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

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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