Question

An amplifier without feedback has a voltage gain of 50, input resistance of 1 k\(\Omega\) and output resistance of 2.5 k\(\Omega\). The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2, is

• A. \(\frac{1}{11} \text{ k}\Omega\)
• B. \(\frac{1}{5} \text{ k}\Omega\)
• C. \(5 \text{ k}\Omega\)
• D. \(11 \text{ k}\Omega\)

Hint:

Remember the simple rule for feedback topology vocabulary: The first word describes the output sampling (Voltage/Current) and the second word describes input mixing (Series/Shunt). Whenever you see Shunt at the input, it always implies parallel division, which drops the input impedance: \(R_{if} = \frac{R_i}{1+A\beta}\). If it says Series at the input, it multiplies: \(R_{if} = R_i(1+A\beta)\).

Answer 1

The Correct Option is A

Concept: Feedback topologies deeply alter the input and output impedances of electronic amplifiers depending on how the signal is sampled at the output and mixed at the input:
• Series Mixing: Increases the input impedance by a factor of \((1 + A\beta)\).
• Shunt Mixing: Decreases the input impedance by a factor of \((1 + A\beta)\). In a current-shunt feedback architecture, the feedback network samples the output current (series configuration at output) and mixes a feedback signal in parallel at the input terminal (shunt mixing). Because it uses shunt mixing at the input, the effective input resistance is substantially reduced.

Step 1: Identify given open-loop and feedback parameters.
From the problem statement, we isolate the basic structural properties of the open-loop amplifier:
• Open-loop gain (\(A\)) = \(50\)
• Open-loop input resistance (\(R_i\)) = \(1 \text{ k}\Omega\)
• Feedback fraction (\(\beta\)) = \(0.2\)

Step 2: Calculate the feedback desensitivity factor.
The amount of feedback, or the desensitivity factor, is defined by the core term \((1 + A\beta)\). Let's evaluate this value precisely: \[ A\beta = 50 \times 0.2 = 10 \] Adding unity to find the full modifier factor: \[ 1 + A\beta = 1 + 10 = 11 \]

Step 3: Compute the modified input resistance with shunt feedback.
As established by feedback network topology theorems, parallel (shunt) connection of the feedback signal at the input node acts to reduce the input resistance. The formula for input resistance with shunt negative feedback (\(R_{if}\)) is: \[ R_{if} = \frac{R_i}{1 + A\beta} \] Substituting our known values: \[ R_{if} = \frac{1 \text{ k}\Omega}{11} = \frac{1}{11} \text{ k}\Omega \] This explicitly aligns with option (A).