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

Relaxation time — where it appears

A relaxation time is the time constant of one slow polarisation process in a dielectric — the delay with which that part of the material follows an applied field. A real dielectric has a distribution of them spanning decades rather than one value, and that spread is what makes a capacitor read as more capacitance the longer it is watched.

Named by 5 essays across one field — each of them below, with the objects they name alongside it.

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 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.

Dielectric absorptionModel rangeMarchingSettling timeMeasurement conditionReal capacitorResiduesSample-and-holdClosed formComponent toleranceCondition numberLoss tangent

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