Question

An op-amp with slew rate \( 0.5 \text{ V/}\mu\text{s} \) is used to generate a sine wave of amplitude \( 10\text{ V} \). Maximum frequency without distortion is approximately

• A. \(8\text{ kHz} \)
• B. \(16\text{ kHz} \)
• C. \(32\text{ kHz} \)
• D. \(50\text{ kHz} \)

Hint:

To quickly find the full power frequency limit, remember the standard formula: \[ f_{\max} = \frac{\text{SR}}{2\pi V_m} \] Make sure your units match! Slew rates are normally written in \(\text{V/}\mu\text{s}\), so don't forget to multiply the numerator by \(10^6\) or keep track of the micro factor in your final frequency unit computation.

Answer 1

The Correct Option is A

Concept: The Slew Rate (SR) of an operational amplifier describes its maximum possible rate of change of output voltage per unit of time. If an input signal requests an output change faster than this structural limit, the output distorts into a triangular waveform instead of tracking the desired wave shape. The maximum frequency at which an op-amp can deliver an undistorted sinusoidal output voltage with a given peak amplitude is termed the Full-Power Bandwidth. It is governed by finding the maximum derivative of the sinusoidal waveform expression:
• Mathematical representation of a sine wave: \( v(t) = V_m \sin(2\pi f t) \)
• Maximum rate of change: \( \left. \frac{dv(t)}{dt} \right|_{\max} = 2\pi f V_m \)
• Slew rate condition to avoid distortion: \( \text{SR} \ge 2\pi f_{\max} V_m \)

Step 1: Converting the given units to SI standard equivalents.
We are provided with the following parameters: \[ \text{Slew Rate (SR)} = 0.5\text{ V/}\mu\text{s} = \frac{0.5\text{ V}}{10^{-6}\text{ s}} = 0.5 \times 10^6\text{ V/s} \] \[ \text{Peak Amplitude } (V_m) = 10\text{ V} \]

Step 2: Setting up the inequality and isolating the maximum frequency \(f_{\max}\).
To avoid any slew-induced distortion, the highest rate of change of our sine wave must be less than or equal to the operational threshold limits of the op-amp: \[ \text{SR} = 2\pi f_{\max} V_m \] Rearranging this relationship to isolate our target parameter, the maximum distortion-free operating frequency \(f_{\max}\): \[ f_{\max} = \frac{\text{SR}}{2\pi V_m} \]

Step 3: Calculating the numeric values carefully.
Substitute our converted numbers back into the rearranged equation: \[ f_{\max} = \frac{0.5 \times 10^6}{2 \times \pi \times 10} \] \[ f_{\max} = \frac{0.5 \times 10^5}{2\pi} = \frac{50000}{2\pi} = \frac{25000}{\pi} \] Using the approximate mathematical value for pi (\(\pi \approx 3.14159\)): \[ f_{\max} \approx \frac{25000}{3.14159} \approx 7957.7\text{ Hz} \] Converting this value from Hertz into kilohertz (kHz): \[ f_{\max} \approx 7.96\text{ kHz} \] Comparing this result against the options provided, it is approximately equal to \(8\text{ kHz}\). Hence, the correct choice is option (1).