Circuit Response
every model has an edge, and the figure carries it
Series

Slew rate — the series

2 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Five steps, each divided by its own size, from an amplifier limited to 0.50 V/µs. A linear circuit would put these five curves exactly on top of each other. The 20.0 mV step is linear; everything above 79.6 mV is not, and the largest step takes 16.0 µs to travel a distance the linear model says takes 0.159 µs.

    The step that is too big

    A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.

    part 3 · transients
  2. Settling to 0.01% takes 1.47 µs for any step up to 79.6 mV, and twice that only from 1.02 V — the slew limit arrives in the settling time gently. computed by solving, not by drawing, marched and in closed form: an amplifier of 1 MHz gain–bandwidth and a 0.5 V/µs slew rate, at unity gain, settling to 0.01% of its step, against the step's size. Up to SR·τ = 79.6 mV it never slews and settles in τ·ln(1/ε) = 1.466 µs whatever the size. Above it, it slews until its error is SR·τ and settles exponentially from there, in (V − SR·τ)/SR + τ·ln(SR·τ/εV): 1.56 µs at 200 mV, 2.01 µs at 500 mV, 2.90 µs at 1 V, 4.79 µs at 2 V, 10.65 µs at 5 V, 20.54 µs at 10 V. The march and the closed form agree to a per cent. The curve leaves the flat line with zero slope and joins the pure slewing line V/SR (dashed) from below; the settling time has doubled only at 1.02 V, 12.8 times the linear boundary.

    How long a large step takes to settle

    An amplifier of 1 MHz gain–bandwidth and 0.5 V/µs slew rate stops being linear above an 80 mV step. Its settling time does not notice for a long while: to 0.01 per cent it is 1.47 µs for any step up to 79.6 mV, 1.56 at 200 mV, 2.90 at a volt, and twice the linear figure only from 1.02 V, thirteen times the boundary. The reason is that a large step slews until its error is SR·τ and then settles exactly as a small one does, so the extra time is τ(x − 1 − ln x) — second order at the boundary and the whole slewing time far beyond it. A data sheet's 20.5 µs for a 10 V step therefore says almost nothing about a 1 V step, which takes 2.9, and nothing about 100 mV.

    part 4 · transients

All series