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Solution Manual Electric Circuit Analysis and Design ch 3, Exercises of Electronic Circuits Analysis

This solution manual was provided by Maria Geven at Assam University for Electrical Circuital Analysis course. It includes: Transistor, Saturation, Cutoff, Active, Power, Nominal, Current, Equivalent, Circuit

Typology: Exercises

2011/2012

Uploaded on 07/06/2012

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Electronic Circuit Analysis and Design, 2 edition Solutions Manual Chapter 3 Exercise Solutions E3.1 0.980 Fora = 0.980, = 98 _ or a oe te 0.995 = 0.995, ¢= 3 For + 8 = Tages = 88 49.6 8 < 199 E33 fe=(l+ ale Se (1+4) = oe = 81.25 > 80.3 4: 212 2 canst Ic = Bla = (80.3)(9.60 gA) = fo = 0.771 mA 0,990 Toa *T-os0 ~ 22% — fe _ 2180 - fe=ay +H) 100 => fg = 21-50 pA fe = alg = (0.990)(2.150) & fo = 213 mA B3.5 Va 50 "Te Te fg =0.1mMA > rp = 1.5 Mi fe = 1.0 mA & rg = 150 kD Te = 10 mA = rp = 1SkR ¥e = (i+ 2) At Ves =1,fe=1 a Va = 73 (1+ x) = fq = 0.9868 mA Ti At¥ee =10. Ie = ca.seesy(1 + ) => ais a) => ly = 0.9934 mA AtVer =10. fe = roomy (2+ +e 150; ic = 1.06 ma E37 Beg = ap a = 40.5 vols E3.8 BV ex, = PM cae vB BV ce, = (U100)(30) = 139 ¥ E39 2 KW =0.2 < Ves(on) > fe = lem, ESV P=0 b, Vi = 3.6 Transistor is driven into saturation. 26-07 . = 26507 =4. Tp = a = Ip 453 mA 3-Veg(sat) 5-02 = So Mes(sat) 2 8202 2p. 210.9 mA fe Re dag Ems Note that 7= fe 103 241 <4 which shows that the mansistor is indeed in saturation. P= icVon+I5Vae = (10.9)(0.2} + (4.53)(0.7) = 218 +917 P2535 mW £3.10 For Vac <0) = 0.7V Then fo = S587 op fe 29.77 mA fe 9.17 ig=>>= = 0.195 and Ja 7 30 9.195 mA ¥; = Je Re + Vee(on} = (0.195)(6.64) + 0.7 = Vy) = 0.825 ¥ Power = leVeg + leVee a (9.77 )(0.7) + (0.195}(0.7) Powe = 8,38 mW