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 21 . 2 2 n n k W K L ⎛⎞ ⎛ ⎞ == = ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ For 0.1 V DS V = Non Sat Bias Region (a) 00 GS D VI = = (b) () ( ) ( ) 2 1 V 0.333 2 1 0.8 0.1 0.1 0.01 mA GS D ⎡⎤ = ⎣⎦ (c) ( ) ( ) 2 2 V 0.333 2 2 0.8 0.1 0.1 0.0767 mA GS D −− = (d) ( ) 2 3 V 0.333 2 3 0.8 0.1 0.1 0.143 mA GS D = 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 nn n n n n KK 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 DD G S G S ii V V 3.4 a. () ( ) 0 02 . 5 2 . 5 GS DS GS TN V Vs a tV V V = − = i. ()( ) 2 0.5 V Biased in nonsaturation 1.1 2 0 ( 2.5) 0.5 0.5 2.48 mA DS V II = = ii. 2 2.5 V Biased in saturation 1.1 0 2.5 6.88 mA DS V = = 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 V = =
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ii. () 2 2.5 V Nonsaturation 1.1 2(2 ( 2.5))(2.5) (2.5) 17.9 mA DS DD V II = ⎡⎤ =− −− = ⎣⎦ iii. () ( ) 2 5 V Saturation 1.1 2 2.5 22.3 mA DS V = = 3.5 02 2 DS GS TN VVV V >−= = Biased in the saturation region 2 2 2 0.080 1.5 0 2 9.375 2 n DG S T N k W IV V L WW LL =⋅ ⎛⎞ = ⎜⎟ ⎝⎠ 3.6 ( ) 14 10 600 3.9 8.85 10 2.071 10 no x nn o x ox ox ox kC tt t μ × × == = = (a) 500 A 2 41.4 A/V n k ′ = (b) 250 2 82.8 A/V n k ′ = (c) 100 2 207 A/V n k ′ = (d) 50 2 414 A/V n k ′ = (e) 25 2 828 A/V n k ′ = 3.7 a. 14 82 8 00 8 2 3.9 8.85 10 7.67 10 F/cm 450 10 2 16 4 650 7.67 10 24 0.399 / ox ox ox xx x n n C C W K L Km A V × ∈∈ × = b. ( ) ( ) 22 3 V Saturation 0.399 3 0.8 1.93 mA GS DS Dn G S T N D VV IK I = = 3.8 2 2 0.08 1.25 2.5 1.2 23.1 m 1.25 2 n S T N k V ω = 3.9 14 8 0 3.9 8.85 10 400 10 8.63 10 F/cm ox ox x C t × ×
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() 8 5 2 1 600 8.63 10 22 . 5 1.036 10 no x n n C W K L W KW μ =⋅ ⎛⎞ ⎜⎟ ⎝⎠ 2 2 35 1.2 10 1.036 10 5 1 7.24 m Dn G S T N IK VV WW −− =− ×= × = 3.10 Biased in the saturation region in both cases. 2 2 p DS G T P k W IV V L + (1) 2 0.040 0.225 3 2 TP W V L =+ (2) 2 0.040 1.40 4 2 TP W V L Take ratio of (2) to (1): 2 2 (4 ) 1.40 6.222 0.225 (3 ) 4 6.222 2.49 2.33 3 TP TP TP TP TP V V V V + == + + + Then 2 0.040 0.225 3 2.33 25.1 2 LL = 3.11 5 V, 0 5 V 0.5 5 0.5 4.5 SGS G TP SD SG TP VVV V s a t V V V = =+= −= a. 2 0 5 V Biased in saturation 2 5 0.5 40.5 mA D DD II = = = b. ( ) ( ) 2 2 V 3 V Nonsaturation 2 2 5 0.5 3 3 36 mA D = = ⎡⎤ = ⎣⎦ c. ( ) () () 2 4 V 1 V Nonsaturation 2 2 5 0.5 1 1 16 mA D = = = d. 5 V 0 0 D D I = = = 3.12 (a) Enhancement-mode (b) From Graph V TP = + 0.5 V 2 2 2 2 2 0.45 2 0.5 2.25 0.20 1.25 3 0.5 6.25 0.20 2.45 4 0.5 12.25 0.20 4.10 5 0.5 20.25 0.202 Avg 0.20 mA/V pp p p kK K K =−= = =− = = (c) 2 2 (sat) 0.20 (3.5 0.5) 1.8 mA (sat) 0.20 (4.5 0.5) 3.2 mA D D i i = =
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3.13 () SD SG TP Vs a tV V =+ (a) () () 12 1 SD SD a t a t V =− + = (b) 02 2 SD SD a t a = (c) 3 SD SD a t a = 2 2 22 pp DS G T P S D kk WW I VV V s a t LL ′′ =⋅ + =⋅⋅ ⎡⎤ ⎣⎦ (a) 2 0.040 6 1 0.12 2 DD I Im A ⎛⎞ = = ⎜⎟ ⎝⎠ (b) 2 0.040 6 2 0.48 2 I A = = (c) 2 0.040 6 3 1.08 2 I A = = 3.14 (sat) 3 08 22V SD SG TP V.
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Ch03s - Chapter 3 Problem Solutions 3.1 W k 10 0.08 2 Kn =...

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