Question

The open-loop voltage gain of an amplifier is 240. The noise level in the output without feedback is 100 mV. If a negative feedback with \( \beta = \dfrac{1}{60} \) is used, the noise level in the output will be:

• A. 1.66 mV
• B. 2.4 mV
• C. 4.0 mV
• D. 20 mV

Hint:

Negative feedback reduces output noise by a factor of $(1 + A\beta)$. Just divide the original noise by this value!

Answer 1

The Correct Option is D

The output noise with negative feedback is reduced by the factor $(1 + A\beta)$, where:
$A = 240$, $\beta = \dfrac{1}{60}$
Calculate the feedback factor:
$1 + A\beta = 1 + 240 \cdot \dfrac{1}{60} = 1 + 4 = 5$
Now compute the reduced noise level:
$\text{Output noise with feedback} = \dfrac{100\,mV}{5} = 20\,mV$
Thus, the noise level in the output with feedback is 20 mV.