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3. In Figure 3, the DC operating point for BJT Q, is Ici = 170 uA and VCE = 4 V, and the DC operating point for BJT 02 is Ic2

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1+ .) * (1 + 8 mm) = 1+/0.10726) x (1 + 29. 25.62 29.2 1.008 N Hence Vo = 14:28 -171476 X 0115143 X 1.008 4.25tT ( CRI=lk) -1so that Ib = 0. Do. kot otvo 4. YB +D ID I-BI Rout Roll R31101 clearly BI b 2 = 0 In pre The – Q3 +D Im Roll23/1m) BIbn-] VoAlso -Brez Toy - R G D Jb2 -BI-I = 0 = -13 Tb (reat R6) + B Ib Rb +IR6 Dinde through out by Rb we get I = ß Ibz (1+rez - BIbiRI Jo I WoM ovo R310|| 19 YB I pter Bb. Rollkilla D) Beton 2. = -- (small signal equivalent circult.) We have Vo = -BI b 2 (RFor a Given B = 100, VA = 75rolt; VCE = 4 volt, Id = 17OMA. IE, = ( B+)) Ic, = (1 + ton X470 MA IE,= 176674A i emitter resist- rez=26mV = 26 IE2 01 2424 - 107.2bln frez= 107,2614 in output resutance for 02, Toa= VA = 75 kn Ira 0.24 - 312.5kn pod = 31oy B Jb2 = Bitby R311 input Circuit, & from the We have V = VIX Rill kill pres Bill koll på + RI But Hence Vi= pre, Ibl. w =

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