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Slew-rate check for 40 dB amplifier at 20 kHz

Short answer

Gain of 40 dB is 100, giving 10 V peak output. Required slew rate 2π×20 kHz×10 V = 1.257 V/µs exceeds the 1 V/µs SR, causing distortion. Remedy: use an op-amp with SR ≥1.26 V/µs or reduce gain to ≤79.6 (output 7.96 V peak).

An amplifier using an op-amp with a slew rate SR = 1V/μs has a gain of 40 dB. Peak value of 100 mV, 20 KHz sinusoidal wave is applied to the input of amplifier. Determine whether the output will be distorted due to slew rate limiting of op-amp. If so, find remedies.

Given: SR = 1 V/µs, gain = 40 dB, input = 100 mV peak, f = 20 kHz.

Gain: 20 log10 Av = 40 dB  →  Av = 10^(40/20) = 100
Output peak: Vm = 100 × 0.1 = 10 V
Required slope of output (sine wave):
  (dVo/dt)max = 2πf·Vm
              = 2π × 20 000 × 10
              = 1.2566 × 10⁶ V/s = 1.257 V/µs

Since 1.257 V/µs > SR = 1 V/µs, the output will be distorted by slew-rate limiting (the peaks are rounded into a triangular shape and the amplitude falls).

Check with full-power bandwidth: fmax = SR/(2πVm) = 10⁶/(2π × 10) = 15.92 kHz < 20 kHz, confirming distortion.

Remedies

  1. Use a faster op-amp: SR ≥ 1.257 V/µs (for margin, 2 V/µs or more, e.g. LF351 with 13 V/µs).
  2. Reduce the gain: the largest undistorted output at 20 kHz is Vm = SR/(2πf) = 10⁶/(2π × 20 000) = 7.96 V, so gain ≤ 7.96/0.1 = 79.6 (about 38 dB). For example, with R1 = 1 kΩ use Rf ≤ 79.6 kΩ (standard 75 kΩ).
  3. Reduce the input amplitude: at gain 100, Vin(peak) ≤ 7.96/100 = 79.6 mV.
  4. Operate below 15.92 kHz if the signal frequency can be lowered.

Answer: output is distorted (needs 1.257 V/µs > 1 V/µs); use SR ≥ 1.26 V/µs or gain ≤ 79.6.

✓ Answer checked by Claude Opus 5.5 (high effort)

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