How to estimate the voltage gain, Input impedance, current gain, power gain, and output impedance for the amplifier in the fig?

Question

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

Answer 1

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.

Answer 2

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.