Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 3 (Q.No. 10)
10.
An R-C coupled amplifier has mid-frequency gain of 200 and a frequency response from 100 Hz to 20 kHz. A negative feedback network with β = 0.2 is incorporated into the amplifier circuit, the Bandwidth will be
Answer: Option
Explanation:
B.W. = fH' - fL' .
Discussion:
8 comments Page 1 of 1.
Leendon said:
6 years ago
Right, thanks @Rajashekar.
Ammar said:
6 years ago
In the case of negative feedback, the new bandwidth is given by;
B.W' = B.W (1+AB).
With B.W = 20k - 100 = 19.9k.
putting values we get;
B.W' = 815.9 kHz.
B.W' = B.W (1+AB).
With B.W = 20k - 100 = 19.9k.
putting values we get;
B.W' = 815.9 kHz.
Akshay said:
7 years ago
It's β not B.
Rajashekar said:
8 years ago
Here, B=0.02.
(1)
Amod kumar Jha said:
8 years ago
No, The -ve feedback lower frequency is decreases and higher frequency is increased by the factor of (1+A * B),
So, that BW will increase.
So, that BW will increase.
Aurnox said:
8 years ago
It's bw = { 820 - 2.44 *10pow(3)} khz = 819.99 khz.
Abhi said:
9 years ago
The method is right, but the answer is wrong.
Shailesh said:
1 decade ago
Fl' = 100/41 = 2.43 & Fl' = 41*20K = 820k.
BW = 820k-2.43 = not 100K.
So answer is wrong or method is wrong that given in explanation?
BW = 820k-2.43 = not 100K.
So answer is wrong or method is wrong that given in explanation?
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