Solution for Question 7 : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper II

Q.7. (a): A two stage amplifier has the following parameters:

         

                                                      First stage                 Second stage 

   Voltage gain                                      12                               20

   Input resistance                             500 ohms                    80 K ohms

   Equivalent Noise Resistance          1500 ohms                    10 K ohms

   Output Resistance                         25 K ohms                     1 M ohms

   Calculate:

   (i)   the equivalent noise resistance of the two stage amplifier;  

 (ii) the-noise figure of the amplifier if it is driven by a generator with, output impedance 50 ohms. 

Solution: The net resistance of the stage 1 is -

R1 = Input resistance + Equivalent Noise Resistance = 500 + 1500 = 2000 Ohms

Now, for the net resistance of resistance 2, the total input resistance will be parallel combination of input resistance of second stage and output resistance of the first stage.

Hence,

R2 = (80000 || 25000) + 10000 = 29047.6 Ohms

Now, the net resistance for any successive resistance is -

R3 = 1 M Ohms ....i.e. the output resistance of the second stage. 

Hence, the equivalent noise resistance of the two stage amplifier is -

Req = R1 + R2/squr(G1) + R3/squr(G1).squr(G2)

Req = 2000 + 29047.6/144 + 1000000/144x400 = 2219.08 Ohm

Now, lets find the value of noise figure,

Req' = Req - Ri = 2219.08 - 500 = 1719.08 Ohm

Noise figure (F) = 1 + Req'/Ra = 1 + 1719.08/50 = 34.38 = 15.36 dB.

Q.7. (b): Explain Neutralization and show how it can be realized.   

The circuit shown has an internal and stray wiring capacitance of 20 pF. If L1 = 80 mH and

L2 = 120 mH, determine to what value the neutralizing capacitance CN should be set so as to neutralize Cinternal.

Solution: