AIPMT 2008 Question on Feed back Amplifier

The following question on negative feed back amplifier is simple even though Plus Two students will normally be unprepared to answer it:

The voltage gain of an amplifier with 9% negative feed back is 10. The voltage gain without feed back will be

(1) 10

(2) 1.25

(3) 100

(4) 90

If the voltage gain without feed back is A_v and the feed back factor (fraction of output voltage fed back to the input) is β, the voltage gain (A_fb) with feed back is given by

A_fb = A_v/(1– βA_v)

In the case of negative feed back the sign of the feed back factor β is negative so that the voltage gain with feed back is given by

A_fb = A_v/(1+ βA_v)

Since A_fb = 10 and the magnitude of β is 9% = 0.09, we have on substituting,

10 = A_v/(1+ 0.09A_v)

This gives A_v = 100.

Here is another question on feed back amplifiers:

Pick out the wrong statement:

When negative feed back is applied in a transistor amplifier

(1) its voltage gain is decreased

(2) its band width is decreased

(3) distortion produced by the amplifier is decreased

(4) the transistor current gain is unchanged

The second option alone is incorrect. The band width of a negative feed back amplifier will be greater than that of the same amplifier without the feed back. Since there is a reduction in the voltage gain consequent on the negative feed back, the 3 dB down frequency on the lower side will be shifted towards lower frequency and the 3 dB down frequency on the upper side will be shifted towards higher frequency.