Electronics and Communication Engineering - Analog Electronics - Discussion

Discussion Forum : Analog Electronics - Section 1 (Q.No. 6)
6.
An RC coupled amplifier has an open loop gain of 200 and a lower cutoff frequency of 50 Hz. If negative feedback with β = 0.1 is used, the lower cut off frequency will be
about 50 Hz
about 5 Hz
about 2.38 Hz
about 70.5 Hz
Answer: Option
Explanation:

New lower cutoff frequency = .

Discussion:
12 comments Page 1 of 2.

Ankush said:   1 decade ago
Why this formula?

Rahul said:   1 decade ago
fc/(1+B*Af).

Vini said:   1 decade ago
It should be higher frequency formula, coupling reduce.

Shaz said:   1 decade ago
At i/p side for CE RC amplier there is a series feedback. I/p resistance must increases by (1+AB).

Swathi said:   10 years ago
Yes @Shaz its absolutely right.

fl' = fl(1+A(beta)).

Approximately answer is 1050.

Chenelyn Cedron said:   9 years ago
Besides having the formula F{new}= f{0}/ (1 + AB), there is another way around.

Remember that we have this Gain-Bandwidth Product. It means that the product of the gain and the bandwidth is always constant. In formula A * BW = constant. Or, BW = {constant}/{A}, where A is gain and BW is bandwidth. So as you can see, BW is inversely proportional to gain A.

If you have a negative feedback, the gain is always less than the open loop gain. It implies that the bandwidth will increase (inversely proportional remember) to compensate for the reduction of the gain.

Now notice from the choices that only letter C is less than 5 Hz. Since 5 Hz is the lower cutoff frequency, an increase in BW means a new frequency less than 5 Hz. Therefore, it is the only choice that indicates an increase in BW.
(2)

Vijay said:   8 years ago
Thank you for giving the explanation.

Pallabi said:   8 years ago
Thank you all for the explanation.

Praveen said:   8 years ago
Thank you all for the explanation.

Ashfaq said:   8 years ago
Thanks to all.


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