If the input is a 1 kHz square wave and the op-amp has finite slew rate, what happens to the output waveform?

• A. Edges remain perfectly square at 1 kHz.
• B. The amplitude remains the same but the edges are slightly slower.
• C. Edges become rounded due to finite slew rate, causing distortion.
• D. The waveform is delayed by a constant phase shift.

Answer: (C)

Slew rate limits are the maximum speed at which an op-amp output can change, defined as dV/dt. A 1 kHz square wave wants instantaneous transitions, which would require an infinite slope. When the op-amp has a finite slew rate, those transitions become ramps with a slope equal to the available SR. The result is edges that are no longer perfectly vertical but rounded, producing distortion of the square shape. The amplitude stays tied to the input and supply rails, so it’s the edges that mainly suffer; if the ramp time becomes a sizable fraction of the half-period, the rounding is even more pronounced. In short, the finite slew rate causes the edges to round off, distorting the waveform.