The gain that stops making a stage faster

The bandwidth the product still gets right put a second pole at $r$ times the closed-loop bandwidth the gain–bandwidth product predicts, and found the half-power bandwidth 41.4 per cent wide at $r = 2$ while the noise bandwidth stayed exactly where the product says it should be. It closed on the third reading of the same second pole, the one a converter’s driver or a sampled system cares about: a wider half-power bandwidth comes with a step that rises faster and overshoots a little, and whether that stage settles sooner or later than the one-pole stage the product describes, and at which $r$, need not agree with either bandwidth.

It does not agree with either, and that is the smaller of the two results here. The larger concerns the product itself. Every statement of the gain–bandwidth product, including the one-line derivation of it, makes a promise that a lower gain gets a faster stage in proportion, and with a real second pole that promise has a floor. Below a definite gain the settling time stops depending on the gain at all. It also stops depending on the gain–bandwidth product, which falls right out of the expression. A stage on that side of the knee is being timed by the second pole alone.

The stage, in one coefficient #

The stage is a non-inverting stage of gain $G$ round an amplifier with a gain–bandwidth product GBW and a second pole at $f_2$. The closed loop is a second-order low-pass:

$$H(s) = \frac{1}{1 + s\tau + s^2\tau^2/r}, \qquad \tau = \frac{G}{2\pi\,\mathrm{GBW}}, \qquad r = \frac{f_2 G}{\mathrm{GBW}}.$$

The whole of the second pole’s effect is in $r$, as the earlier essay found, and time scaled by $\tau$ makes the step a function of $r$ alone. Its damping ratio is $\sqrt r/2$, so $r = 4$ is critical damping, where the two poles coincide, and $r = 2$ is the 0.707 that gave the widest flat band. The poles are

$$s = -\frac{r}{2\tau}\left(1 \mp \sqrt{1 - \frac{4}{r}}\right).$$

Settling is measured from the exact step, as the last time the output is outside a band about its final value, and bisected to the digit it needs. Every number is checked against a netlist: the stage built round a two-pole amplifier model and solved by residues. The two agree within a part or two in a thousand at every $r$ and gain used, which is the floor a transient grid sets on the netlist’s side. The hero figure’s stage at $r = 3$ reads 5.055τ and 5.057τ.

Sooner, and by how much #

The one-pole stage takes $\ln(1/b)$ time constants to come inside a band $b$, which for 0.1 per cent is 6.91τ. That is the number the product implies, and it is what a data sheet’s quick estimate of settling is built from. The second pole, if it were only a lag, could only make things worse. It does not, because with the second pole the rise is faster, and for a time the loop’s overshoot is doing useful work.

At $r = 1$ the stage takes 12.7τ within 0.1 per cent, nearly twice the one-pole time, because it is ringing and each swing has to die away. At $r = 2$, where the half-power bandwidth is at its widest, it takes 7.24τ, which is a little more than one pole’s. So the bandwidth reading and the settling reading already part company at the first number a designer would reach for: 41 per cent more half-power bandwidth is 5 per cent more settling time.

Going further gets better very quickly. At $r = 3$ the time is 5.06τ, at $r = 3.32$ it reaches a best of 3.44τ, and then past $r = 4$ it climbs: 4.62τ at critical damping, 6.06 at $r = 8$, 6.70 at $r = 30$, approaching the one-pole figure from below for as long as the second pole contributes anything. The minimum is half of what the product predicts, a stage that is better than its own model by a factor of two for a reason a designer would ordinarily think of as a defect.

The curve has steps, particularly on the left of the minimum, and the cliff before the fastest settling measured exactly this structure against damping: a ring’s successive peaks each define a settling time until they drop inside the band, so the time jumps by half a ring period at each one. The best damping is not the one to build worked out what that means for a stage whose value is not exactly known, and the same warning carries over, with a difference. There the knob was damping. Here the same fragile optimum is expressed as a position of the amplifier’s second pole, which is the least well-controlled number on its data sheet.

What the best pole is worth #

The minimum moves with the band. For a coarse band, a designer can land the first overshoot just inside it, and a fine band has no such option.

At 10 per cent the best $r$ is 2.43 and the saving is modest, 19 per cent. At 1 per cent it is $r = 2.73$ with a ratio of 0.551. At 0.1 per cent it is $r = 3.32$ and 0.498. At a thousandth of a per cent it is $r = 3.73$ and 0.480, and the trend has nowhere further to go. Two equal poles at $r = 4$ are each at $2/\tau$, twice as fast as the single pole, so a fine band arrives in about half the time. At exactly $r = 4$ it does worse than that, 0.669 at 0.1 per cent, because the double pole’s extra term $(1 + 2t/\tau)$ adds a logarithm. The best position stays a little below four, where the pair’s last swing just stays inside the band, and that is why the ratio sits slightly under a half.

There is a neat way to put that. Holding GBW and $G$ fixed, moving the second pole from infinity to $r = 4$ doubles the settling speed of a fine band. A second pole that is taken as a blemish on the amplifier is, at the right position, doing the same job as a compensation network’s zero. The widest flat band is at $r = 2$, and the fastest fine settling is at nearly twice that, and for a stage that must both pass a band and settle quickly the designer’s target is the higher of the two.

The sum of the poles #

So far the second pole has been placed in units of the closed-loop bandwidth, which is the convenient way to draw a curve and the wrong way to think about an amplifier, since a real amplifier has its second pole fixed in hertz and a designer varies the gain. Written for that case, the real part of a complex pole pair is

$$\sigma = \frac{r}{2\tau} = \frac{f_2 G}{\mathrm{GBW}} \cdot \frac{2\pi\,\mathrm{GBW}}{2G} = \pi f_2.$$

Both $G$ and GBW have cancelled. That cannot be an accident of the algebra, and it is not. Closing a loop around a constant moves the poles of an open loop without changing their sum, because the sum is the coefficient of $s$, which the loop gain does not touch. The open loop has poles at nearly zero and at $2\pi f_2$. Their sum is $2\pi f_2$. Once the closed-loop poles have met and become a complex pair, each has half that sum as its real part, for every gain.

The netlist does not need telling. Fitting the decay of the ring’s extrema at each gain below the knee gives flat lines at 1.57, 4.71 and 9.42 million per second, within four per cent at every combination, and for most of them within a part in a thousand. The result is the same one where the Q comes from and one margin, and the gain under it lean on from other directions, that a pair’s real part is a sum and is the same for every loop gain: the root locus of a two-pole loop is a vertical line, drawn here in the time domain.

The ring’s frequency is what the gain moves. A lower gain puts the pair higher up that vertical line, ringing faster and with less damping, while its envelope falls at the same rate. So a designer who lowers the gain gets a faster first rise and more cycles of ringing inside exactly the same envelope. For settling to a fine band, which is a statement about the envelope, nothing has changed.

The knee, in seconds #

Put that on the axis a designer actually has, gain, and the floor is obvious. A 1 MHz amplifier, with a second pole at 0.5 MHz or at 3 MHz, settling within 0.1 per cent:

With a 0.5 MHz second pole the stage takes 4.10 µs at unity, 4.04 µs at a gain of two and 4.22 µs at five. The product predicts 1.10, 2.20 and 5.50. So at unity gain the prediction is low by a factor of nearly four, and lowering the gain from five to one buys essentially nothing, against a promised factor of five. The floor’s value is what the sum of the poles predicts: $\ln(1000)/(\pi f_2)$ = 4.40 µs, with a little better than that available depending on where the last peak falls.

The knee is at $r = 4$, that is $G = 4\,\mathrm{GBW}/f_2$: a gain of eight for this pole. Above it the closed loop’s poles are real and split, the slower one tends to GBW/G, and the product comes back: 8.51 µs at ten against 11.0, and 108 µs at a hundred against 110. Just past the knee the stage is actually faster than the product says. In the last decade of gain the two curves are indistinguishable, which is the region in which the product is the right description of the amplifier.

With a 3 MHz second pole the knee is at 1.33, so a stage at unity sits barely on the floor, and the product is a good guide everywhere from a gain of two upward. That is the amplifier a stable unity-gain part is designed to be, with its second pole safely above its crossover. The case with the second pole at half the product’s crossover, on the other hand, is what happens when the amplifier is decompensated: a part sold with a larger gain–bandwidth product and a stated minimum stable gain. Its GBW is not available below the knee, whatever the headline figure says, and a user who picks the part for its speed at a gain of two is paying for a number the stage cannot use. And it is not a tiny difference in practice. A decompensated part at a gain of two is timed by $f_2$, and a part with a three times lower GBW and the same $f_2$ settles identically.

What this does to a converter’s driver #

The case the question usually comes from is a converter’s input driver, a gain-of-one or gain-of-two stage whose whole job is to settle between samples. The rule in common use comes from the product: pick the part’s GBW so that the required number of time constants fits the acquisition window. The edge of a chain and its neighbour showed that this reasoning already goes wrong for several stages together. Here it goes wrong for one, and it does so in a way the data sheet’s headline hides.

If the second pole is well above the crossover, the rule is conservative, and it can be very conservative. A stage at $r = 3.3$ needs half the time, so a part with half the GBW would do, provided its second pole is where this says. If it is below the crossover, on a decompensated part used near its minimum stable gain, the rule is optimistic by a factor of four, and more GBW will not help at all. In both cases the number that should be on the data sheet, and is at best given as a phase margin at one gain, is the second pole’s frequency. Everything else about settling can be calculated from that and the product. The Q the amplifier decides found a filter section’s behaviour tied to the gain–bandwidth product alone; a section at low enough gain in a fast enough circuit is timed by this pole as well.

Why a phase margin does not say this #

A second-order loop’s phase margin and its $r$ are one number: the margin is a function of $\sqrt r/2$, and two measurements of one margin checks that identity on an actual loop. So a stage’s margin in principle says where it sits on the first figures here. It does not say the thing that matters in seconds, since a phase margin of 60° at one gain is a different second pole from the same margin at another, and the settling time in seconds depends on the second pole’s frequency and nothing else on this side of the knee. A designer comparing two parts by phase margin at their respective gains has compared them on the wrong axis. Two stages with equal margins, at different gains, can settle at different speeds in proportion to their gains. Below the knee, two stages at very unequal margins settle at the same speed.

What the two-pole model leaves out #

An operational amplifier has further poles and usually a zero, and it slews. The slew limit in particular would replace the small-signal envelope entirely for a large step, and the step that is too big is where that boundary is drawn; everything here is true for steps small enough to stay on the linear side of it. A third pole near the second adds lag and changes both the sum of the poles and the shape of the settling tail, and a doublet would add a slow tail no second-order picture contains, as the tail that settling leaves measures. Both of these make the settling worse in ways this essay’s figures do not show, and neither moves the structure: a second pole’s position still sets the fastest stage and the floor.

The direct-current gain here is high enough to ignore. At a gain of a hundred with an open-loop gain of a million, the final value is a hundredth of a per cent short, which is normalised away so that the settling is measured to the final value. On a stage with only 60 dB of open-loop gain that would be a static error larger than the band, and it would be a different question.

The numbers to carry #

A stage’s closed-loop poles are a complex pair below $r = 4$, and their real part is $\pi f_2$ whatever the gain and whatever the product, so a 1 MHz amplifier with a 0.5 MHz second pole takes about 4 µs to reach 0.1 per cent at every gain from one to five, where the product promises 1.10 to 5.50. The knee is at $G = 4\,\mathrm{GBW}/f_2$. In units of $\tau$, the fastest stage for a 0.1 per cent band has its second pole at $r = 3.32$ and settles in 3.44τ, half the one-pole stage’s 6.91, and the widest flat band at $r = 2$ takes 7.24τ. The half-power bandwidth, the noise bandwidth and the settling time therefore each answer to the second pole differently, and only the last has a floor.

Still open: the third pole, a zero, and a decompensated part used below its gain #

A third pole. Adding a pole changes the sum of the closed loop’s poles by itself, so the floor moves up by half of it, but the dominant pair then no longer takes half of the total. Where the third pole starts to erode the floor, and whether a fast third pole helps settling as a second one does, is the measurement that would say how far the two-pole picture can be used for a real amplifier.

A zero in the loop. A capacitor across the feedback resistor adds a zero, which alters the sum of the closed-loop poles rather than leaving it fixed. This is the standard cure for a decompensated part, and whether it moves the floor or only reduces the ring within the same envelope is exactly the difference between buying time and buying appearance.

A large step. A real amplifier slews before it settles, and a slew limit makes the first part of the response independent of the small-signal poles. Whether a stage on the floor spends its budget in the slew or in the ring, and so whether more gain–bandwidth product helps a large step where it could not help a small one, is the question a converter’s driver actually asks.