in nonsaturation 1.1 2 0 ( 2.5) 0.5 0.5 2.48 mA DS D D V I I = = − − = ii. ( ) ( ) ( ) 2 2.5 V Biased in saturation 1.1 0 2.5 6.88 mA DS D D V I I = = − − = iii. V DS = 5 V Same as (ii) 6.88 mA D I = b. V GS = 2 V ( ) ( ) sat 2 2.5 4.5 V DS V = − − = i. ( ) 2 0.5 V Nonsaturation 1.1 2(2 ( 2.5))(0.5) (0.5) 4.68 mA DS D D V I I = = − − =

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