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Maximum Vp for undistorted output (1 kΩ/33 kΩ)

Short answer

Gain = 33. For undistorted output, 2πf·33Vp ≤ SR ⇒ Vp ≤ 0.6×10⁶/(2π×1000×33) = 2.89 V. For Vp = 5 V, reduce Rf to ≤ 19.1 kΩ (gain ≤ 19.1) or use an op‑amp with SR ≥ 1.04 V/µs.

In a closed-loop inverting configuration of an op-amp, the input resistance and the feedback resistance are 1 KΩ and 33 KΩ respectively. (i) Determine the value of Vp for an undistorted output if an input to the circuit is Vin = Vp sin(2πft), where f = 1 kHz and slew rate = 0.6 V/μs. (ii) If the value of Vp is assumed to be 5V, state two possible remedies in the circuit to have an undistorted output.

Given: inverting amplifier, R1 = 1 kΩ, Rf = 33 kΩ, f = 1 kHz, SR = 0.6 V/µs.

Closed-loop gain |Av| = Rf/R1 = 33
Output: vo = -33 Vp sin(2πft),  peak Vm = 33 Vp

(i) Maximum Vp for undistorted output

For a sine wave the maximum slope of the output is 2πf·Vm. To avoid slew-rate distortion:

2πf·Vm ≤ SR
Vm(max) = SR/(2πf)
        = 0.6 × 10⁶ / (2π × 1000)
        = 95.49 V
Vp(max) = Vm/33 = 95.49/33 = 2.894 V

Answer: Vp ≤ 2.894 V (about 2.89 V) as far as slew rate is concerned.

Note: with a ±15 V supply the output cannot exceed about ±13 V anyway, so in practice clipping (saturation) limits Vp to about 13/33 ≈ 0.39 V before the slew-rate limit is reached.

(ii) Vp = 5 V: remedies

With Vp = 5 V, Vm = 33 × 5 = 165 V and the required rate is 2π × 1000 × 165 = 1.037 V/µs > 0.6 V/µs, so the output is distorted. Two remedies:

  1. Reduce the closed-loop gain: gain ≤ 95.49/5 = 19.1, i.e. Rf ≤ 19.1 kΩ (with R1 = 1 kΩ); in practice the gain must also be small enough to keep the output within the supply (e.g. gain ≤ 2.6 for ±13 V swing, Rf ≤ 2.6 kΩ).
  2. Use an op-amp with a higher slew rate: SR ≥ 1.037 V/µs (e.g. LF351, 13 V/µs).

Other options: reduce the input amplitude (attenuate before the amplifier) or reduce the frequency.

✓ Answer checked by Claude Opus 5.5 (high effort)

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