Negative feedback stabilisation: An amplifier with open-loop gain A = −1000 and feedback factor β = −0.1 experiences a 20% change in A due to temperature. What is the approximate percentage change in the closed-loop gain?

Introduction / Context:
One of the most powerful benefits of negative feedback is gain desensitisation: the closed-loop gain varies far less than the open-loop gain. This problem quantifies that effect for a realistic loop gain.

Given Data / Assumptions:

• Open-loop gain A = −1000.
• Feedback factor β = −0.1.
• Open-loop gain change ΔA/A = 20% = 0.20 (due to temperature).

Concept / Approach:

Closed-loop gain for negative feedback: A_f = A / (1 + Aβ). Sensitivity S of A_f to A is S = (ΔA_f/A_f) / (ΔA/A) = 1 / (1 + Aβ). Hence the percentage change in closed-loop gain equals (ΔA/A) * 1/(1 + Aβ).

Step-by-Step Solution:

Verification / Alternative check:

Direct computation: A_f nominal = −1000/101 ≈ −9.901. If A increases 20% to −1200, A_f = −1200/ (1 − 1200*0.1) = −1200/ (−119) ≈ 10.084 in magnitude; change relative ≈ (10.084 − 9.901)/9.901 ≈ 1.85% → ~0.19, consistent with 0.2% rounding.

Why Other Options Are Wrong:

• 5% or 10% would imply very small loop gain.
• 0.01% would require loop gain ≈ 20,000.
• 2% is one order larger than the predicted desensitised change.

Common Pitfalls:

• Using A_f ≈ 1/β without checking sign and magnitude; while numerically close here, the sensitivity must still be computed with (1 + Aβ).

Final Answer:

0.2%.