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HW#3: Apart from Cei and Ce2 (whose values are mentioned), all coupling/bypass capacitors are large enough in the circuit. For the BJTS, B-150 is given (V₁-25.9mV @T-300K). Rc2518 →+Vcc=15V 150 RCR12 R₂1 >kΩ Co 180 12k2 +Vo ΚΩ 7719 T₁ 10μF R22 R12< RL VBE 0.7V can be used in DC. a) Calculate Ici and Ic2 at DC. b) Determine the type of feedback, draw the AC open-loop equivalent (with loading effect) and find the open-loop gains Avk-Vo/Vg and Av=Vo/Vel for medium frequencies. Also determine the open-loop resistances R₁ and Ro. (note: Vel: AC emitter voltage of Ti) REL 18 15 RE2 k2 820 56k2 1.22 ΚΩ ΚΩ ¥ HH Rof Rif RF 10K22 c) For closed-loop, find the closed-loop gains Avk,vo/vg and Av.Fro/vel for medium frequencies. Then, determine the low frequency poles of Avkr(S)=vo/vg. d) Now, let's assume that the coupling capacitor connecting collector of T₁ to base of T₂ (Let's name it as Cex) is not large, but has a value of Ca-1μF. So, Avkr(s) vo/vg will have another low frequency pole due to this capacitor. By considering that this capacitor is located inside the feedback loop, calculate the low frequency pole of Avkr(s)=vo/vg due to this coupling capacitor. Cel 22μF Swit |100Ω Rg, T₂