Question

For an op-amp having a slew rate of \(4 \text{ V}/\mu s\), what is the maximum closed-loop voltage gain that can be used when input signal varies by \(0.1V\) in \(10\mu s\)?

• A. \(40\)
• B. \(4\)
• C. \(400\)
• D. \(4000\)

Hint:

For slew-rate based op-amp questions, always use: \[ \text{Slew Rate}=A_v\left(\frac{dV_i}{dt}\right) \] Thus, \[ A_v=\frac{\text{Slew Rate}}{dV_i/dt} \] The calculated gain gives the maximum value possible without introducing slew-rate distortion.

Answer 1

The Correct Option is C

Concept: The slew rate of an operational amplifier defines the maximum rate at which the output voltage can change with time. Mathematically, \[ \text{Slew Rate}=\frac{dV_o}{dt} \] For distortion-free amplification, the rate of change of output voltage must never exceed the slew rate specification of the op-amp. If: \[ A_v=\text{closed-loop voltage gain} \] then: \[ V_o=A_vV_i \] Differentiating with respect to time: \[ \frac{dV_o}{dt}=A_v\frac{dV_i}{dt} \] Therefore: \[ A_v=\frac{\text{Slew Rate}}{dV_i/dt} \] This relation is extremely important in determining the maximum allowable gain without slew-rate distortion.

Step 1: Write the given data. Given: \[ \text{Slew Rate}=4\text{ V}/\mu s \] The input signal changes by: \[ 0.1V \] in: \[ 10\mu s \]

Step 2: Calculate the rate of change of input voltage. The rate of change of input voltage is: \[ \frac{dV_i}{dt}=\frac{0.1}{10\mu s} \] \[ \frac{dV_i}{dt}=0.01\text{ V}/\mu s \]

Step 3: Use the slew-rate relation to determine the maximum gain. We know: \[ \frac{dV_o}{dt}=A_v\frac{dV_i}{dt} \] For maximum undistorted output: \[ \frac{dV_o}{dt}=\text{Slew Rate} \] Hence, \[ A_v=\frac{\text{Slew Rate}}{dV_i/dt} \] Substituting the values: \[ A_v=\frac{4}{0.01} \] \[ A_v=400 \]

Step 4: Write the final answer. Therefore, the maximum permissible closed-loop voltage gain is: \[ \boxed{400} \] Hence, the correct option is: \[ \boxed{(C)\ 400} \]