How long a large step takes to settle

The step that is too big found the amplitude at which a real amplifier stops being linear: about eighty millivolts for a part with a megahertz of gain–bandwidth and half a volt a microsecond of slew rate. Below it, a step twice as large produces an output exactly twice as large; above it, the output can only climb at its fixed rate and the step response stops being a scaled copy of itself. The essay located the boundary and drew what crossing it does to the shape. It did not ask what crossing it does to the number most specifications actually quote for a step, which is how long it takes to settle.

That question has a surprising answer. The slew limit, which changes the shape of a large step completely, barely changes its settling time until the step is many times the boundary, and when it does change it, the change has a closed form with a single dimensionless variable in it. This essay marches the rate-limited amplifier for steps from ten millivolts to ten volts, measures how long each takes to settle, and finds a curve that leaves the linear settling time with zero slope.

Flat, then a tangent #

The amplifier is the earlier essay’s: a single pole whose loop crosses at 1 MHz, so the small-signal time constant is $\tau = 1/(2\pi \cdot 1\,\text{MHz})$ = 159 ns, with its output’s rate of change clamped at 0.5 V/µs. The step response is marched by a fourth-order Runge–Kutta rule at sixteen thousand steps, and the settling time is the last instant at which the output is further than a band $\varepsilon$ of the step from its final value, interpolated within its step.

To 0.01 per cent, every step up to 79.6 mV settles in the same 1.466 µs, which is $\tau\ln(1/\varepsilon)$ — the linear answer, independent of size, as a linear circuit’s must be. That boundary is exactly the earlier essay’s: $\mathrm{SR}\cdot\tau$, the step at which the loop’s own initial rate, the step over $\tau$, first equals the limit. Above it the settling time grows: 1.56 µs at 200 mV, 2.01 at 500 mV, 2.90 at a volt, 4.79 at 2 V, 10.65 at 5 V and 20.54 at 10 V. And it grows from the flat line with zero slope, joining the pure slewing line $V/\mathrm{SR}$ from below. The settling time has doubled only at 1.02 V, a step 12.8 times the linear boundary.

Move the slider to a coarser band and the pattern holds with different numbers. To 1 per cent the linear settling time is 0.733 µs, and it has doubled at 608 mV, 7.6 times the boundary.

A delay in front of the same exponential #

The shape comes from what the slewing does to the error.

The 20 mV step never reaches the limit and its error falls as a straight line on the logarithmic scale from the start: the exponential, at the time constant. The 500 mV and 5 V steps first fall linearly in volts — the slew — which on a logarithmic scale is a curve bending downwards, until their error has fallen to $\mathrm{SR}\cdot\tau$. From that moment the loop’s own demanded rate is below the limit, the amplifier is linear again, and the error falls along exactly the same exponential as the small step’s, measured on all three at 159.2 ns to two per cent.

So slewing is a delay placed in front of an unchanged linear settling. A step of size $V$ slews for $(V - \mathrm{SR}\,\tau)/\mathrm{SR}$ and then settles exponentially from an error of $\mathrm{SR}\,\tau$ down to $\varepsilon V$, which takes $\tau\ln(\mathrm{SR}\,\tau/\varepsilon V)$. The total is

$$t_s = \frac{V - \mathrm{SR}\,\tau}{\mathrm{SR}} + \tau \ln\frac{\mathrm{SR}\,\tau}{\varepsilon V},$$

and the marched settling times match it to within a per cent at every step size. The first term grows with the step; the second shrinks with it, because a larger step’s settling band, $\varepsilon V$, is larger in volts, so the exponential tail from $\mathrm{SR}\,\tau$ down to it is shorter. At the boundary the two changes cancel to first order, which is why the curve leaves with zero slope.

The penalty, as a function of one number #

Written in the step over the boundary, $x = V/\mathrm{SR}\,\tau$, the settling time’s excess over the linear value is a function of $x$ alone.

The excess is

$$\frac{t_s - \tau\ln(1/\varepsilon)}{\tau} = x - 1 - \ln x,$$

with no settling band, gain–bandwidth or slew rate left in it. The marched values follow it: 0.095 of a time constant at 1.5 times the boundary, 0.307 at twice, 0.901 at three times, 2.39 at five, 6.70 at ten and 94.4 at a hundred.

Near the boundary the function is $(x-1)^2/2$ — second order. A step fifty per cent past the point at which the amplifier stops being linear costs a tenth of a time constant of extra settling, fifteen nanoseconds here, which no settling measurement would distinguish from the part-to-part spread of the time constant itself. Far beyond it the function is $x$, the pure slewing time in units of $\tau$, and the linear tail becomes the negligible part.

That is the reconciliation between the earlier essay and a data sheet. The earlier essay was right that 80 mV is where linearity ends: the shape of every larger step is different. And a designer who measures settling times is right that nothing much happens at 80 mV: the settling time does not care about the shape, only about when the error last crosses a band, and the slewing delays that moment only by a second-order amount at first.

What a quoted settling time is measuring #

A settling time is usually quoted once, for one step size and one band, and the two choices decide which property of the amplifier it reports.

The share of the settling time spent slewing rises from zero at the boundary towards one for large steps, and where it passes a half depends on the band. To 1 per cent it passes at 332 mV; to 0.01 per cent at 646 mV; to 0.0001 per cent at 979 mV. The exponential tail’s length grows as $\ln(1/\varepsilon)$ and the slewing time does not depend on $\varepsilon$ at all, so a tighter band spends proportionally longer settling linearly and needs a larger step before the slew rate dominates.

So a settling time quoted to a fine band for a moderate step is a statement about the loop — its time constant and, in a real amplifier, its secondary poles and doublets, the things the cliff before the fastest settling found deciding the tail. One quoted to a coarse band for a large step is a statement about the slew rate, and little else. Both are honest; they measure different parts.

Scaling a data sheet’s number #

The practical failure is to take one quoted figure and scale it.

The 10 V step settles to 0.01 per cent in 20.54 µs. Read as a rate — “two microseconds a volt” — it predicts 10.27 µs for 5 V, 4.11 for 2 V, 2.05 for 1 V and 0.21 for 100 mV. The marched values are 10.65, 4.79, 2.90 and 1.47: close at 5 V, forty per cent out at 1 V, and seven times out at 100 mV, where the step never slews at all and settles in the linear time the 10 V figure does not contain. Read the other way — as a constant — the 10 V figure overstates the 100 mV step’s settling time fourteenfold.

Settling time is neither proportional to the step nor independent of it. It is a delay that scales plus a tail that does not, and a single figure measured at one size mixes the two in a ratio that belongs to that size. An analogue-to-digital converter’s driver that sees full-scale steps between channels and small steps between samples of one channel has two settling times, and the data sheet usually gives the one it does not need.

When the band is in volts #

Everything above measures the band as a fraction of the step, which is how data sheets quote it. A converter does not. An analogue-to-digital converter needs its input within half a least-significant bit of the final value, and that is a fixed number of volts whatever the step was: for a 4 V full scale, 488 µV at twelve bits and 30.5 µV at sixteen.

With the band fixed in volts, the formula rearranges into something simpler. Below the boundary the settling time is $\tau\ln(V/\delta)$ for a band $\delta$, which grows with the step, since a larger step has further to fall through the same exponential. Above the boundary it is $(V - \mathrm{SR}\,\tau)/\mathrm{SR} + \tau\ln(\mathrm{SR}\,\tau/\delta)$, and the second term no longer depends on the step at all: every slewing step hands over at the same error and falls through the same number of time constants to the same band. So the settling time is a logarithm below the boundary and a straight line of slope $1/\mathrm{SR}$ above it, and the two meet with the same slope, $\tau/\mathrm{SR}\,\tau$.

For the sixteen-bit case the numbers are 0.92 µs for a 10 mV step, 1.18 for 50 mV, 1.49 for 200 mV, 3.09 for a volt and 9.09 for the full 4 V. At twelve bits the full-scale step takes 8.65 µs and a volt takes 2.65, and the difference between them is exactly 6.00 µs — three volts at half a volt a microsecond — at both resolutions. Four more bits of accuracy cost the full-scale step 0.44 µs, which is $\tau\ln 16$, and nothing else. A multiplexed converter’s worst case is the channel-to-channel full-scale step, and on an amplifier like this one, 7.84 µs of its 9.09 is slewing that no amount of resolution changes.

Which specification to buy #

The closed form also answers a question a designer choosing between two amplifiers actually asks: whether a faster settling time is cheaper to get from more gain–bandwidth or from more slew rate. The two enter the formula in different places. Gain–bandwidth sets $\tau$, which is the whole of the linear tail and a factor in the boundary; slew rate sets only the slewing term and the boundary. And doubling the slew rate at a fixed gain–bandwidth is, by the formula, exactly the same as halving the step: $t_s(V, 2\,\mathrm{SR}) = t_s(V/2, \mathrm{SR})$, because the slewing time and the boundary both scale with $V/\mathrm{SR}$ and nothing else in the expression does.

So the question has a step-size answer, and the numbers are these, to 0.01 per cent. At 100 mV, the original part settles in 1.47 µs; doubling its gain–bandwidth brings that to 0.78, and doubling its slew rate instead leaves it at 1.47, because a 100 mV step with twice the rate is a 50 mV step with the old rate, and that one never slewed in the first place. At 1 V the order reverses: twice the gain–bandwidth gives 2.40 µs and twice the slew rate gives 2.01. At 10 V the gain–bandwidth is nearly worthless — 20.21 µs against 20.54 — while twice the slew rate halves the time to 10.65.

The step at which the two purchases are worth the same is 652 mV to 0.01 per cent, 337 mV to 1 per cent and 985 mV to a part per million, and those are within a few per cent of the steps at which slewing is half the settling time in the figure above. That is not a coincidence of these values. At the step where slewing and settling share the time equally, halving either one’s governing constant saves about the same amount — which is the whole content of the share figure, read as a purchasing rule.

Past that step, more gain–bandwidth approaches a floor it cannot pass. At a volt, a 2 MHz part settles in 2.40 µs, a 10 MHz one in 2.05 and a 100 MHz one in 2.002 — the floor is $V/\mathrm{SR}$ = 2 µs, the time to slew the whole step, and nothing that shortens the linear tail can shorten the slew. The ideal amplifier and its bandwidth treated gain–bandwidth as the number that decides an amplifier’s speed; for large steps it decides only the part of the settling time that is already small.

A threshold decides what is measured #

A settling time is a crossing time, and the threshold chooses which part of the waveform it reports. That is the same observation the threshold a delay is quoted at made about propagation delay, where a delay quoted at half the swing and one quoted at ten per cent describe different things about the same edge. Here a coarse band ends the measurement early, while the slewing is still most of the time elapsed, and a fine band pushes it far into the exponential where only the time constant matters. Neither is the settling time; each is a settling time.

It also explains why the essays on damping found cliffs in settling time that this one does not. Three cliffs and where they are found the settling time of a second-order loop jumping discontinuously as the damping changed, because a ringing response crosses a band several times and the last crossing can move from one peak to the next. A single-pole loop does not ring, so its error falls through the band exactly once, and the settling time is a smooth function of the step. The slew limit adds a delay and removes nothing smooth. With a second pole it would hand over to a ringing tail instead, and the cliffs would return, now moved by the step size as well as the damping — which is the first question left open below.

And it is why the tail is worth checking separately. The slowest one decides it found that when a small, slow tail is split among several pole–zero doublets, the slowest of them decides a fine settling band however small its own share, while the step response’s shape is unchanged to the eye. Nothing about the slewing interacts with that tail: it is linear, it starts where the slewing ends, and it adds to the settling time for every step size alike. A fine-band specification on a real part is therefore two measurements summed — the slewing delay drawn here and a linear tail this model contains in its simplest form.

The march itself is the usual check against a formula, and the reason to trust the law above rather than the figure alone. As in one step, computed twice, the numerical solution is not drawn from the closed form and the closed form is not fitted to the numerical solution; they are computed independently and required to agree, here to a per cent at every step size from ten millivolts to ten volts, and to two per cent on the decay rate of every trace.

How the numbers were obtained #

The amplifier is the earlier essay’s model: the output’s rate of change is the loop’s demanded rate, $\omega_t(v_{in} - v_{out})$ at unity gain with $\omega_t = 2\pi \times 1\,\text{MHz}$, clamped at ±0.5 V/µs. It is marched by the fourth-order Runge–Kutta rule at sixteen thousand steps over 1.6 times the expected settling time plus five time constants; with the clamp removed the same integration reproduces the linear exponential. The settling time is the last sample at which the error exceeds the band, interpolated linearly to the crossing. The closed form is computed from the slew rate, the time constant and the step, and compared, never drawn in place of the march; the share and the doubling step are bisected on the closed form. The decay rate after slewing is fitted on each trace between half the linear boundary and a millionth of the step.

What it leaves out #

A second pole. A real amplifier’s loop has more than one pole, so its linear settling rings and its tail is set by a pair or a doublet rather than a single exponential. The slew-then-settle structure survives — the slewing still ends when the demanded rate falls below the limit — but the linear part it hands over to is the ringing one, and where in the ring the slewing ends changes the tail’s length.

Asymmetric slewing. Many amplifiers slew faster in one direction than the other, so a positive and a negative step of the same size settle in different times, and a settling specification quoted for one polarity is half of one.

Recovery from saturation. A step large enough to drive internal stages to their rails adds a recovery time before slewing begins, which no rate-limit model contains and which can exceed everything drawn here.

Still open: the second pole, the load, and the step the loop overshoots #

Slewing into a ringing tail. With two poles, the moment the slewing ends sets the phase at which the linear ringing starts, and the settling time can then depend on the step size non-monotonically — a slightly larger step ending its slew at a better phase of the ring and settling sooner. Whether that happens for the loop of thirty-five degrees of margin that two measurements of one margin read from both domains is a march with one more state.

A capacitive load. The step too large to have an impedance found a capacitive load’s impedance ceasing to exist above a current fixed by the input pair’s own nonlinearity — the same limit that sets the slew rate. How a load capacitance changes the tail this essay found unchanged by slewing, when the load also takes part of the current the slewing needs, is the combination of the two.

Steps between steps. A converter’s driver sees a sequence of steps whose sizes are random, and its settling error at each sample is the tail of the previous step. The distribution of that error over a random sequence, given the $x - 1 - \ln x$ law, is a closed-form question about the distribution of step sizes.