The slowest one decides it

The cancellation that leaves a tail measures one pole nearly cancelled by one zero and reduces the whole of it to a closed form: the settling time is $\tau_s\ln(A/B)$, containing neither the fast circuit nor the doublet’s separation as such. Three things are read off that expression and two of them are comforting.

The fast circuit does not appear, so a stage that is not fast enough cannot be fixed by making it faster. That one is the essay’s headline and it is uncomfortable.

The dependence on the miss is logarithmic, so halving the mismatch buys $\tau_s\ln 2$ and no more — “real, but not the order-of-magnitude win the trim looks like it should be.”

And the boundary is blunt: if the tail is already smaller than the band, the logarithm is negative and the doublet costs nothing at all.

All three are statements about one time constant. The arrangements that produce doublets without anybody building them produce many, and that essay names the case itself: the dielectric absorption of a film or ceramic part is a set of slow time constants in parallel with the intended one — a doublet, or several, at milliseconds to seconds, with amplitudes of a few tenths of a per cent.

The spread, and what it costs #

Hold the total tail at one per cent, hold the fastest time constant at a tenth of a millisecond, and spread the other seven out:

The first row is the check rather than a result. With no spread the eight doublets coincide, the arrangement is one doublet of the whole amplitude, and the settling is exactly $\tau\ln(A/B) = 0.2303$ ms — the essay below’s closed form, reproduced to a part in $10^9$, which is what says this is the same object spread out rather than a different one.

The last row is the finding. Four decades of spread multiply the settling by two thousand, and nothing about the tail’s total size has changed. Each of the eight doublets is an eighth of a per cent, which is a quarter of the band being settled to — a tail that, on its own, that essay’s boundary would dismiss as costing nothing at all.

And what the number costs, which is nothing #

The obvious companion question is whether it is the number of doublets that hurts, and the answer is no.

Holding the spread at two decades and varying how many doublets fill it, from one to thirty-two, changes the settling by less than a factor of two. Sixteen times as many doublets, each a sixteenth the size, in the same two decades — and the settling barely moves.

That is the shape of the whole result. The settling is governed by the slowest time constant, and its own share enters only through a logarithm. Adding doublets divides the amplitude among them, which reduces the slowest one’s share, which helps — logarithmically. Spreading them multiplies the slowest time constant, which hurts — proportionally. The two are not comparable, and a design that worries about how many absorption mechanisms a dielectric has is worrying about the wrong one.

Bounded rather than exact, the settling is between the two closed forms either side of it:

$$\tau_\mathrm{max}\ln\frac{A/N}{B} \;\le\; t_\mathrm{settle} \;\le\; \tau_\mathrm{max}\ln\frac{A}{B}$$

the first being the slowest doublet settling alone and the second the whole tail at the slowest time constant. At two decades of spread that bracket is 2.23 to 23.0 milliseconds and the measurement is 7.28, because near $t \approx \tau_\mathrm{max}$ several of the slowest doublets are still contributing and how many depends on how densely the decades are filled.

That bracket is the honest statement and it took a wrong one to find. The figure as first written required the lower bound as an equality — the slowest doublet settling alone — which is out by a factor of three at two decades of spread, and out in the direction that makes the arrangement look better than it is.

What a design can actually do about it #

The arithmetic says the slowest time constant decides everything and the amplitude hardly matters, so it is worth listing what that leaves available — because the list is short and it is not the list a designer would assemble from the measurement before it.

Choose the dielectric on its slowest relaxation, not its total absorption. Polypropylene and polystyrene have absorption an order below polyester and two below a class II ceramic, and the figure of merit that matters is where their relaxations sit rather than how much charge they hold. Since no data sheet prints the second, the ratio test above is the only route to it.

Do not divide the capacitance. Two capacitors in parallel of half the value each have the same total absorption and the same relaxation spectrum, so the settling is unchanged — which is worth knowing because paralleling is the usual remedy for a great many capacitor problems and is worth nothing at all against this one.

Reduce the voltage swing across it. Absorption is proportional to the charge that was held, so the tail scales with the swing — and the tail enters the settling logarithmically, so halving the swing buys $\tau_\mathrm{max}\ln 2$. Ten times less swing buys $\tau_\mathrm{max}\ln 10$, which at ten milliseconds is twenty-three. Real, and not a factor.

Or arrange for the capacitor not to be in the settling path at all. A sample-and-hold’s capacitor is; an integrator’s is; a feedback network’s is. A decoupling capacitor’s is not — two capacitors in parallel is where that one’s own problem lives, and it is a different one. The arrangement that avoids the settling problem is a topological choice, and it is the only item on this list that buys a decade.

That last one is the honest answer and it is the same shape of answer these essays keep reaching. The cancellation that leaves a tail’s own conclusion is that settling to a part in ten thousand needs a cancellation good to a part in ten thousand, however fast the amplifier in front of it is — a requirement on a component rather than on a stage. This essay’s is a requirement on a material, and materials have fewer grades than components do.

What this does to that essay’s trim #

That essay’s second reading is that trimming a doublet is worth little: the dependence on the miss is logarithmic, so halving the mismatch buys $\tau_s\ln 2$, which on its circuit is seventy fast time constants — real, and not the order-of-magnitude win the trim looks like it should be.

Spread over decades, that arithmetic survives its form and changes its size. Halving the whole tail still buys a time of order $\tau\ln 2$; the $\tau$ is now the slowest one rather than the doublet’s own, which at two decades of spread is a hundred times larger.

So a trim is worth a hundred times more than that essay’s arithmetic suggests, in absolute time, and the same fraction of the settling. Which is the practical inversion worth carrying: the useful action is not trimming the tail down, it is moving the slowest time constant. It is the same kind of redirection the cliff before the fastest settling makes about damping — the parameter that looks adjustable is not the one the answer depends on. Halving the whole tail buys $\ln 2$ of $\tau_\mathrm{max}$; removing the slowest decade buys a factor of ten of it. A design that can identify which mechanism is slowest and eliminate it has gained a decade; one that halves everything has gained seventy per cent.

And that essay’s third reading — the blunt boundary, that a tail already smaller than the band costs nothing — needs restating rather than survives. Each doublet here is an eighth of a per cent against a band of a tenth, so each one individually is above the band by a factor of 1.25 and the boundary is barely crossed. Eight of them together are ten times the band. A set of doublets each of which individually matters little can collectively dominate, because the amplitudes add and the times do not.

Why the logarithm is so weak, and what would make it strong #

The result that the amplitude barely matters rests on a logarithm, and it is worth seeing how weak a dependence that is, because it is the reason the whole arrangement behaves as it does.

Settling to a band $B$ from a tail $A$ takes $\tau\ln(A/B)$. The factor between a tail of one per cent and a tail of a tenth of one — ten times smaller, which on any other measurement would be a transformation — is $\ln(10)/\ln(100) = 0.5$. Half. Ten times less absorption buys a factor of two in settling time.

Whereas a decade of slowest time constant buys a factor of ten, exactly, because it multiplies the whole expression.

So the two parameters a dielectric has differ by a factor of about five in what they are worth per decade, and only one of them is on the data sheet. That asymmetry is not a subtlety; it is the reason a capacitor chosen on its absorption percentage can be the wrong capacitor by a decade of settling time.

What would make the amplitude matter is the boundary the essay before it identifies: if the tail is smaller than the band, the logarithm is negative and the doublet costs nothing. That is a threshold rather than a gradient, and crossing it is worth everything where approaching it is worth almost nothing. For a tail of one per cent settling to a tenth, each of eight equal doublets is at 1.25 times the band — so the arrangement sits just on the wrong side of the threshold, and a tail four times smaller would put every branch under it and the settling would collapse to the fast circuit’s own.

Which gives the one case where reducing the absorption is transformative rather than logarithmic: when it takes the slowest branch below the band being settled to. Above that it buys a logarithm and below it buys everything, and the crossing is at an amplitude a design can compute from the band it needs and the number of branches the material has. That is the sharpest form of that essay’s blunt boundary, and it is the only place on this page where the amplitude is the variable.

Why a dielectric is exactly this #

The arrangement measured here is not a hypothetical: it is what a capacitor is.

The capacitor that remembers measures dielectric absorption directly — the charge a capacitor gives back after being discharged, from a set of slow relaxations in the dielectric running from milliseconds to seconds. Those relaxations are exactly a set of series R-C branches in parallel with the intended capacitance, and a series R-C branch across a capacitor is a pole and a zero close together. A doublet.

So a film capacitor in a settling circuit is not a capacitor with a small defect; it is a capacitor with eight or twenty doublets on it, spread over three or four decades. Which means the table above is the capacitor’s own contribution to settling, read directly, and the two thousand is what a design pays for choosing a dielectric with a wide relaxation spectrum over one with a narrow one.

That is a different way of choosing a capacitor from the usual one. The figure of merit a data sheet prints is the total absorption — a percentage, which is $A$ in the arithmetic above — and the quantity that decides the settling is the slowest relaxation time, which no data sheet prints. Two capacitors with the same absorption percentage and relaxation spectra a decade apart differ by a decade in settling time, and nothing on either part says so.

The measurement that finds the slowest one #

If the slowest time constant decides everything, the useful measurement is the one that finds it — and the essay below supplies it without naming it as such.

Its ratio test: settling to $B$ and to $B/10$ should differ by $\tau\ln 10 = 2.30$ time constants of whatever is limiting, so if the two settling times differ by 230 fast time constants rather than 2.3, the thing limiting is not the fast pole and its time constant is the difference divided by $\ln 10$.

Applied to a spread, that test returns $\tau_\mathrm{max}$ — but only once the bands are deep enough for one branch to dominate, and how deep that is is itself informative:

against a slowest time constant of 10.000 milliseconds. The first pair of bands, which is the pair anybody would use, returns it eighteen per cent low; the third returns it to half a per cent.

The shortfall is not an error in the test — it is the spread showing. At a band of a tenth of a per cent the second-slowest and third-slowest branches are still contributing, so the decay is not a single exponential and the difference of two settling times is not one time constant’s $\ln 10$. The convergence of that number as the bands deepen is a measurement of how spread the spectrum is, which is more than the test was being asked for.

So one extra settling measurement identifies the slowest relaxation in a dielectric to within tens of per cent, and three identify it to a per cent and say how crowded the decade below it is — with no knowledge of how many branches there are or how the amplitude is divided among them. That is a better instrument than the absorption percentage a data sheet quotes, it needs the equipment a settling measurement already needs, and it returns the quantity that actually decides the answer.

The other half of the diagnosis is that essay’s own error view — plotting the error rather than the response, with the vertical axis expanded by a hundred — and on a spread tail it looks different from a single doublet’s in a way worth knowing. A single doublet’s error is a straight line on a log-linear plot. A spread one’s is curved, concave, because it is a sum of exponentials with different rates: each decade of time is dominated by a different branch. A straight line says one mechanism and a curve says several, before any number is read off either.

Where this leaves the two settling results either side of it #

This field has three measurements of settling time now and they limit it in three different ways, so it is worth putting them together — because a design that is slow to settle has to know which of the three it is suffering from, and the three have nothing in common but the symptom.

The poles, which is where a designer looks first. The best damping is not the one to build measures the settling of a second-order circuit against its damping and finds it moving by a third across five thousandths of it — so the settling is decided by where the poles are, and the repair is to move them.

A residue, which is the essay before it. The poles do not move at all; one of them has a residue that is small instead of zero, and the settling multiplies by thirty-six while the magnitude response moves by 0.078 decibels. Nothing that looks at the poles finds it, and the repair is to match two components.

And a spectrum, which is this essay, and which one step, computed twice is the arithmetic for: a residue expansion against a marched network, agreeing to the integration error and no further. Many residues at many poles, none of them individually large, with the settling set by the slowest pole present at any amplitude at all. Neither of the other two repairs touches it: the poles are where they should be, no two components are mismatched, and the dielectric has what the dielectric has.

The diagnostic that separates the three is the error view the essay before it prescribes — the error rather than the response, on a logarithmic vertical axis — and the three look different on it.

A pole problem gives a decaying oscillation. A single doublet gives a straight line whose slope is the slow time constant. A spectrum gives a curve, concave, because each decade of time is dominated by a different branch. One plot, three shapes, and the shape says which repair is available before any number is read off it.

That is worth having as the practical output of all three essays, because settling time is the one specification in this collection that three unrelated mechanisms limit and that no frequency-domain measurement sees at all.

What is not here #

The doublets are independent. The arrangement sums exponentials with separate amplitudes and time constants, which is what a residue expansion of a parallel R-C network gives. A real dielectric’s relaxations are not independent in that way — their amplitudes and times come from one physical process and are related — so the spread and the amplitudes are not free parameters the way this figure sweeps them. The coefficient that is about one reading is where the same complaint is made about a capacitor’s printed temperature coefficient — one number that cannot determine the two parameters the part actually has.

The amplitudes are equal. They are not, in a dielectric: the slow relaxations usually carry less than the fast ones. Since the slowest one’s share enters only logarithmically, that matters less than it might — which is itself a finding, and it is the reason the equal-amplitude case is a fair stand-in.

And nothing here is a solved network. The essay before it marches its netlist and compares against a residue expansion; this one is the residue expansion alone, checked against that essay’s closed form at one doublet. The two routes agree where they overlap and the spread case has one route.

Still open: the network marched, and the relaxation spectrum measured #

The same arrangement as a netlist. A capacitor with eight R-C branches across it, marched, would give the same tail from the other direction and would carry the thing the residue sum cannot: what the branches do to the circuit’s response as well as to its settling, since each of them loads the node it is on. The dielectric field’s own tools builds exactly that network and this figure does not use it.

The spectrum, read off a settling measurement. The ratio test above returns the slowest time constant. Repeating it at several bands should return the whole spectrum — each decade of band probing a different branch — which would turn a settling measurement into a dielectric characterisation. Whether the inversion is well conditioned is the question, and it is the same question every inversion of a sum of exponentials has.

And the amplitude distribution that actually occurs. Everything above sweeps the spread with equal amplitudes. What a real dielectric has is a distribution, and the settling depends on the product of the slowest time constant and the logarithm of its own share — so a spectrum weighted towards the fast end is much better than one weighted evenly, by an amount that is one measurement away and is not measured here.

What is checked #

At one doublet the sum reproduces that essay’s closed form, $\tau\ln(A/B)$, to a part in $10^9$. That is the check that this is the same object spread out rather than a different arrangement, and it is the only place the two essays can be compared directly.

At no spread the settling is required to equal the single doublet’s exactly, and the trim to buy exactly $\tau\ln 2$ — both to a part in $10^6$. A threshold like “five times slower than one doublet” holds from two decades of spread upwards and fails at nought, correctly, since at nought there is nothing to be slower than.

The settling is required to lie between the two closed forms rather than to equal either. Requiring the lower one, which is what the figure did first, is out by a factor of three at two decades of spread and in the flattering direction.

And the two sweeps are held against each other: four decades of spread must multiply the settling by more than a hundred while sixteen times the count must change it by less than two. The pair is the finding, and either alone is a curve.