Ch04p - Chapter 4 Exercise Solutions EX4.1 g m = 2 K n (VGS...

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Chapter 4 Exercise Solutions EX4.1 () ( ) 2 2 2 a n d 0.75 0.5 0.8 2.025 2 0.5 2.025 0.8 1.22 / 1 1 (0.01)(0.75) 133 133 mn G S T N D n G S T N GS GS mm o DQ o g K VV IK V g gm A V r I k rk λ =− = = = = = EX4.2 ( ) 2 2 0.5 0.4 0.8944 mA/V 0.8944 10 8.94 vm D D Q v Ag R gK I A == = EX4.3 (a) 2 12 2 320 51 . 9 0 5 V 520 320 0.20 1.905 0.8 0.244 mA 2 2 0.2 0.244 0.442 mA/V GS DD DQ D Q o R RR I I r ⎛⎞ = ⎜⎟ ++ ⎝⎠ = = =∞ (b) ( ) 0.422 10 4.22 D R (c) 520 320 198 K i RRR = && (d) 10 K OD EX4.4 ( ) 2 2 2 At transition point, 1 mA sat 1 0.2 sat sat 2.236 V D Dn G S t T N n D S DS DS I K V = = = = 2 2 1 5 2.236 Want 2.236 3.62 V 2 53 . 6 2 2.76 K 0.5 0.5 0.2 0.8 2.38 V 1 DSQ D GSQ GSQ GSQ DD DD V R R R R V R =+ = = + & ( ) 1 1 So 2.38 200 5 420 K and 382 K R =
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() ( ) 2 2 0.2 0.5 0.6325 mA/V 0.6325 2.76 1.75 vm D mn D Q v Ag R gK I A =− == = EX4.5 (a) ( ) ( ) 2 12 2 2 2 2 2 250 10 5 10 5 3 V 250 1000 5 2 35 2 0 . 50 . 6 2 1.2 0.36 0.2 1.64 0 0.2 0.04 4 1.64 1.385 V 2 0.5 1.385 0.6 0.308 mA 10 0.3 G GG S Dn G S T N GS GS GS GS GS GS GS GS DQ DQ DSQ R V RR VV IK V V V V II V ⎛⎞ = = ⎜⎟ ++ ⎝⎠ −− + = −=− + = ±+ = 08 10 2 6.30 V DSQ V + = (b) ( ) ( ) 22 0 . 5 0 . 3 0 8 1 0.7849 mA/V 0.7849 10 3.05 1 0.7849 2 mD n D Q mS m vv gR K I g AA = + = = + EX4.6 ( ) 2 2 0 53 3 V and 0.5 mA 4 k 0.5 0.5 1 1 1.71 V 5 1.71 3.29 V 1 0 . 5 1.414 mA/V 1.414 4 5.66 0.46sin 5.66 0.0813sin SDQ DQ D D DQ P SG TP SG SG GG GG D mP D Q m sd vi ii i VI R R V V V V R I g t Av v ω = = = = = = = t EX4.7 a. 2 2 2 9, 921 . 2 2 92 . 4 4 4 SG DQ S DQ P SG TP SG SG SG SG R I K V V = + 2 2.4 8.6 0.6 0 SG SG −+ =
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() () () ( ) ( ) 2 2 8.6 8.6 4 2.4 0.6 22 . 4 3.51 V, 2 3.51 2 4.57 mA 9 9 1.2 1 18 4.57 2.2 7.95 V SG SG DQ DQ SDQ DQ SDQ V VI I V ±− = == = =+− + = − = b. ± ² R D V i ± ² V 0 R S C S g m V SG V SG ( ) ( ) 0 2 2 2 4.57 6.046 mA/V 6.046 1 6.05 mP D Q mS G D vm D v gK I Vg V R Ag R A = = =− EX4.8 ( ) 12 5 10 1.5 3.33 k 1.5 1 0.8 2.025 V 5 7.025 V 10 400 DSQ DD DQ S SS DQ n GS TN GS GS G S G G DD VV I R RR IK V V V V V V V ⎛⎞ = = ⎜⎟ + ⎝⎠ So 21 281 k , 119 k ( ) ( ) ( ) 0 0 1 1 0 0 Neglecting , 1 0.015 1.5 44.4 k 3.33 44.4 3.1 k 2 2 1 1.5 2.45 / 2.45 3.1 0.884 1 2.45 3.1 Si v DQ S mn D Q vv gRr RA rI Rr g KI mAV AA λ = + ⎡⎤ = Ω ⎣⎦ Ω = = = + & & && EX4.9 ( ) 2 2 1 1 0 32 2 3 . 2 2 V 5 53 . 2 2 3 0.593 k 0.02 3 16.7 k 2 3 4 . 9 / DQ P SG TP SG SG SG DQ S DQ D Q V V V IR I m A V = = = = Ω === For 0 0 1 Lv grR g rR =∞ = + & & 0 16.7 0.593 0.573 k S Ω
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() ( ) 4.9 0.573 0.737 1 4.9 0.573 vv AA = = + 0 0 If is reduced by 10% 0.737 0.0737 0.663 1 mS L v L grRR A =− = = + Let 0 SL rRR x = && ( ) ( ) 4.9 0.663 0.663 4.9 1 0.663 14 . 9 0.402 0.573 0.573 0.402 0.573 0.402 0.402 0.573 1.35 k 0.573 L L LL L x x x xR R RR R = + == = −= + EX4.10 2 12 22 9.3 5 70.7 9.3 0.581 V 5 GD D DQ p SG TP P S G TP S S R VV IK V V K V V V V R ⎛⎞ ⎜⎟ ++ ⎝⎠ = = = ( ) 2 2 2 2 Then 0.4 5 0.581 0.8 5 2 1.381 5 2 2.762 1.907 5 2 4.52 1.19 0 SS S V −−= −+ = −− = ()( ) ( ) ( ) 2 4.52 4.52 4 2 1.19 52 . 5 2.50 V 0.5 mA 5 2 2 0.4 0.5 0.894 / 1 0.894 5 70.7 9.3 0.770 1 0.894 5 70.7 9.3 0.5 S SD Q mP D Q v S i v V VI g KI mAV gR A gR R R R A ±+ = = = =⋅ = & & & & Neglecting , Si R A v = 0.817 00 11 5 5 1.12 0.915 k 0.894 S m R g = & EX4.11 ± ² V 0 R D R L g m V sg V i ± ² R S V sg
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() ( ) 0 2 2 2 2 2 a n d 5 51 4 0 . 8 54 1 . 6 0 . 6 4 4 5.4 2.44 0 5.4 (5.4) 4 4 2.44 24 1.71 V . 7 1 0.822 mA 4 2 2 1 0.822 1.81 / ms g D L s g i vm DL SG DQ p SG TP S SG SG SG SG SG SG SG SG SG DQ mp D Q v Vg VRR VV AgRR V IK V V R VVV V V I gK I m A V A == = −= + −− = ±+ = = = & & ( ) 1.81 2 4 1.81 1.33 2.41 11 4 4 0.552 0.485 k 1.81 v in S in m A RR R g = === & & EX4.12 ( ) 2 2 2 22 2 1 0.015 2 0.030 / 2 63 6 36 0.030 1.08 / 1.08 0.015 72 no x n nn v n C W K mA V L KK A Km A V WW LL μ ⎛⎞ =⋅= = ⎜⎟ ⎝⎠ =− = ⎛⎞ ⎛⎞ = = ⎜⎟ ⎜⎟
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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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Ch04p - Chapter 4 Exercise Solutions EX4.1 g m = 2 K n (VGS...

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