Question

A negative feedback amplifier has an open-loop gain of 100 and a feedback factor \(\beta = 0.09\). If the open-loop gain changes by 10% due to aging, what is the percentage change in the closed-loop gain of the amplifier?

• A. \(1 \ \% \)
• B. \(10 \ \% \)
• C. \(0.1 \ \% \)
• D. \(0.9 \ \% \)

Hint:

Negative feedback reduces gain sensitivity by the desensitivity factor: % Change in \(A_f\) = \(\frac{\text{\% Change in } A}{1 + A\beta}\). It instantly simplifies feedback sensitivity problems.

Answer 1

The Correct Option is A

Concept: Negative feedback is widely used in amplifier design because it stabilizes the overall gain against changes caused by temperature fluctuations, aging, or component variations. The mathematical expression for the gain of an amplifier with negative feedback is: \[ A_f = \frac{A}{1 + A\beta} \] Where:
• \(A_f\) represents the closed-loop gain with feedback.
• \(A\) represents the open-loop gain of the internal amplifier block.
• \(\beta\) represents the feedback factor of the attenuation loop.
• \((1 + A\beta)\) is the desensitivity factor of the feedback network. To understand how variations in \(A\) affect \(A_f\), we differentiate \(A_f\) with respect to \(A\), which yields the sensitivity relationship: \[ \frac{dA_f}{A_f} = \frac{1}{1 + A\beta} \cdot \frac{dA}{A} \] This expression states that the percentage variation observed in the closed-loop gain is equal to the percentage variation of the open-loop gain divided by the desensitivity factor.

Step 1: Extracting given parameters and computing the desensitivity factor.
The problem provides the following initial conditions:
• Initial open-loop gain, \(A = 100\)
• Feedback factor, \(\beta = 0.09\)
• Percentage change in open-loop gain, \(\frac{dA}{A} \times 100\% = 10\%\) Let us calculate the desensitivity factor \((1 + A\beta)\): \[ 1 + A\beta = 1 + (100 \times 0.09) = 1 + 9 = 10 \]

Step 2: Calculating the percentage change in closed-loop gain.
Substitute the calculated desensitivity factor and the given open-loop percentage change into our sensitivity equation: \[ \frac{dA_f}{A_f} \times 100\% = \frac{1}{1 + A\beta} \cdot \left( \frac{dA}{A} \times 100\% \right) \] \[ \frac{dA_f}{A_f} \times 100\% = \frac{1}{10} \cdot (10\%) = 1\% \] Thus, the resulting percentage change in the closed-loop gain is exactly \(1\%\). This corresponds perfectly to Option (A).