Question:
If an amplifier with gain of \(-1000\) and feedback of
\[
\beta=-0.1
\]
has a gain change of \(20%\) due to temperature, calculate the change in gain of the feedback amplifier.
If an amplifier with gain of \(-1000\) and feedback of
\[
\beta=-0.1
\]
has a gain change of \(20%\) due to temperature, calculate the change in gain of the feedback amplifier.
Show Hint
Negative feedback reduces gain sensitivity by the factor
\[
1+A\beta.
\]
A large loop gain greatly improves amplifier stability.
Updated On: Jun 25, 2026
- \(0.2%\)
- \(0.1%\)
- \(0.001%\)
- \(0.0001%\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
For a feedback amplifier,
\[
A_f=\frac{A}{1+A\beta}.
\]
The sensitivity of gain to parameter variations is
\[
\frac{\Delta A_f}{A_f}
=
\frac{1}{1+A\beta}
\frac{\Delta A}{A}.
\]
Step 1: Calculate the loop gain.
\[ A=-1000 \] \[ \beta=-0.1 \] \[ A\beta = (-1000)(-0.1) = 100. \] Therefore, \[ 1+A\beta = 101. \]
Step 2: Apply sensitivity relation.
Given \[ \frac{\Delta A}{A}=20%. \] Hence, \[ \frac{\Delta A_f}{A_f} = \frac{20}{101}%. \] \[ = 0.198%. \]
Step 3: Approximate the value.
\[ \boxed{ \frac{\Delta A_f}{A_f} \approx0.2% } \] Thus the gain variation after feedback is only \(0.2%\). Therefore the correct option is \[ \boxed{(A)\;0.2%} \]
Step 1: Calculate the loop gain.
\[ A=-1000 \] \[ \beta=-0.1 \] \[ A\beta = (-1000)(-0.1) = 100. \] Therefore, \[ 1+A\beta = 101. \]
Step 2: Apply sensitivity relation.
Given \[ \frac{\Delta A}{A}=20%. \] Hence, \[ \frac{\Delta A_f}{A_f} = \frac{20}{101}%. \] \[ = 0.198%. \]
Step 3: Approximate the value.
\[ \boxed{ \frac{\Delta A_f}{A_f} \approx0.2% } \] Thus the gain variation after feedback is only \(0.2%\). Therefore the correct option is \[ \boxed{(A)\;0.2%} \]
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