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

This solution manual was provided by Maria Geven at Assam University for Electrical Circuital Analysis course. It includes: DC, Biasing, Minimum, Collector, Voltage, Parameters, Condition, Open, Circuit, Load

Typology: Exercises

2011/2012

Uploaded on 07/06/2012

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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