The spread a winding resistance holds

The walk the core sees drove a transformer from a relaxation oscillator with noise on its thresholds and found the core’s volt-seconds walking. Each decision is a fresh draw, the draw that ends one half cycle also starts the next, and in the difference between the high half and the low half that shared draw enters with the coefficient $B - A$ rather than $A + B$. The summed imbalance therefore spreads at $\sigma\sqrt{2N}\,|B - A|$ after N periods, exactly β times the rate at which the timing walks, and at β = 0.5 with five millivolts of noise that was 1.89 nanoseconds per root period.

That core was ideal. Its winding had no resistance, so whatever volt-seconds a period left behind stayed behind for ever, and the only question the essay could ask was how long the walk took to reach saturation. The walk that stops asked the same question of a deterministic imbalance and found that the question changes once resistance is in the loop: a fixed excess in one half cycle no longer ramps the flux without end, because the magnetising current it builds drops a voltage across the resistance that opposes it, and the flux settles at an offset. The boundary there stopped being a time and became a resistance — 0.1014 ohms, below which a one per cent imbalance still saturated its core.

A random imbalance under the same resistance should settle too, not to an offset but to a spread. This essay measures that spread, says how much of it the anticorrelation between consecutive duty errors removes, and turns it into the boundary the previous essay pointed at: a contour in millivolts and ohms.

A branch that forgets #

The magnetising branch is an inductance L in series with a resistance R — the winding’s own, plus whatever output resistance the driver has, because the current that builds the flux flows through both. Between two flips the oscillator holds a constant ±V across that branch, so the flux linkage $\lambda = Li$ obeys $d\lambda/dt = v - \lambda/\tau$ with $\tau = L/R$, and over a half cycle of length h the answer is exact:

$$\lambda \leftarrow \lambda\,e^{-h/\tau} + v\tau\left(1 - e^{-h/\tau}\right).$$

Nothing has to be marched. The event map already produces the flip instants — each half cycle is one logarithm of where the capacitor started and where its noisy threshold sits — and carrying the flux from flip to flip through that expression gives its value at the end of every period with no step size and no discretisation error. Setting τ to infinity recovers the ideal core: the flux is then V times the running sum of high time less low time, which is the walk the earlier essay measured.

The same oscillator throughout: a microsecond of RC, five-volt rails, β = 0.5 and five millivolts of threshold noise, so the period is 2.197 microseconds and one half cycle swings the flux by V·T/2 = 5,493 volt-nanoseconds. Every spread below is quoted as a fraction of that swing, because that is the unit a core’s headroom is designed in: a core that has been given twenty per cent of margin above its nominal peak has twenty per cent of this number to spend on everything the nominal drive does not do.

The three seeded drives above show the whole change at once. Each feeds two cores with the same flips, one ideal and one leaking with a time constant of a hundred periods. The ideal traces are walks: after two thousand periods they have wandered by several per cent of the swing in directions that have nothing to do with one another, and they would wander further given longer. The leaky traces are not. They fluctuate just as quickly, but about zero, inside a band that does not widen.

Where the walk settles #

How wide the band is follows from linearising one period about the noiseless one. With $q = e^{-h/\tau}$ for a nominal half cycle h, a small error $\delta_+$ in the high half and $\delta_-$ in the low half move the period-end flux by

$$\delta\Phi' = q^2\,\delta\Phi + \frac{2V}{1+q}\left(q^2\delta_+ - q\,\delta_-\right).$$

The first term is the leak — the flux remembers a fraction $q^2$ of where it was a period ago — and the second is the period’s own contribution, with each half weighted by how much of it the leak has already forgotten by the time the period ends. As τ grows, q tends to one and the contribution becomes $V(\delta_+ - \delta_-)$, the imbalance times the rail, which is the ideal core’s step.

The two half-cycle errors are the event map’s three draws. The high half starts from a draw (sensitivity A) and ends on one (sensitivity B); the low half starts from that second draw and ends on a third. So each period’s input to the flux is a weighted sum of three draws, consecutive inputs share one of them with opposite signs, and the flux is a first-order filter with pole $q^2$ driven by that correlated input. Its stationary variance is

$$\operatorname{Var}\Phi = \frac{\operatorname{Var}u + 2q^2\operatorname{Cov}(u, u')}{1 - q^4},$$

and the hero figure draws its square root against L/R from a third of a period to a thousand. Three seeded runs of sixty thousand periods each sit on it at every leak tried: 0.146% of the swing at one period (0.145% from the closed form), 0.401% at ten, and 1.25% at a hundred against 1.22%.

For a slow leak the expression simplifies to something worth carrying in one’s head:

$$\sigma_\Phi \to V\,\sigma\,|B - A|\,\sqrt{\tau/T},$$

the ideal core’s walk rate, $V\sigma\sqrt2|B-A|$ per root period, times $\sqrt{\tau/2T}$. A core that remembers τ/T periods holds a spread equal to the distance the walk would have covered in half that many. It grows as the square root of L/R — ten times the inductance, or a tenth of the resistance, makes the spread 3.16 times larger — and it never stops growing as the resistance falls, which is the ideal core’s unbounded walk recovered as the limit.

What the anticorrelation is worth, and when #

The covariance term in that variance is the anticorrelation the earlier essay found between consecutive duty errors. A draw that ends one period’s low half lengthens the half being subtracted there and the high half being added in the next, so a period whose high half ran long tends to be followed by one whose high half runs short. Drop that term — treat the duty errors as independent, which is what a designer does who reads a pulse-width spread off an oscilloscope and feeds it into a time constant — and the answer is the upper curve in the hero figure: 1.61% of the swing at a hundred periods rather than 1.22%.

The figure above divides one by the other and shows that the discount is not a constant. It depends on how many periods the core remembers. A leak of a period or less forgets each period’s error before the next one arrives, so there is nothing for the anticorrelation to act on and the ratio is one: at a third of a period the two estimates agree to two parts in a thousand. A slow leak sums many consecutive errors, the shared draws telescope exactly as they did in the ideal core, and the ratio falls to

$$\sqrt{\frac{2(B - A)^2}{A^2 + (B - A)^2 + B^2}},$$

which is 0.378 at β = 0.2, 0.756 at 0.5 and 0.936 at 0.8. The transition is quick. Each curve is halfway between one and its limit by about two periods — 2.2, 1.8 and 1.6 — so for any real transformer, whose magnetising time constant is hundreds or thousands of the drive’s periods, the limit is the number that applies.

That makes the oscilloscope estimate wrong in a direction worth knowing. At β = 0.5 it overstates the spread by a third; at β = 0.2 it overstates it by a factor of 2.65, because with little hysteresis the two sensitivities A and B are close, the shared draw almost cancels in the difference, and the per-period duty error is mostly the two draws at the ends of each period, which a slow core never sees. A designer sizing headroom from the cycle-to-cycle duty jitter is sizing it for a core that does not exist — one that integrates each period’s error and then, somehow, forgets that the next period partly undoes it.

The other side of the same fact is that the slow leak is the only case in which the ordering of the divider ratios is what the ideal core suggested. At β = 0.2 the leaky core settles at 1.06% of the swing for a hundred-period leak, against 1.22% at β = 0.5 and 2.02% at 0.8, the spread rising with the hysteresis just as the walk’s rate did. In a core fast enough to forget each period the per-period duty error is what matters, and that does not vanish with the hysteresis at all; there the ordering by β is much weaker. The two regimes are separated by a time constant of a couple of periods, which no transformer is near, but an RC filter averaging a pulse-width-modulated output can be.

A ramp stops at an offset, a walk at a spread #

The comparator in the delay that is two delays had a skew — its upward decision later than its downward one — and the walk essay found that skew and noise compete. A skew adds the same imbalance to every period, so in an ideal core its volt-seconds ramp linearly while the noise’s spread grows as a square root, and the ramp overtakes the walk after

$$N^* = \frac{2\sigma^2(B - A)^2}{r^2}$$

periods, for an imbalance r per period. Before $N^*$ the core wanders at random about a drift too small to see; after it, the drift dominates.

With a leak the competition is settled differently, because neither one grows without limit. The skew’s ramp stops at the offset the walk that stops found for a deterministic imbalance, $V r\,\tau/T$, proportional to L/R. The noise’s walk stops at the spread above, proportional to the square root of L/R. So the offset beats the spread at long time constants and loses at short ones, and the two are equal where

$$\frac{\tau^*}{T} = \frac{\sigma^2(B - A)^2}{r^2} = \frac{N^*}{2}.$$

The leak at which a skew’s offset equals the noise’s spread is half the count at which the ideal core’s ramp overtook its walk. It is the same comparison read in the other variable: an ideal core run for N periods and a leaky core that remembers N/2 of them see the two mechanisms in the same proportion.

The figure checks the offset against the event map’s own flux at two leaks before drawing it, because the skew’s effect on the halves is not obvious. A comparator that is 300 picoseconds late going up lengthens the high half by exactly 300 picoseconds, and the capacitor, which kept charging during the delay, starts the low half further from its threshold, so the low half is longer too — by 100 picoseconds. The imbalance per period is the difference, 0.2 nanoseconds, which is the skew times 2β/(1 + β). That factor of two-thirds at β = 0.5 is why a skew is less damaging to a core than its size suggests, and it is the r in the expressions above.

With that r the crossing is at 46.1 periods, against the 44.4 that N*/2 gives. The difference is the leak acting inside each half cycle, which the slow-leak limit ignores; at a thirtieth of the skew the crossing moves to 4,450 periods and the limit predicts 4,445. At a nanosecond of skew the crossing is at 5.5 periods and the halving no longer holds at all, because a leak of five periods forgets a measurable part of each half before the period ends — a regime no transformer is in.

The boundary in millivolts and ohms #

The point of a spread rather than a walk is that it can be compared with a headroom once and for all. An ideal core with noise saturates eventually, however small the noise; a leaky one either fits inside its margin or it does not, and the answer is a contour.

Take a magnetising inductance of a millihenry, driven by this oscillator through a series resistance R. L/R is then 455/R periods — a hundred periods at 4.55 ohms, a thousand at 0.455. Give the core a fifth of its swing as headroom, and require the skew’s offset plus three standard deviations of the noise’s spread to fit inside it. Because the spread is proportional to the noise, the largest noise that fits is the headroom left after the offset, divided by three spreads per millivolt — one division at every resistance.

With no skew the allowance rises as the square root of the resistance: 4.07 millivolts at a tenth of an ohm, 12.9 at one ohm and 40.4 at ten. That is the $\sqrt{\tau}$ law read as a design rule. A winding resistance that a power designer would count only as copper loss is, for a drive like this, the thing that sets how noisy the oscillator may be — and the relationship runs the uncomfortable way, since the better the winding, the quieter the drive has to be.

A skew changes the shape. Its offset grows as L/R, linearly in 1/R, so at low resistance it uses the whole headroom by itself and no noise at all is allowed. Fifty picoseconds of skew needs at least 0.069 ohms, a hundred needs 0.138 and two hundred needs 0.276 — the least resistance is proportional to the skew, because the offset is. Above that resistance each skew’s curve rises to meet the no-skew curve, since the offset falls as 1/R and the spread only as $1/\sqrt{R}$: at ten ohms all four skews allow nearly the same noise, and the skew has stopped mattering.

The figure is the boundary the walk essay could not draw. Its ideal core had a count and nothing else: any noise saturated it eventually, and the question was only when. Here the same noise and skew either fit or do not, and what decides it is a number on the schematic — the resistance in series with the magnetising inductance — rather than a duration.

How far the three-sigma rule can be trusted #

Three standard deviations is the conventional allowance and it deserves a sentence on why it is reasonable here rather than merely conventional. The stationary flux is a weighted sum of a great many independent normal draws — with L/R at a hundred periods, roughly two hundred of them carry most of the weight — so it is normal to a very good approximation, and three standard deviations is exceeded with a probability of about 0.27% at any one period-end. That is not the probability of saturating over a lifetime. A stationary process visits its tails repeatedly, and a core that runs for a billion periods will pass three standard deviations millions of times.

What protects it is that a period-end flux three standard deviations out is not yet saturation; the headroom was set at a fifth of the swing, and the figure asks that offset plus three spreads fit inside it. How often a stationary process with this correlation time exceeds a level further out is a level-crossing question, and its answer depends on how quickly the flux decorrelates — on τ again. For the purpose of drawing a boundary three standard deviations is a reasonable convention, and the curves scale exactly: a designer who wants four simply reads each allowance as three-quarters of what is drawn.

Where the parts of this model come from #

Three assumptions bound what the numbers say, and each is a place the model could be made more real.

The noise is a held draw. Each threshold carries one normal sample, armed when the threshold is armed and held until it is crossed, which is the model under every result since the decision taken where the ramp is slowest. The anticorrelation, and therefore the discount, depends on it: a draw appears in two consecutive half cycles precisely because it is held across the flip. A comparator whose noise is a wideband floor would decorrelate the end of one half from the start of the next, and the discount would shrink toward one — the bandwidth noise sees is where that kind of noise is priced.

The branch is linear. The inductance does not depend on the flux. That is right inside the headroom and wrong at its edge, where a real core’s permeability falls and its magnetising current rises faster than the flux; a boundary in volt-seconds is where that edge is drawn, and the leak there is not a constant L/R but one that shortens as the core approaches it.

The resistance is fixed. A winding warms as it carries current, and copper’s resistance rises about 0.39% a kelvin, so a transformer that runs hot has a shorter L/R, a smaller spread and a larger allowance than the same transformer cold. The cold start is the case the boundary should be drawn for, which is the opposite of the usual habit of designing for the hot condition.

Still open: a level crossed rather than a spread, the count without a leak, and a real core’s edge #

How often the stationary flux crosses a given level. The spread is a standard deviation, and the design question is a rate: how many times per second a core with this spread, and this correlation time, reaches a headroom four or five standard deviations out. For a first-order process that rate has a known form in terms of the level and the time constant, and the anticorrelation should enter it through the curvature of the flux’s autocorrelation at zero lag — which is exactly where a held draw and a wideband floor differ most.

The count in the ideal core, as a distribution rather than a comparison. Without a leak, the skew’s ramp and the noise’s walk together reach a headroom at a count that is random, and the walk essay compared only their sizes. The distribution of that count — when the first of many drives saturates, not when the average one does — is a first-passage problem with drift, and its answer is a contour in skew and noise for a burst of stated length, the ideal core’s counterpart of the figure above.

A core whose leak depends on the flux. Near its edge the magnetising inductance falls, the leak shortens, and the restoring pull on the flux strengthens exactly where it is needed. Whether that self-limiting effect widens the safe region measurably, or only at flux levels already beyond the headroom, is a march with the core’s own curve in place of a fixed L.