Solution for Question 2 : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper II

Q.2. (a):  Calculate Ri, Av and Ai  for the circuit shown.  Use hie = 100 Ohms, hfe = 99, hie = hoe = 0.

Solution:  The ac base current through ri (hie) is

Ib = (Vi - V0). ri

Since, V0 = (1 + hfe).Ib.RE

Substituting V0 in Ib,

Ib.ri = Vi - V0 = Vi - (1 + hfe).Ib.RE

=> Vi = Ib (ri + (1 + hfe).RE) ~= Ib (ri + hfe.RE)

hence, the ac input impedance looking in to the transistor base is -

 Vi/Ib = (ri + hfe.RE)

=>  Ri  = RB || (ri + hfe.RE)

Here RB  = 0,

Ri  = (ri + hfe.RE) = 1000 + (100)(1K) = 101 KOhms

Now, 

Av = V0/Vs = 1  (approximate)

and  

Ai = hfe1 . hfe2 . (Zin.rE / Zin(base).RL)

=>   Ai = hfe1 . hfe2       ....(approximate)

=>      Ai = 10e+6

Q.2. (b):  An amplifier with open loop voltage gain A = 1000 ± 100 is available. It is required to have an amplifier whose gain varies by no more than ± 0.2 %. Find (i) reverse transmission factor B of the feedback network (ii) gain with feedback. Derive the formula you used.

Solution: The given data is-
            open loop voltage gain, A = 1000 ± 100 
            variation in gain = ± 0.2 %
i.e.

We know, the relation between the variation in gain after and before application of the feedback can be expressed as-

from given data, it emplies-

hence, we can obtain value of  reverse transmission factor as - 

Now, to find the gain with the effect of the feedback as -