Question = 1. Consider the CE amplifier under the following conditions: Rsig = 10012, R2 = 33KS2, R2 = 22KN, Rc = 4.7k 2, RL = 5.6k12, the DC emitter current le = 0.3mA with B = 120, Cei = Cc2 = 1uF, and CE = 20uF. Find (6-points) Vcc T w اله R اللي Ro C2 OV Rsig Cct RL 4 Vsig R2 CE W1 RE wth (a) fp1 (6) fp2 (c) fpE (d) fl (estimation)

V2HEW4 The Asker · Electrical Engineering

Transcribed Image Text: = 1. Consider the CE amplifier under the following conditions: Rsig = 10012, R2 = 33KS2, R2 = 22KN, Rc = 4.7k 2, RL = 5.6k12, the DC emitter current le = 0.3mA with B = 120, Cei = Cc2 = 1uF, and CE = 20uF. Find (6-points) Vcc T w اله R اللي Ro C2 OV Rsig Cct RL 4 Vsig R2 CE W1 RE wth (a) fp1 (6) fp2 (c) fpE (d) fl (estimation)
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Transcribed Image Text: = 1. Consider the CE amplifier under the following conditions: Rsig = 10012, R2 = 33KS2, R2 = 22KN, Rc = 4.7k 2, RL = 5.6k12, the DC emitter current le = 0.3mA with B = 120, Cei = Cc2 = 1uF, and CE = 20uF. Find (6-points) Vcc T w اله R اللي Ro C2 OV Rsig Cct RL 4 Vsig R2 CE W1 RE wth (a) fp1 (6) fp2 (c) fpE (d) fl (estimation)
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BOGAF4

8Sol. Civen a CE Amp., R_(r)=100 OmegaB=120 OmegaquadI_(t)=0.3mA.quadC_(c_(1))=C_(c_(2))=1HF", "{:[C_(E)=20MF","S_(0)r_(re)=(26mV)/(IE)=(26(m))/(0.3(m))=86.6mOmega],[R_(Th)=R_(1)||R_(2)=((33k)(22k))/(33k+22(K))=13.2kOmega],[R_(in)=R_(m)11(B+1)r_(e)=(13.2k)11(121 xx86.67)],[R_(in)=5844 Omega],[f_(P_(1))=(1)/(2pi(R_(s)+R_(in))C_(c_(1)))=(1)/(2pi(100+5844)xx10^(-6))],[f_(P_(1))=26.78Hz","] ... See the full answer