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