Solution Manual Electric Circuit Analysis and Design ch 6, Exercises of Electronic Circuits Analysis

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Electronic Circuit Analysis and Design, 2™ edition Solutions Manual Chapter 6 Exercise Solutions 56.1 Su 2K, (Vor -Vry) and 15> K,(Vos—Vew) 0.75 =O.5(Veg —08)" = Veg = L025 V fq = 2{05)(2.025-08) => g,=122 mA/V os 1, = [4K Vor —Vre)| =[(oo1xo5\2025-08)'] = 3M r, E6.2 8, = 2K (Var Vow) and ty © Ky(Var —Vin 145 = coors) => we 80.6 L L = [2K Wor —You} | = [Mao] 7, =((0.015\2)]"' =», =3330 563 8 tag = KVas Vin)” 0.4 = 0.5(Ves — 27 = Vos = 2.89 V Vasa = Von - log Rp = 10 — (0.4)(10) => Vosg = 6 V 0. 8_=2K (Vag — Vow} = 2{05\(289~2) > gen = 0.89 MALY ro = (Mpg), 4 =0 > ro S00 w= =~ gn Ra = —(0.39)(10) mA, 2 cys Ossinut & ve, = ~(8-91(0-4)sinwe Ya, = 3.56 sinwt (AD Vpsy 26-436 = 2.44 Ves1 = 2.89404 = 3.29 Vos! ~Viy = 329-2 = 129 So Vo > Veg, —Vny 2% Biased in saturation region E64 a. Vsog = Veo ~ [ogo T= 12 — Ipal6) + fog = 0.833 mA f Fog = K,(Vse -Wrel) 0.833 = 1Vse - 1) = Vso = 1.65 ¥ Boba = 2K (Mog Wel) =202KL05-1) Be gm S26 mA/V. to = vo. Avs T= ~9mRo = ~(2.6)(6) > Ap = -15.6 E65 f, Tog = Falta ~Vrw)? Vor “Vo - (= f, 8a = 2K (Vas Vou) = 2K, So gn =2YKilog 153