Question

A voltage follower is as shown in figure. Given that the open loop gain of an op-amp (A = 99), then what is the value of the output voltage $V_o$. Assume that the op-amp is ideal except that it is having a finite open loop gain.

• A. $1\text{ V}$
• B. $0.09\text{ V}$
• C. $0.99\text{ V}$
• D. $0.9\text{ V}$

Hint:

An ideal op-amp has $A = \infty$, giving $V_o = V_{in} = 1\text{ V}$.
With finite gain, the output is always slightly less than the input.
The fractional error is approximately $\frac{1}{A} \times 100\% = \frac{1}{99} \approx 1\%$, giving $V_o \approx 0.99\text{ V}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a voltage follower (non-inverting amplifier with unity feedback factor $\beta = 1$) implemented with an op-amp having a finite open-loop gain $A = 99$.
We need to find the actual output voltage $V_o$ when the input voltage is $1\text{ V}$.

Step 2: Key Formula or Approach:
For any feedback amplifier, the closed-loop gain $A_f$ is given by:
\[ A_f = \frac{V_o}{V_{in}} = \frac{A}{1 + A\beta} \]
For a voltage follower, the feedback factor is $\beta = 1$. Thus, the gain formula reduces to:
\[ A_f = \frac{A}{1 + A} \]

Step 3: Detailed Explanation:

• Given parameters:
Input voltage, $V_{in} = 1\text{ V}$.
Open-loop gain, $A = 99$.
Feedback factor, $\beta = 1$.

• Substituting the values into the closed-loop gain formula:
\[ A_f = \frac{99}{1 + 99} = \frac{99}{100} = 0.99 \]

• Calculating the output voltage $V_o$:
\[ V_o = A_f \cdot V_{in} = 0.99 \times 1\text{ V} = 0.99\text{ V} \]

Step 4: Final Answer:
The output voltage is $0.99\text{ V}$, which corresponds to Option (C).