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Solution Manual Electric Circuit Analysis and Design ch 5, Exercises of Electronic Circuits Analysis

This solution manual was provided by Maria Geven at Assam University for Electrical Circuital Analysis course. It includes: Saturation, Nonsaturation, Region, Transistor, Bias, Point, Load, Driver, Drain, Current

Typology: Exercises

2011/2012

Uploaded on 07/06/2012

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Electroni rd Chapter 5 Exercise Solutions ES.1 (@) Vay =L2V . Ves Vag (sat) = Vag —Vyy = 2-12 508 (i) Vos = 0.4 = Nonsaturation (ii) Vos 3 l= Saturation Git) Vs =5.= Saturation (b) Vy ==12V, Vag 32 Vae(sat) =Vo5 —Vey = 2-(-12)=32V i) Vpy =0.4=9 Nonsaturation Gi) Vo, = 1 => Nonsaturation Gil) Vip =S=> Saturation E52 @) K,= WC. 2k (39)(8.85x10""} =7.67x10* F fom be 450x10* 100)(500}(7.67x10" 4 = WOONSOO} 7.6710") K, 20274 mA /v? 27) TO () Viv =12¥, Voy =2V a) Vy5 = 0.4 V => Nonsaturation £5 =(0274][2(2-12Y04)~(04)'] = Ip = 0.132 mA G@) Vj, =1¥ = Saturation I = (0.274212) = [, = 0175 mA Gil} -V,, =5V = Saturation 1, =(0274\(2-12)' = 1, = 0175 mA Vyy =ALZV, Veg = 2 @ Vp = 0.4 => Nonsaturation 1, (0.274 2(2+12X04)-(04)'] => Ty = 0.658 mA ‘ (ii) Vos =1¥ => Nonsaturation Ins (0.274)2(2+ 12\(1)-()']= to (ili) Vgg = SV => Saturation 1, =(0274\(2+12)' => 1, =281mA ES.3 Vig = Vee = 3V Woe = 45V Vag = 45> Veg( sat) =Vos -Vy = 3-1=2V Transistor biased in the saturation region 1 = K,(Vag —Vpy)" =9 08 = K,(3—1)" => K,=02marv? Solutions Manual (a) Vag =2V, Vog =45¥ Saturation region: 1, = (022-1) => f,=02mA {b) Vas =3V, Vg =i Nonsaturation region: 1, =(02[23-1)0)-()']> fo ES.4 @) Vp S-2¥, Vyg =3¥ Vep(sat) =Voq + ¥yp =3-2=1¥ (i) Vip = OSV 29 Nonsaturation Gi} «Vy, = 2 => Saturation Gi} Vi, =5V = Saturation (b) Vp =O5¥, Wi =3V Vig ( sat) =Voq +Vyp = 3405 = 35¥ fi) Vag = 05 = Nonsaturation (ii) Vig =2V = Nonsaturation Gil) V,) = 5V = Saturation B55 (@) A=0, Vps(sar}=25-08=L7V For Vpg =2V, Vog =i0¥ => Saturation Region 1, = (0125-08) = fy =0.289 mA (b) 4=002V" 1g = KAVax— Ven) (1+ a¥ ng) For Vp5 =2¥ 7, = (01X25~-08)'[1+(0.02\2)] => Ip = 0300 mA V5 = 10¥ 1,5 (01 (25-08) (1+ (om2X10}}] = Ty =O347 mA (c} For part (a), 4-0 ,=00 For part (b), A=0.02V, 6. =[AK,Wes Yoo] =[(ooayonyas- oxy] or 1,217