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Maximum peak input for undistorted output

Short answer

Slew rate is the maximum output voltage rate of change, SR=(dVo/dt)max. For a sine wave, Vm=SR/(2πf); with SR=1 V/µs and f=20 kHz, Vm=7.958 V. Dividing by closed-loop gain 20 gives maximum peak input 0.398 V.

Define Slew Rate of an operational amplifier. An amplifier has a bandwidth of 20KHz and closed loop gain of 20. Find the maximum peak input signal that can be applied to obtain undistorted sine wave output. Assume SR = 1V/μs.

Slew rate

Slew rate (SR) is the maximum rate of change of output voltage per unit time that an op-amp can produce, SR = (dVo/dt)max, usually in V/µs (0.5 V/µs for the 741). For a sine output Vm sin 2πft, the maximum slope is 2πf·Vm, so the output is undistorted only if 2πf·Vm ≤ SR.

Maximum peak input

Given: bandwidth (highest signal frequency) f = 20 kHz, closed-loop gain = 20, SR = 1 V/µs = 10⁶ V/s.

Maximum undistorted output peak:
  Vm = SR / (2πf)
     = 10⁶ / (2π × 20 × 10³)
     = 7.958 V
Maximum peak input:
  Vin(peak) = Vm / gain = 7.958 / 20
            = 0.398 V

Answer: maximum peak input ≈ 0.398 V (398 mV), giving a 7.96 V peak output at 20 kHz. A larger input would make the output slew-rate limited (triangular) at the top of the band.

✓ Answer checked by Claude Opus 5.5 (high effort)

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