Solution Manual Electric Circuit Analysis and Design ch 11, Exercises of Electronic Circuits Analysis

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ic Circuit Analysis and Design, 2™ editi Solutions Manual Chapter 11 Exercise Solutions Ell. Veo - Ve = 2+ 0,008 sin wt — (0.5 ~ 0.005 sin wt} => Gm 15 +0010 sin wt (V) Va Vem = 2 __ 240.005 sin wt + 0.5 - 0.005 sinwt ~ 2 => Vom = 1.25 V E1L2 ve = —Vee(on) > vg = 0.7 ¥ fo =ioz = OS mA ver = ver = 10 ~ (0.5)(10) = VoL = veg = 5 V EIL3 For my = vp = +4 => Minimum vey = ¥e2 = 4V t fer = fers SR = mA Ro = Meh Rowen Elid icz 1 => vg = 118.5 mV ELS & Wend ee SOV Vac = (0.25)(8} = 2 ¥ ove Seq = -3V Sup STV boy em SBS Ve op OV => ype) = 6.2 V © 2.5 V > vp = -L3V = vec = BLL7 Let fg =Lmd, then oq = Top, = 05 md 05 = Sqr = 19.23 ed (PF Sn = Su Qo96 So, 150= 1929) R = Ray = 156 4Q At y, cas atasl iar SFr So, -100 = ~ F092), => Ry, =104kQ TEV* =+10F and Vo =—i0F, de biasing is OK, B18 a. Diff. Gain As = fake 4Vr For vy = = 5 V => Minimum collector voltage vor = 5 V = 4B. Re sis-s=10¥ or Ig Ro = 20 ¥ for max. Aa Then Ag = pt o Aa(max) = 192 <= Spoaay > Aaland = 292 b. [lg = 058 mA, Re = 40 ko nen. act) 1 + 281}L6-5) 09} (6.028)(200) => Acm = = 9.199 and CM RRap = 20log,0 (3) => CM RRep = 59.7 AB