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β of the BJT is 100 and from dc analysis it is found that the quiescent point is ICQ=4.12mA and VCE = 6.72V. If the emitter resistor RE is subdivided into RE1=100 Ω and RE2=900 Ω, with bypass capacitor in parallel with RE2. How to calculate the answers, How this circuit is affected by this change?

Ans: By Ac equivalent circuit, I have calculated input impedance Zi=(beta+1)re || (R1||R2) = 6.770Kohm/5.36Kohm = 1.263 Kohm

Where re is the dynamic emitter resistance = 26mv/Ie and to calculate Ie, I did DC analysis and by Thevenin equivalent ckt (Vth=5/15*15 =5V & Rth=R1||R2= 3.33 Kohm) then by KVL , Ib=Vth-Vbe/Rth+(beta+1)*1Kohm= 0.127microA therefore Ic=100*Ib=1.27mA and Ie=Ib+Ic=1.291mA. so re=26mv/1.291mA=20.13ohm

then output impedance is Zo=ro||Rc...... by ac equivalent circuit ro is not given so I assumed ro>> very high therefore Zo=Rc=1Kohm

Av= -betaIbRc/Ib(beta+1)re assuming beta+1 ~ beta Av= -Rc/re = -49.67

I am stuck with Ai(current gain) calculation and Power gain calculation now

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From the problem description looks like two problems. First with Re fully bypassed and Second with only part of Re bypassed. Start with fully bypassed. re=.026/4.12e-3=6.25 Ohms making hie about 635 Ohms. This will make Zi (looking to the right of the coupling capacitor) 635||5k||10K~533 Ohms and not what you calculated. Vs sees 500 Ohms in series with Zi, a voltage division of Zi/(500+Zi)--> 0.53*Vs at the base of the transistor. This makes base current Vb/hie=Vb/((beta+1)*re) and collector current -Beta times this or ic~-vb/re=-0.53*Vs/6.25=-Vs*84.8e-3 and the voltage transfer function H=ic*Rc=-84.8*Vs

So comparing your results with these I see 1)re incorrect and 2) voltage gain is from vs, not vb so that there is a voltage division factor missing in your Av equation.

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  • \$\begingroup\$ Ok. Thank you for your clarification. How should I proceed and calculate Power gain and Current gain in this situation? \$\endgroup\$ – Prajakta Bhutkar May 15 '20 at 6:41
  • \$\begingroup\$ Proceed in the same manner. Define what you want. Current gain is by definition iout/iin, etc. Write nodal relations and eliminate intermediate variables. Simplify result to conditions asked for... \$\endgroup\$ – Agustin Ochoa May 16 '20 at 15:23
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You want an estimate? OK - here is it (based on Ic=4mA and beta=100):

1.) From this: Voltage gain (ref. to the base) A=-gm*Rc||RL=(Ic/25mV)0.6k=-96

2.) Input resistance rin=R1||R2||rbe

with rbe=beta/gm=625 ohms (beta=100 assumed) >> rin=3.3k||0.625k=0.53k

3.) r_out=1k

4.) Total gain: Ao=-96[0.53/(0.53+0.5)]=-49,3

5.) Simulation results (BC108, B=200): rin=1.3k ; Gain A=-105; Total gain Ao=-70

Comment: Without knowing in advance the collector current, we could start at the base divider (and ignoring the small unknown base current) with Vb=15(5/15)=5V and Ve=5-0.7=4.3V. This will give us an emitter current Ie=4.3mA.

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  • \$\begingroup\$ A transistor circuit like that at fairly high levels distorts the signal like crazy. The top of the waveform will be squashed (even when it is far from clipping) making the voltage gain difficult to measure., \$\endgroup\$ – Audioguru Jan 21 at 18:45
  • \$\begingroup\$ Yes - but the question was not if the quality of the output signal would be good or bad... \$\endgroup\$ – LvW Jan 22 at 8:51

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