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

Dielectric absorption — where it appears

A dielectric's polarisation following the field with relaxation times spread over decades, so some charge returns after the part is discharged. It makes the capacitance depend on the frequency it was measured at and the settling depend on how long it is watched, with no time constant after which either is over.

Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.

Three capacitances, all correct: 2.000 µF, 5.814 µF and 2.105 µF at 5 V. computed by solving, not by drawing. The charge is C∞·v + Q_s·tanh(v/V_k) — a linear backbone and a polarisation that saturates — with both parameters pinned by the capacitance at zero volts and at the rated voltage, so there is no third degree of freedom to tune the answer with. The three curves are three questions. The small-signal value is the slope at the bias, which is what a ripple sees. The charge-average is the total charge moved from zero divided by the voltage, which is what a reservoir or a hold capacitor obeys. What a bridge reads is neither: it is the fundamental of the charge waveform under a one-volt test, which is a measurement condition. At 5 V they are 2.000 µF, 5.814 µF and 2.105 µF — a factor of 2.91 between the extremes, and every one of them is the capacitance.

The capacitance that is not one number

A ten-microfarad ceramic at its rated five volts is 2.000 µF as a slope, 5.814 µF as a charge average and 2.105 µF as a bridge reads it — three answers to three different questions, all correct, all called the capacitance. At zero bias the standard one-volt test alone reads 5.2 per cent low. Marched in a circuit the same part distorts as the square of the drive with no bias and in proportion to it with a bias, because the bias is what puts a second harmonic there.

frequency · Real capacitor The charge that comes back: a 0.2% dielectric, 10 s shorted, read at 900 s. computed by solving, not by drawing. The capacitor is charged to 10 V until every relaxation is complete, shorted for 10 seconds, then opened and watched. It climbs back to 20.00 millivolts — 0.2000 per cent of where it was — and the shape is the finding: it is a straight line on a logarithmic time axis, gaining 0.097 per cent of the charging voltage per decade. There is no time constant after which it is over, because there is no single time constant: one branch of the model comes to equilibrium per decade, for as many decades as the dielectric has. A decade before the reading it was at 0.1033 per cent.

The capacitor that remembers

Charge a capacitor, short it for ten seconds, open it, and it climbs back to a fifth of a per cent of where it was. Nothing leaked and nothing was gained: some of the dielectric had not finished discharging. The same defect measured as an admittance says the part is 0.593 per cent more capacitance at a tenth of a millihertz than at a kilohertz, and measured in a sample-and-hold it says a millisecond of hold costs a hundred parts per million — thirteen bits, on a part specified at nothing.

transients · Dielectric absorption The reading is a count of decades: 1.0288 parts per thousand of them. computed by solving, not by drawing. 17 marched tests, three families of absolute time — a tenth of a second, one second and ten seconds of short — plotted against the number of decades between the short and the reading. The families lie on one another, which is the finding: the answer is not a property of the part alone and not a property of either duration, it is a count of the decades of relaxation time the test leaves in. The line is a least-squares fit through the origin at 1.0288e-3 per decade; the model's own capacitance per decade of relaxation time is α = 1.0343e-3, which nothing in the fit was told — the fit sits 0.53 per cent under it, because the charge that comes back is shared with the slow branches it came off. The worst residual is 4.80 per cent, at the narrowest ratio drawn, and 1.23 per cent over the 8 tests that are two decades wide and read before the slowest relaxation the model has; the 3 read after it fall away to 4.49 per cent, which is where the law ends. Families a hundred times apart in absolute time differ by at most 0.84 per cent, which is the whole of the collapse.

Ten seconds, and fifteen minutes

A data sheet's dielectric absorption is quoted as a property of the part. It is not: the same modelled capacitor reads 0.4050 per cent with a tenth-of-a-second short and 0.0305 per cent with a thousand-second one, and 0.0047 against 0.2948 depending on when the reading is taken. Seventeen marched tests collapse onto one line — the recovery is 1.0288 parts per thousand for every decade between the two durations — and four dielectrics the specified test declares identical read a factor of 3.31 apart one decade away from it.

transients · Dielectric absorption 8 doublets over 2 decades settle 31.6 times slower than one. A tail of 1.0 per cent split into 8 doublets whose time constants are spread over 2 decades from 0.100 ms, each carrying 0.125 per cent. Settling to 0.10 per cent takes 7.278 ms against 0.2303 for the same total tail at one time constant — a factor of 31.6 — and it is between τₘₐₓ·ln((A/N)/B) = 2.231 and τₘₐₓ·ln(A/B) = 23.03 ms, which says the SLOWEST doublet decides it whatever its share. The spread is what costs and the number is not: four decades of spread multiply the settling by 2.04e+3 while sixteen times the count changes it by 0.386. And the trimming result, halving the tail for τ·ln 2, now buys 3.86 ms — the slowest time constant's ln 2 rather than the fastest's.

The slowest one decides it

One doublet's settling is its own time constant times the log of the tail over the band, so halving the mismatch buys ln 2 of it and no more. Split the same one-per-cent tail into eight doublets spread over four decades and the settling goes from 0.23 milliseconds to 470 — a factor of two thousand — while sixteen times as many doublets over the same spread changes it by less than two. It is the spread that costs and not the number, because the slowest doublet decides the settling however small its own share, and its share enters only through a logarithm.

transients · Doublet Driven through 1kΩ, the absorbing capacitor's tail is 2.6e-5 of the step, not 1%. computed by solving, not by drawing. The error at a capacitor's terminal after a step applied through 1k ohms, where the capacitor is 1 nF with eight absorption branches spread over two decades from 0.1 ms, together holding one per cent of its capacitance. The poles are found by bisection and the residues in closed form, and a trapezoidal march of the same netlist agrees. The tail the branches leave is 2.58e-5 of the step — the source resistance over the branch resistances in parallel — and settling to 1e-5 takes 173µs against 11.5µs with no branches; the model that weights each branch by its share of the capacitance says 48.4ms.

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.

transients · Doublet 3 relaxations over two decades come back within 0.22% from readings good to 1e-5. computed by solving, not by drawing. A one-per-cent tail divided equally among 3 branches spread over two decades from 0.1 ms, read at 80 times with 1e-5 of seeded noise on each reading, and the amplitudes recovered by least squares with the time constants given to the fit. Dots: one recovery; bars: the rms spread over 200. The condition number of the fit is 8.29e+0, and the worst branch comes back with an rms error of 0.224 per cent.

How many relaxations a settling curve holds

A settling curve read at eighty times, each good to ten parts in a million of the step, returns the amplitudes of three relaxations spread over two decades to 0.22 per cent and of eight to 46 — even when the fit is handed every time constant exactly. The condition number of the inversion grows by a factor of 7.7 for each extra relaxation packed into a decade, so each tenfold improvement in the reading buys a little more than one more relaxation a decade and no more. A settling measurement can count the dielectric's slowest relaxations; it cannot resolve a crowded spectrum, and how crowded is readable before any amplitude is.

transients · Doublet Weighted towards the fast end as τ^(−1), the tail falls as a power of time and each decade of band costs 9.9× the time. computed by solving, not by drawing. A one-per-cent tail spread densely over five decades of time constant from 0.1 ms to 10 s, with each branch's amplitude weighted as τ to the power −m, for m = −0.5, 0, 0.5 and 1. The drawn weighting is m = 1: it settles to 10⁻³, 10⁻⁴, 10⁻⁵ and 10⁻⁶ in 972µs, 9.71ms, 96.3ms, 891ms. With m above zero the fast end carries more and the error between the ends is a power of time; at or below zero the slow end decides.

The tail with no slowest relaxation

Spread a one-per-cent tail densely over five decades of time constant and weight each relaxation as τ to the power −m. With m = 1 the tail falls as one over time and settles to 10⁻⁴ in 9.71 ms and to 10⁻⁶ in 891 ms — each tenfold tighter band costing ten times the time, not the fixed τ·ln 10 one relaxation charges. With even weighting it takes 14.4 s to reach 10⁻⁴; weighted towards the slow end, 26.5 s. The exponent decides whether any relaxation is the slowest one that matters: above zero none is, the settling is a power of the band with slope −1/m, and the spread's far end is never reached.

transients · Doublet A 0.2% part sampling 0 V after long at 10 V, holding 1 ms: the first reading is 109 ppm of the step, and the hundredth still 4.5 ppm, every one of them made from the same branches. computed by solving, not by drawing: the reading at each sample, as a fraction of the step, when a sample-and-hold with a 1 µF part of 0.2% absorption samples 0 V after a long time at 10 V, acquiring for 100 µs and holding for 1 ms. The first reading agrees with a march of the netlist. It is 109 ppm, the tenth 46.7 ppm, the hundredth 4.5 ppm and the four-hundredth 0.904 ppm. The earlier samples are still being read, and for far longer than an acquisition suggests, because each acquisition can reach a branch only in proportion to the fraction of its time constant it lasts.

The sample before the sample

A sample-and-hold whose capacitor has 0.2 per cent of dielectric absorption reads a 10 V step 109 parts per million wrong on the first sample after it, 46.7 on the tenth, and 4.5 on the hundredth, because every earlier sample is still in the dielectric. That error has been described as distortion, since it grows with the signal. It is not. The state is linear and one period is a single matrix, so the reading is a fixed weighted sum of earlier samples, a filter with a very long tail: a sine reads 28.1 parts per million low and its largest harmonic is 1.5 × 10⁻¹² of the fundamental. A filter can be undone. Subtracting fifty earlier samples takes a stepped signal from 51.6 parts per million to 1.61, and with the absorption known to 20 per cent it stops at 5.47.

transients · Dielectric absorption

Named alongside it

The objects these essays reach for when they reach for this one.

Model rangeMarchingRelaxation timeSettling timeResiduesMeasurement conditionReal capacitorSample-and-holdTransient responseTime constant matchingClosed formComponent tolerance

All concepts