HW#3: Apart from Cei and Ce2 (whose values are mentioned), all coupling/bypass capacitors are large enough in the circui
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HW#3: Apart from Cei and Ce2 (whose values are mentioned), all coupling/bypass capacitors are large enough in the circui
HW#3: Apart from Cei and Ce2 (whose values are mentioned), all coupling/bypass capacitors are large enough in the circuit. For the BJTs, ß-150 is given (V₁-25.9mV @T-300K). 150 +Vcc=15V Ro2)18 VBE=0.7V can be used in DC. RCR11k R21 >ΕΩ C $180 12ΚΩΣ +10 ΚΩ HF 10μF T₁ T -2 R22 R12 RL a) Calculate Ici and Ic₂ at DC. b) Determine the of type 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 Vg frequencies. Also determine the open-loop resistances R₁ and Ro. 18 156kQ2 ΚΩ Ro,f Rif RF 10kΩ (note: Vel: AC emitter voltage of T₁) c) For closed-loop, find the closed-loop gains Avk,vo/vg and Avvo/vel for medium frequencies. Then, determine the low frequency poles of Avk,f(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 Cex-1μF. So, Avk,f(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 Avk,f(S)=vo/vg due to this coupling capacitor. Cel Rg 22µF PWHE 10092 RE1 1.2 ΚΩ] 15 RE2 ΚΩ] 820 22 HH
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