Ch12p - Chapter 12 Exercise Solutions EX12.1 a. A 1 + A A A...

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Chapter 12 Exercise Solutions EX12.1 a. 4 1 11 111 1 0.05 0.0001 20 10 f ff f A A A AA β ββ = + += =− = = 0.0499 b. () 20 20 0.998 1/ 1/ 0.0499 20.040 f A == = EX12.2 6 6 1 11 1 1 0.01 10 100 10 0.009999 f f A A A = + = = = 6 1 1 100 20 % 0.002% 10 f dA A dA dA A A A dA dA ⎛⎞ =⋅ = ⎜⎟ + ⎝⎠ = = EX12.3 0 5 0 4 Bandwidth 1 10 21 0 100 2 10 rad / sec 10 kHz H H t A A A f ωβ ωπ =+ = = EX12.4 (a) ( ) () () 12 2 100 10 10 OA i n in vA A v A v vv 1000 100 10 oi i on i Sv S Nv N (b) 2 5 5 5 10 10 1 0.001 10 10 . 0 0 0 OC i n A v ΑΑ 22 ++ + + 3 4 10 10 0.1 i i SvS N
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EX12.5 a. 100 99 1 m V Sf b VVV ε =− = −= () ( ) 0 0 0 5 5000 V/V 0.001 0.099 0.0198 V/V 5 5000 50 V/V 1 1 0.0198 5000 vv v fb fb v vf vf v VA V A A V VV V A AA A ββ β = = = = == = = ++ b. ( ) ( ) ( ) ( ) 0 00 1 5 1 0.0198 5000 500 k 4 40 1 1 0.0198 5000 if i v if ff v RR A R R A =+ = + ⎡⎤ ⎣⎦ ⇒⇒ Ω EX12.6 a. 100 99 1 A b III μ ( ) 0 0 5 5000 A/A 0.001 0.099 0.0198 A/A 5 5000 50 A/A 1 1 5000 0.0198 ii fb i if if i I I I I A A = = = b. ( ) ( ) ( ) 0 5 50 1 1 0.0198 5000 1 1 0.0198 5000 4 400k i if if i fi f R A RA R R Ω EX12.7 4 4 21 5 5 10 3.9984 10 1 1 1/ 13 0 / 1 0 10 3.99984 10 1 0 / 1 0 3.99984 3.9984 100% 0.0360% 3.9984 v vf vf v vf A A A = + + + + + + × EX12.8 Use a non inverting op-amp. 22 11 5 1 4 += = Let 2 1 140 K 10 K R R = = 2 1 1 0.066667 1 R R +
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Input resistance. () 3 5 0.06667 5 10 1.67 M if R Ω 3 50 0.15 0.066667 5 10 of R =≡ Ω × EX12.9 ( ) 1 1 80 0.026 4.16 0.5 FE E oi k FE E FE k hR ii r h R h rk ⎛⎞ ⎜⎟ =⋅ + ⎝⎠ + + == Ω Then 4.16 0.0514 18 1 k FE r k h Ω + Then we want 80 0.95 81 0.0514 o E iE i R iR + or 0.9619 0.0514 E E R R = + which yields min 1.30 E Rk and () ( ) 81 0.7 0.5 1.3 0.7 min 1.36 80 EE VI R V V ++ =+ = + = EX12.10 Use the configuration shown in figure 12.20. 500 , 200 SL RR =Ω =Ω Let 1 11 5 F R R += For example, let 1 2 K 28 K F R R = = EX12.11 a. 2 12 2 20 10 5 10 5 20 30 1 V 1 5 G G SG S S Dn G S T N S R V V VV V IK V V R =− =− − −− 2 15 1 . 5 0 . 4 2 GS GS
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() 2 40 . 64 4 GS GS GS VV V −= + 2 0.6 1.4 1.6 0 GS GS −− = () ( ) 2 1.4 1.4 4 0.6 1.6 20 .6 3.174 V GS GS V V ±+ = = ( ) ( ) 22 1 . 5 3 . 1 7 2 3.52 mA / V mn G S T N m gK V V g =− = = 0 0 2 3.52 1.76 D mg s g s DL gs R I gV V RR IV ⎛⎞ ⎜⎟ ++ ⎝⎠ 1 i ig sm g s S g s mS V VV g VR V g R =+ = + ( ) 0.4153 1 3.52 0.4 i g si V == + ( ) 0 0 1.76 0.4153 0.731 mA / V f i I A V b. For 2 1/ n K mA V = From dc analysis: ( ) 2 41 0 . 4 2 GS GS 2 . 44 4 GS GS GS V + 2 0.4 0.6 2.4 0 GS GS = () () 2 0.6 0.6 4 0.4 2.4 .4 GS V = 3.31V GS V = 2 1 3.31 2 2.623 mA / V m g = 0 2 2.623 1.311 g sg s I + ( ) 0.488 12 . 6 2 30 . 4 i gs i V + ( ) 0 1.31 0.488 i I V
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0 0.6398 mA / V gf i I A V == 0.731 0.6398 % change 12.5% 0.731 = EX12.12 Use the circuit with the configuration shown in Figure 12.27. The LED replaces . L R 3 1 10mS 10 10 100 gf E E AR R × = EX12.13 dc analysis: 00 10 4.7 47 20 D VV I =+ + (1) () 2 Dn G S T N IK =− (2) 20 0.2985 20 47 GS V ⎛⎞ ⎜⎟ + ⎝⎠ (3) ( ) 0 2.13 0.213 0.0149 2.13 0.2279 D D VI V IV −= + (1) From (2): ( ) 2 2 0 2 2.13 0.2279 1 0.2985 1.5 2.13 0.2279 0.0891 0.8955 2.25 0.0891 0.6676 0.12 0 V ⎡⎤ ⎣⎦ + −+ = ( ) 2 0 0 0.6676 0.6676 4 0.0891 0.12 2 0.0891 7.31 V V V ±− = = 10 7.31 7.31 0.572 0.109 4.7 67 0.463 mA D D I I = = 0.463 1.5 2.18 1 GS V = a.
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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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Ch12p - Chapter 12 Exercise Solutions EX12.1 a. A 1 + A A A...

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