Before the steady state

The tail the source decides

A capacitor with one per cent of dielectric absorption does not leave a one-per-cent tail in the circuit it is in. Driven through a kilohm, the eight slow branches that model the absorption leave 2.58×10⁻⁵ of the step at its terminals — the source resistance over the branch resistances in parallel — and settling to 10⁻⁴ takes 9.49 µs against 9.21 with no absorption at all. The share of the capacitance is not the amplitude; it is a ceiling the circuit reaches only when the source's own time constant falls inside the spread of the branches, and there the cost is 2.2 times the settling at 10⁻⁴ and 16 times at 10⁻⁵.

19 min read 5 figures Every model has an edge Two routes to a number The parasitic is the component

Assumes: The cancellation that leaves a tail · The capacitor that remembers

The slowest one decides it took a one-per-cent tail, split it among eight doublets spread over two decades, and found the settling governed by the slowest time constant however small its share. It gave each doublet an amplitude — an eighth of a per cent — and it said where that amplitude came from: the dielectric absorption of a film or ceramic part is a set of slow time constants in parallel with the intended one. A capacitor with one per cent of absorption, on that reading, is a capacitor that puts a one-per-cent tail into whatever it is settling.

That essay flagged its own gap, in its last section: nothing here is a solved network. The amplitudes were assigned, not computed. This essay builds the network and solves it, and the network disagrees with the assignment by between two and five orders of magnitude, in a direction that depends entirely on what is driving the capacitor.

The capacitor as a netlist

The capacitor that remembers gives the model: the intended capacitance $C_0$ with a set of series R–C branches across it, each branch a relaxation of the dielectric with its own time constant $\tau_i = R_iC_i$ . Here $C_0$ is a nanofarad and there are eight branches spread over two decades from 0.1 to 10 milliseconds, each with $C_i = 1.25$ pF, so that together they hold one per cent of the part’s capacitance — the same one per cent, spread the same way, that the essay on the slowest doublet assigned as a tail.

The branch resistances follow from the time constants: $R_i = \tau_i/C_i$ , from 80 MΩ for the fastest branch to 8 GΩ for the slowest. Those are the numbers that matter below, and nothing in the data-sheet figure of “one per cent absorption” mentions them.

The capacitor is driven by a unit step through a source resistance $R_s$ and the error at its terminal is followed until it is inside a band.

Fig. 1 The error at the terminal of a 1 nF capacitor with one per cent of absorption in eight branches, after a step applied through a kilohm. Solid: the network, by residues and checked by a march. Dotted: the same node with no absorption. Dashed: each branch carrying its share of the capacitance, which is the tail the model of a spread of doublets assigns. The slider is the source resistance.

Driven through a kilohm, the terminal settles to 10⁻³ in 6.93 µs, to 10⁻⁴ in 9.49 µs and to 10⁻⁵ in 173 µs. With no absorption at all the same node takes 6.91, 9.21 and 11.5 µs. The share-weighted tail says 7.28 ms to 10⁻³ and 48.4 ms to 10⁻⁵.

So the absorption is invisible at a tenth of a per cent, costs three per cent of the time at a hundredth, and multiplies it by fifteen at a thousandth. A model that put a one-per-cent tail on this node would have been out by a factor of a thousand at the first band and nearly three hundred at the last, and out in the direction that condemns a capacitor that is doing its job.

Why the tail is the source resistance’s

The reason is a sentence of circuit analysis, and it is worth having because it replaces the share with a number a designer controls.

While the source’s own time constant $R_sC_0$ — a microsecond here — is short against every branch, the source charges the main capacitance long before any branch has moved, and from then on it holds the node. Each branch then charges through its own resistance from a node sitting at nearly the full step, drawing a current $V/R_i$ that decays with the branch’s own time constant. That current has to come from the source, and the only way it can is by dropping a voltage across $R_s$ . The error at the node is that drop:

e(t) \approx \sum_i \frac{R_s}{R_i}\, e^{-t/\tau_i}, \qquad e(0^+) \approx \frac{R_s}{R_\parallel}

where $R_\parallel$ is the branch resistances in parallel — 38.7 MΩ here. The tail is $R_s/R_\parallel$ : 2.58×10⁻⁵ at a kilohm, and it scales with the source.

Two things have vanished from the expression. The branch capacitance $C_i$ appears only through $R_i$ , so a larger absorption with the same time constants puts in a proportionally larger tail but a tail still divided by the ratio of a small resistance to a very large one. And $C_0$ has gone altogether. The amplitude is a ratio of resistances, and one of the two is on the circuit diagram rather than in the capacitor.

The weighting inside the tail is also different from the one the spread-doublet model assumed. Every branch there carried an equal share. Here each branch carries $R_s/R_i = R_sC_i/\tau_i$ , so with equal branch capacitances the slow branches carry less in proportion to their slowness — the slowest a hundredth of the fastest. The essay on the slowest doublet noted that a spectrum weighted towards the fast end is much better than one weighted evenly; a driven capacitor weights its own spectrum towards the fast end, automatically, in exact proportion to $1/\tau$ .

Fig. 2 The tail left at the terminal once the source’s own pole has gone — read at twenty of its time constants — against the source resistance, with

R_s/R_\parallel

dashed. Wherever the source settles before the fastest branch moves the two agree to 2.0 per cent, rising a decade per decade of source resistance, and at no source resistance does the tail approach the one per cent that the share would give it.

The closed form holds to two per cent across every source resistance at which the source’s own pole is gone before the fastest branch has moved, and at a kilohm the residue sum’s slow part is 2.614×10⁻⁵ against the formula’s 2.580, 1.3 per cent apart. The small excess is the branches’ own interaction: a branch that has partly charged changes the node voltage the next branch sees, which is second order in a ratio that is already 10⁻⁵.

A march of the same netlist

The residues are computed in closed form, and that makes it worth saying how they are found, because a closed form is exactly the kind of result a second route should check.

The driving-point admittance of the node is $Y(s) = sC_0 + \sum_i sC_i/(1 + s\tau_i)$ and the step error is $E(s) = R_sY/\bigl(s(1 + R_sY)\bigr)$ . It is an RC network, so every pole is real and negative, and the poles interlace the branch corners: exactly one below $1/\tau_\mathrm{max}$ , one between each adjacent pair of corners, and one above $1/\tau_\mathrm{min}$ that is the source’s own. Each is found by bisection in its own interval, which cannot miss one or find one twice. The residue of the error at a pole $s = -\sigma$ is then

r = \frac{1}{\sigma R_s\, Y'(-\sigma)}, \qquad Y'(s) = C_0 + \sum_i \frac{C_i}{(1 + s\tau_i)^2}

which is positive, so the error is a sum of decaying exponentials with positive weights, monotone, and a settling time is a single bisection.

The residues must sum to exactly one — the node starts at nought, so the whole step is error at $t = 0$ — and they do, to a part in $10^{12}$ . That is the check that every pole was found once. The second route is the netlist itself, the source, the capacitor and all sixteen branch elements, marched with the trapezoidal rule over twenty thousand steps — the same pairing of a residue expansion against a march that one step, computed twice uses to measure the march. Read at the instant the closed form puts the error through a tenth of a per cent, the march and the residues agree to two parts in a million of the step, which is the march’s own step error.

The share is not wrong. It is the amplitude in a different circuit, and the network says which.

Raise the source resistance and $R_sC_0$ climbs towards the branches. The source stops holding the node before the branches have finished, the two sets of poles begin to interact, and the tail stops being the drop across a stiff source. Past the far end of the spread, when $R_sC_0$ is longer than every branch, the branches charge with the node rather than after it: each one is a small capacitor in parallel that has always finished before the node has, and the network is one capacitor of 1.01 nF charging through a large resistor. The tail vanishes again, for the opposite reason.

Fig. 3 Driven through 100 kΩ, the source’s own time constant is 100 µs — the fastest branch’s — and the node no longer settles before the absorption begins. Settling to 10⁻⁵ takes 9.78 ms against 1.15 ms with no absorption: 8.5 times as long. The closed form for a stiff source no longer applies and the network is solved as it stands.

Between those two limits there is a source resistance at which the branches cost the most, and it is where the source’s time constant sits inside the spread. The settling time at three bands, divided by the same node’s settling with no absorption, measured against the source resistance from ten ohms to a gigohm:

The branches cost 2.2× the settling at 20.5kΩ and nothing at either end. computed by solving, not by drawing. The time for a 1 nF capacitor with one per cent of absorption, spread over eight branches from 0.1 to 10 ms, to settle to 10⁻³, 10⁻⁴ and 10⁻⁵ of a step applied through a source resistance, divided by the same node's settling with no branches. At 1e-4 the branches cost nothing below about 2.74kΩ, cost most — 2.23 times — at 20.5kΩ where RsC₀ is 20.5µs and inside the spread, and cost nothing again once the source is slower than every branch. — Fig. 4 The time for the absorbing capacitor to settle to 10⁻³, 10⁻⁴ and 10⁻⁵ of a step, divided by the same node’s settling with no absorption, against the source resistance. The shaded band is where the source’s time constant $R_sC_0$ lies inside the spread of the branches. At 10⁻⁴ the absorption costs nothing below about 2.7 kΩ, costs 2.23 times at 20.5 kΩ, and costs nothing again past the spread.

source	$R_sC_0$	to 10⁻⁴, against no absorption
10 Ω	10 ns	1.0003
1 kΩ	1 µs	1.03
10 kΩ	10 µs	2.08
100 kΩ	100 µs	2.06
1 MΩ	1 ms	1.33
100 MΩ	100 ms	1.01

The worst case at 10⁻⁴ is 2.23 times and at 10⁻⁵ it is 16.2 times; at 10⁻³ it never reaches 1.07. None of these is the factor of two thousand the spread-doublet model found for the same capacitor, and all three are reached only in the band of source resistances where the circuit’s own time constant falls among the dielectric’s.

The one circuit in which the share is the amplitude is the one the capacitor that remembers is defined by: charge the part, then take the source away. With no source the node is floating, the charge on it is conserved, and the branches that have not finished charging draw their charge out of $C_0$ instead of out of a source. The terminal voltage then falls by the share — each branch’s capacitance times what it still lacks, divided by the total — and that is the recovery test, and the error a sample-and-hold makes when its switch opens.

So the two models are the two ends of one question — who supplies the branch current? A stiff source supplies it and the node barely moves. Nothing supplies it and the node gives it up in proportion to capacitance. The share-weighted tail is the hold’s, and the amplifier inside the sample is where a sampling circuit meets it. What the network adds is that the acquisition phase of the same circuit, driven by an amplifier with a few ohms of output resistance, sees almost none of it. The absorption percentage itself is measured in exactly the open-circuit arrangement, and ten seconds and fifteen minutes is why the number depends on how long the part was charged and shorted before it was read.

That gives the practical reading of a data sheet’s absorption figure. It is a statement about a capacitor that has been disconnected. In a circuit where the capacitor stays driven through a low impedance — a filter behind a buffer, a reservoir behind a regulator, a timing capacitor charged from a current source that is close to ideal — the figure overstates the settling error by the ratio of the share to $R_s/R_\parallel$ , which for a kilohm is a factor of nearly four hundred.

Fig. 5 Through a hundred ohms the tail is 2.58×10⁻⁶ and the node settles to 10⁻⁵ in 1.18 µs against 1.15 with no absorption at all. At this drive the capacitor’s one per cent of absorption is two parts in a million, and every band a converter up to about seventeen bits — six decibels a bit is the conversion — would settle to is reached as if the dielectric were perfect.

What a designer does with a ratio of resistances

Because the tail is $R_s/R_\parallel$ , the design moves available are not the ones the share suggested.

The source impedance is the knob. A source is itself a model with a resistance in it — the source that is not a source is where that resistance is found — and halving it halves the tail, which is linear, not the logarithm the essay on the slowest doublet found for halving an assigned amplitude. It is linear because the amplitude itself is now proportional to something the circuit sets. A buffer’s output resistance, including its behaviour at the frequencies the branches live at — which for a feedback amplifier is its closed-loop output impedance at a few hundred hertz and is usually milliohms — is what goes in the numerator.

The capacitor’s value hardly matters. A larger $C_0$ of the same dielectric has branch capacitances scaled with it, and so branch resistances scaled down, and the tail scales up with the capacitance. That looks bad until it is combined with the fact that a larger capacitor behind the same source charges more slowly: the settling that matters is the ratio to the bare node’s, and the table above shows that ratio depending on where $R_sC_0$ sits in the spread, not on either factor.

The worst arrangement is a medium impedance. Tens of kilohms and a nanofarad put the node’s own time constant among the dielectric’s, and that is the arrangement of an integrator’s input resistor and capacitor, of a slow RC filter in front of a precision converter, and of a sample-and-hold driven from a high-impedance source. Those are the circuits in which the absorption genuinely costs, and it is not because of the dielectric alone.

Why the charge model and the voltage model agree about energy

One more check on the result, and it is independent of the algebra. The charge that ends up in the branches is fixed by the final voltage: $\sum C_i \cdot 1$ , one per cent of the capacitor’s charge. Whatever circuit drives the node, that charge has to arrive, and the source supplies it through $R_s$ .

In the stiff case it arrives over the branches’ own time constants as a current $V/R_i$ , so the charge each branch takes is $\int (V/R_i)e^{-t/\tau_i}dt = V\tau_i/R_i = VC_i$ — the right amount, delivered slowly and through a small resistance, which is why its trace at the node is small. In the hold case there is no source and the same charge has to come from $C_0$ , which moves its voltage by the share. The one per cent is conserved in both circuits; what the circuit decides is whether it is paid for in volts at the node or in current through the source. That is the whole of the difference between the two models, and it is the same bookkeeping the half that never arrives does for the energy a charge costs.

Checking the figures against each other

The residues sum to one to a part in $10^{12}$ : the node starts at nought, so the whole step is error at $t = 0$ , and a missing or duplicated pole would show here first.

A trapezoidal march of the same sixteen-element netlist leaves the same error at the node as the residue sum, to two parts in a million, read at the instant the error crosses a tenth of a per cent.

The stiff-source closed form $R_s/R_\parallel$ is required to match the slow residues to five per cent wherever $R_sC_0$ is a thirtieth of the fastest branch or less, and matches to 1.3 per cent at a kilohm. Outside that range it is not required, because it is not true there and the figures say so.

The settling cost is required to be one at both ends of the source range — below a hundredth of a per cent of excess at ten ohms and within five per cent at a gigohm — and at least one everywhere, since no source resistance can make absorption help. The worst cost is required to fall where $R_sC_0$ lies inside the spread.

What an equal, linear, resistive source leaves out

The branches are equal in capacitance and spread geometrically, which is the arrangement the essay on the slowest doublet swept and the one that makes the two essays comparable. A real dielectric’s branches follow its relaxation spectrum, and that changes the weights $R_s/R_i$ but not the form.

The source is a pure resistance. An amplifier driving the node has an output impedance that rises with frequency and a loop that can ring against a capacitive load — the load that gets inside the loop is where that happens — and the relevant impedance for the tail is the one at the branches’ frequencies, hundreds of hertz, where most amplifiers are at their stiffest.

And nothing here is non-linear. A ceramic dielectric’s absorption depends on the voltage across it, so its branches are different at different operating points; the film capacitor this models is linear, and that is the case the arithmetic covers.

Still open: the integrator, the hold that follows a drive, and the current source

The capacitor in a feedback path. An integrator holds its capacitor between a virtual earth and an output that is free to move, so neither end is a stiff source and neither is open. The branch currents flow through the input resistor and the tail is the charge they take, integrated — which should make the error proportional to the share again, but in charge rather than in volts, and with a time dependence the feedback reshapes. Solving it would say whether an integrating converter’s absorption error is the hold’s or the drive’s.

A hold that follows a drive. A sample-and-hold is driven through a low impedance during acquisition and then opened, so the branches start the hold at whatever state the acquisition left them in. The error in the held value depends on how long the acquisition lasted in units of the branch time constants — too short and the branches have not charged, and the hold gives back the difference. That is a boundary in acquisition time, and it is where the two circuits on this page meet.

A current source instead of a resistor. An ideal current source has an infinite $R_s$ and charges the node as a ramp; the branches then lag the ramp by their own time constants and the voltage error is a constant offset rather than a tail. A real current source’s output resistance puts a ceiling on that, and where the ceiling lies against the branch resistances is the same ratio this essay found, read from the other side.