Ch04s - Chapter 4 Problem Solutions 4.1 (a) C W g m = 2 K n...

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Chapter 4 Problem Solutions 4.1 (a) () 22 2 0.5 2 0.040 0.5 3.125 μ ω ⎛⎞ == ⎜⎟ ⎝⎠ = = no x mn D D C W gK I I L W LL (b) ( ) ( ) 2 2 2 0.5 0.04 3.125 0.8 2.80 V x DG S T N GS GS C W IV V L VV =− = 4.2 (a) 2 2 50 20 100 0.3125 2 po x mp D D C W g KI I L WW = = (b) ( ) 2 2 2 100 20 0.3125 1.2 5.2 V x DS G T P SG SG C W V L =+ = 4.3 ( ) [][ ] 2 1 1 1 11 0 1 3.4 13 . 0 1 5 3.4 1 5 3.0 1 10 3.4 3.0 3 10 3.4 5 0.0308 5 12.5 k 0.4 D n GS TN DS DS D DD S DS o D IK V V I V r I λ λλ =−+ + + = = ++ += + −= = Δ Δ 4.4 ( ) 0 0 1 0.833 mA 0.012 100 D r I II r = = 4.5 ( ) 2 1 1 1 1 1 1 12 0.20 0.22 1 4 0.20 1 4 0.22 1 2 0.8 0.44 0.22 0.20 0.0556 D nG S T N D S DS D S V V I V + = + + = + + =
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2 100 K 0.02 DS oo D V rr I Δ == = Δ 4.6 (a) (i) () 11 1000 K 0.02 0.05 o D r I λ = (ii) ( ) 1 100 K 0.02 0.5 o r (b) (i) 1 0.001 mA 1.0 A 1000 DS D o V I r μ Δ Δ= = = = 1.0 2% 50 D D I I Δ = (ii) 1 0.01 10 A 100 10 2% 500 DS D o D D V I r I I Δ = = Δ = 4.7 ( ) 1.0 mA 100 K 0.01 1 D o D I r I = = 4.8 ( ) 0 || 1 50 ||10 8.33 vm D v Ag r R A =− 4.9 a. 10 6 8 k 0.5 DD D SQ DD DQ VV RR I 2 2 For 2 V, for example, 2 0.5 0.030 2 0.8 11.6 = ⎛⎞ ⎜⎟ ⎝⎠ = GSQ no x DQ GSQ TN V C W IV V L WW LL b. ( ) ( ) ( ) 22 2 0.030 11.6 2 0.8 0.835 mA/V 133 k 0.015 0.50 mn D Q n G S Q T N mm DQ gK IK V V gg I ==− = c. ( ) 0 0.835 133 8 6.30 D v r R A && 4.10 [] 2 2 2 2 2 sin sin 1 sin 1 cos2 2 So 1 cos2 2 ωω ω ⎡⎤ ⎣⎦ ng s n g s s s s K vK V t K V t tt KV Kv t
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() 2 cos2 2 Ratio of signal at 2 to that at : 2s i n The coefficient of this expression is then: 4 ng s n GSQ TN gs gs GSQ TN KV t K VV V t V ω ωω −⋅ 4.11 ( ) 0.01 4 So 0.01 4 3 1 0.08 V = =− = gs GSQ TN gs gs V 4.12 ( ) 12 50 0.9615 50 2 0.9615 1 0.9615 50 10 8.01 om g s o D gs i i i Si vm o D v Vg V r R RR V V RR R Ag r R A ⎛⎞ =⋅ = = ⎜⎟ ++ ⎝⎠ 4.13 0 || 10 100 || 5 2.1 mA/V D mm r R gg −= = 4.14 a. () () 2 2 2 2 2 2 200 12 4.8 V 200 300 4.8 1 2 4 4 273 . 2 0 77 4 2 3 . 2 2.96 V 22 1 2.96 2 0.920 mA 12 0.92 3 2 7.4 V = + == + + −+ = ±− = + + = GD D G GG S Dn G S T N S GS GS GS GS GS GS DD DS DD D D S DS R V IK V V R V V II VVI R R V (b) 1 mg D L o mS gVRR V gR = + where 300 200 120 0.9836 300 200 2 120 2 g ii i i Si VVV V V = = = + Then
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() ( ) ( ) ( ) ( ) 0.9836 || 1 21 2 .96 2 1 .92 / 1.92 0.9836 3|| 3 So 0.585 11 . 9 22 mD L v mS m vv gR R A gm A V AA = + =− = = + c. 0.92 7.4 12 AC load line Slope ± ² 1 3 ±± 3 ³ 2 ² 1 3.5 K ± ( ) 1 i 3.5 k i 0.92 mA 0.92 3.5 3.22 6.44 V peak-to-peak Dd s s v v Δ= Δ Ω = 4.15 a. ( ) 2 2 2 12 1 1 1 2 2 2 3 mA 3 0.5 1.5 V 3 2 2 3.225 V 3.225 1.5 4.725 V 1 1 4.725 200 15 635 k 635 200 292 k 635 = == = = =+ = + = ⎛⎞ ⎜⎟ + ⎝⎠ = = + DS D Q S Q DQ n GS TN GS GS GG SS GD D i n D D IV I R IK V V VV VV V R R V RR R R R R R R b. ( ) || 1 2 2 3.225 2 4.90 mA / V 4.90 2 || 5 2.03 1 4.90 0.5 mD L v m gR R A g = + = = + 4.16 ( ) From Problem 4.14: 0.920 mA 7.4 V 1.92 mA/V || || || 200 || 300 1.92 3|| 3 2.88 0.9836 200 || 300 2 2.83 D DS m vm D L Si v I V g Ag R R RR R A = = = + +
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AC load line Slope ± ² 1 3 ±± 3 ² 1 1.5 K 0.92 7.4 12 ± () ( ) 1 , 0.92 mA 0.92 1.5 1.38 1.5 k 2.76 V peak-to-peak output voltage swing Δ= Δ = Ω DD S D D S iv i v 4.17 (a) 15 2 7.5 K vm D Ag R RR =− −= = (b) ( ) 1 27 . 5 51 K 12 mD v mS S S gR A R R = + = + 4.18 (a) 1 v g R A g R = + (1) 8 11 m g R g + (2) 16 g R Then 16 81 m A / V 16 K m m D g g R = = + = (b) 6 10 0.6 K v S S A R R = + = 4.19 a.
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Ch04s - Chapter 4 Problem Solutions 4.1 (a) C W g m = 2 K n...

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