Circuits that do a job, and the range they do it over

The count a core survives

A relaxation oscillator with a 300 ps comparator skew and 5 mV of threshold noise saturates an ideal core with 100 ns of volt-second headroom after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. The mean does not see the noise at all until the noise is large enough to saturate the core the other way first; the early tail sees nothing else. Every skew and noise level falls on one curve in one ratio, the skew's count over the count at which its ramp overtakes the walk. Read as a burst, a thousand periods with one failure in a thousand allows 150 ps of skew with no noise, 93.5 ps at a millivolt, and none at all above 2.41 mV.

17 min read 6 figures Two routes to a number Every model has an edge Measured, not tabulated

Assumes: Two thresholds because there is a floor · A boundary in volt-seconds

A transformer driven by a square wave from a relaxation oscillator collects whatever volt-seconds the two half cycles fail to cancel. The flux that walks measured that accumulation for a fixed imbalance: one per cent of excess in one half saturates its core in 64 cycles, and the count is simply the headroom divided by the excess. The walk the core sees added threshold noise and found the imbalance walking as well as ramping — a spread of $\sigma\sqrt{2N}|B - A|$ after N periods, β times the timing walk — and compared the two by asking when the ramp’s mean overtakes the walk’s standard deviation. That count, $N^*$ , was a comparison of sizes. It did not say when any particular core saturates.

That is a different question, and for anything that runs in bursts it is the one that matters. A gate drive that switches a thousand times and then rests, a pulse transformer reset between packets, a converter that is started and stopped — each of these runs an ideal-enough core for a finite count and needs to know the probability that it reaches its headroom within that count. With a winding resistance the spread a winding resistance holds found the walk settling to a stationary spread, and the core either fits or it does not. Without one, every core saturates eventually, and the answer is a distribution of counts.

The distribution has a name. A random walk with a drift, started at zero and stopped the first time it reaches a level, takes a random time that is the first-passage time, and for a continuous walk its law is known in closed form. This essay measures it on the event map and holds it against that law.

Two thousand drives into one core

The oscillator is the one every essay on this circuit has used: a microsecond of RC, five-volt rails, β = 0.5 and five millivolts of noise on each threshold, a period of 2.197 microseconds. Its comparator is 300 picoseconds late going up and on time coming down. The core is ideal and starts demagnetised, and it saturates when the summed high-minus-low time leaves ±100 nanoseconds — 500 volt-nanoseconds of flux, 9.1% of the 5,493 that one half cycle swings.

Run the oscillator from two thousand seeds, each until its core saturates, and record the count.

Fig. 1 Two thousand seeded drives with 300 ps of skew and 5 mV of noise, each run until the summed imbalance leaves ±100 ns. The counts on logarithmic bins against the first-passage law of a walk with the same drift and spread: mean 502 against 500, one drive in a hundred saturated by 184 periods. The slider is the skew.

The skew adds an imbalance r to every period, and it is not the skew itself: a comparator late by d going up lengthens the high half by d, and because the capacitor kept charging through the delay it also lengthens the low half by a third of that, so the imbalance is d·2β/(1 + β), 0.2 nanoseconds a period. Alone, that ramp reaches 100 nanoseconds in exactly 500 periods. The noise adds a walk of spread $s = \sigma\sqrt2|B - A|$ = 1.89 nanoseconds per root period.

The two thousand counts have a mean of 502 periods, against the 500 the skew alone gives. Their median is 463, one in ten has saturated by 275 and one in a hundred by 184. The histogram and the first-passage law agree everywhere: the largest gap between their cumulative distributions is 0.014, inside the sampling error two thousand drives allow.

The law is the method of images for a drifting walk between two absorbing barriers — the core saturates on whichever side it reaches first. A change of measure carries the drift, $e^{\mu x/s^2 - \mu^2 t/2s^2}$ times the driftless density, and the driftless density between barriers at ±h is a sum of Gaussians reflected through both. Integrated over the interval it gives the probability that a core has survived N periods as a sum of differences of normal integrals, each weighted by an exponential in the drift. It converges in a handful of images at every count drawn here. A textbook alternative, the eigenfunction series, is exact too and useless in practice when the drift dominates: its terms reach $e^{rh/2s^2}$ — $e^{235}$ for a millivolt of noise and a nanosecond of skew — and cancel to a probability.

The mean does not see the noise

The first thing the distribution says is surprising enough to have a name. For a walk with a positive drift and a single barrier, the mean first-passage time is the distance over the drift, h/r, whatever the noise. It is Wald’s identity: the walk’s expected position after N steps is rN whether it wanders or not, and the passage happens, on average, when that expected position reaches the barrier.

With 5 mV of threshold noise, a 1 ns skew's volt-second ramp overtakes the random walk after 7.99 periods at β = 0.5. A closed form for the walk and the noiseless event map for the ramp. A comparator whose rising decision is later than its falling one by a skew adds a fixed amount to every period's high-minus-low time, measured on the event map; threshold noise adds a walk of σ√(2N)|B − A|. The two are equal, the walk's standard deviation against the ramp's mean, after N = 2σ²(B − A)²/ramp² periods. At β = 0.2: 0.1 ns → 312, 0.3 ns → 34.7, 1 ns → 3.11, 3 ns → 0.344, 10 ns → 0.0302. At β = 0.5: 0.1 ns → 800, 0.3 ns → 88.9, 1 ns → 7.99, 3 ns → 0.887, 10 ns → 0.0795. At β = 0.8: 0.1 ns → 5.00e+3, 0.3 ns → 556, 1 ns → 50.0, 3 ns → 5.55, 10 ns → 0.499. Below N a core driven by the oscillator walks at random; above it, it ramps; and N* falls as the square of the skew. — Fig. 2 From the walk essay: the count after which a skew’s ramp overtakes the noise’s walk, N*, at three divider ratios. It falls as the inverse square of the skew. Its comparison of a mean with a standard deviation is what the distribution below replaces.

So the comparison the walk essay drew above, between a ramp’s mean and a walk’s standard deviation, never moves the average saturation count at all. With 300 picoseconds of skew the measured mean is 501 periods at one millivolt of noise, 501 at two and 503 at five; what changes is how the counts are spread around it.

With a 300 ps skew the mean count stays at 500 from 1 mV of noise to 5, while one drive in a hundred saturates by 410 periods at the first and 184 at the second. Seeded, 1,000 drives at each noise level, against the closed-form first passage. β = 0.5, a skew of 300 ps, 100 ns of headroom. 1 mV: mean 501 (500), 10% by 449 (447), 1% by 407 (410); 2 mV: mean 501 (500), 10% by 399 (398), 1% by 336 (334); 5 mV: mean 503 (500), 10% by 272 (273), 1% by 184 (184); 10 mV: mean 440 (443), 10% by 147 (144), 1% by 88 (77); 20 mV: mean 179 (169), 10% by 49 (45), 1% by 23 (22). While rh/s² is large the mean is h/r = 500 and the noise moves only the spread, and the early tail with it. At 10 and 20 mV the walk can reach the opposite barrier first, and the mean falls to (h/r)·tanh(rh/s²). — Fig. 3 A thousand seeded drives at each of five noise levels with 300 ps of skew. The mean stays at h/r = 500 periods until the noise is large; the counts by which one drive in ten and one in a hundred have saturated fall at every step. The curves are the two-barrier law; the dots are measured.

The early tail falls at every step. One drive in a hundred has saturated by 410 periods at a millivolt, 334 at two and 184 at five. The noise does not shorten the average life of the core, but it brings the first failures forward, and a burst-mode design lives or dies by its first failures.

Past five millivolts the mean starts falling as well, and the reason is the second barrier. A core saturates in either direction, and with enough noise a walk can reach the far side, against the drift, before the drift carries it to the near one. The two-barrier law has a mean in closed form,

\bar N = \frac{h}{r}\tanh\!\left(\frac{rh}{s^2}\right),

which is Wald’s h/r when $rh/s^2$ is large and the walk’s own $h^2/s^2$ when it is small. At ten millivolts $rh/s^2$ is 1.41, the hyperbolic tangent is 0.887 and the mean is 443 against 440 measured; at twenty it is 0.352, the mean is 169 against 179 measured. The noise moves the mean only once it is able to saturate the core the wrong way.

One honest discrepancy shows in the figure. At ten and twenty millivolts the seeded early tail sits about a tenth later than the law predicts — 88 periods for one drive in a hundred against 77, 49 for one in ten against 45. The law is for a continuous walk, which can touch a barrier between two period-ends and come back; the core here is read once a period, so a drive whose imbalance crossed and recrossed inside a period has not saturated by this definition. That matters only when the earliest failures come within a few dozen periods, which is exactly the regime of large noise, and it errs on the safe side.

Why the mean is the wrong number to quote

If the noise leaves the mean alone, what it changes is the spread, and the law gives that too. Far from the second barrier the count has a variance of $hs^2/r^3$ , so its standard deviation relative to its mean is

\frac{\sigma_N}{\bar N} = \sqrt{\frac{s^2}{rh}},

one over the square root of the same ratio that decided the mean’s survival. With 300 picoseconds of skew and five millivolts that is 211 periods on a mean of 500 — a coefficient of variation of 0.42. A core specified as “lasting 500 periods” lasts anywhere from about 200 to about 900 on a typical run, and the specification has said nothing about which.

The variation also shrinks only slowly as the headroom grows. Doubling h doubles the mean and multiplies the standard deviation by $\sqrt2$ , so the relative spread falls by $1/\sqrt2$ , not by half. A larger core makes saturation later and more predictable, but at the rate of a square root; the earliest failures come proportionally closer to the mean only as the core grows by factors of four.

That is the practical reason to describe a burst-mode core by a quantile — the count by which one drive in a hundred, or one in a thousand, has saturated — rather than by the mean count, which is the number the ramp alone suggests and the one most easily computed. The two-barrier law supplies the quantile directly; the seeded drives check it where they can. Two thousand drives resolve the one-in-a-hundred count to a few per cent and the one-in-a-thousand count hardly at all, since only two drives fall beyond it; a direct measurement of the one-in-a-million count, which is where a production volume of gate drives would put the requirement, would need a few hundred million drives. Below the resolution of the seeds the law is the instrument, and the agreement above it, across ten combinations of skew and noise, is what licenses using it there.

Headroom is worth its square against noise

With no skew there is no drift, and the walk reaches a barrier by diffusion alone. The mean count is then $h^2/s^2$ — 2,812 periods for this oscillator, against 2,850 from the seeded drives — and every quantile scales the same way. Doubling the headroom quadruples the life of a core whose only enemy is noise.

Doubling the headroom quadruples a noisy core's life with no skew (533 → 2130 periods) and barely doubles it with 300 ps of skew (212 → 460). Seeded, 800 drives at each headroom, against the closed-form median first passage. β = 0.5, 5 mV of threshold noise. 25 ns: no skew 141 (133), 300 ps 93 (88); 50 ns: no skew 555 (533), 300 ps 215 (212); 100 ns: no skew 2174 (2130), 300 ps 466 (460); 200 ns: no skew 8341 (8522), 300 ps 953 (958). The walk's count rises as the square of the headroom, slope 2.00; the ramp's as its first power, slope 1.15 across the range drawn. — Fig. 4 The median count to saturation against the headroom, from 800 seeded drives at each of four headrooms, with no skew and with 300 ps. With no skew the count rises as the square of the headroom; with the skew, nearly as its first power.

With 300 picoseconds of skew the median rises much more slowly: 212 periods at 50 nanoseconds of headroom, 460 at 100 and 958 at 200. Across the range drawn the slope is 1.15 rather than 2, and it heads towards one as the headroom grows and the ramp takes over. The noise-only median is 533, 2,130 and 8,522 at the same headrooms, and the seeded medians sit on both curves to within a few per cent.

The two slopes are a design rule in miniature. Against a skew, headroom buys life linearly, and trimming the skew buys it linearly too: halve the skew and the core lasts twice as long. Against noise, headroom buys life quadratically, and reducing the noise buys it quadratically: halve the noise and the core lasts four times as long. A designer deciding whether to spend effort on the comparator’s skew or its noise, or on a larger core, needs to know which regime the drive is in, and the next figure says how to tell.

One ratio decides how early the first core goes

The first-passage law has three parameters — the drift r, the spread s and the headroom h — but measured in units of h/r its shape depends on only one combination of them, $rh/s^2$ . That ratio has a meaning in the walk essay’s terms. $N^*$ was $s^2/r^2$ , the count at which the ramp’s mean overtakes the walk’s standard deviation; h/r is the count the skew alone would take; so

\frac{rh}{s^2} = \frac{h/r}{N^*}.

Whether a drive is skew-limited or noise-limited is decided by whether the skew’s own count is longer or shorter than $N^*$ — not by comparing the skew with the noise directly, but by comparing the count at which the ramp alone saturates with the count at which the ramp first stands out of the walk.

Ten skews and noise levels fall on one curve: the first drive in a hundred lasts half as long as the skew alone allows only once h/r is 12 times N*. Seeded, 1,000 drives at each of ten combinations of skew (30 ps to 1 ns) and noise (1 to 20 mV) at β = 0.5 and 100 ns of headroom. For each, the count by which one drive in a hundred has saturated, divided by h/r — the count the skew alone would give — against h/r divided by N = s²/r², the count at which the skew's ramp overtakes the noise's walk. Scaled that way, the first-passage distribution depends on that one ratio, rh/s², and the curve is drawn from it: 100 ps, 1 mV → 46.9, 0.703; 100 ps, 2 mV → 11.7, 0.477; 100 ps, 5 mV → 1.88, 0.186; 300 ps, 5 mV → 5.63, 0.368; 300 ps, 10 mV → 1.41, 0.176; 300 ps, 20 mV → 0.352, 0.046; 1000 ps, 5 mV → 18.8, 0.587; 1000 ps, 20 mV → 1.17, 0.140; 30 ps, 2 mV → 3.52, 0.277; 30 ps, 10 mV → 0.141, 0.019. When h/r is many times N the drive saturates near h/r and the skew is the whole story; when it is less than N* the early drives saturate a small fraction of the way there. — Fig. 5 One drive in a hundred’s count to saturation, as a fraction of the skew’s own count h/r, against the ratio of h/r to N*. Ten combinations of skew and noise, a thousand seeded drives each, fall on the one curve the two-barrier law gives.

Ten combinations — skews from 30 picoseconds to a nanosecond, noise from one millivolt to twenty — land on one curve. When h/r is many times $N^*$ the first drive in a hundred saturates close to the skew’s own count: at a ratio of 47 it lasts 0.70 of it, at 19 it lasts 0.59. When h/r is comparable with $N^*$ the early drives go a small fraction of the way: at a ratio of 1.9 the first in a hundred lasts 0.19 of the skew’s count, at 0.35 it lasts 0.046. The first drive in a hundred lasts half as long as the skew alone allows only once h/r is twelve times $N^*$ .

That threshold is worth having as a number. A design whose skew-limited count is less than about twelve times $N^*$ is, for its earliest failures, a noise-limited design, whatever the comparison of sizes says, and trimming the skew will buy much less than the ramp suggests. The reverse holds above it.

A burst as a boundary

The last step is the contour the walk essay could not draw. Fix a burst length and a tolerable probability of failure — here, one drive in a thousand saturating within the burst — and ask, for each level of noise, what is the largest skew the core can stand.

For one drive in a thousand to saturate within a burst, 100 ns of headroom allows 150 ps of skew with no noise and none at 2.41 mV — over a thousand periods. computed by solving, not by drawing, from the first-passage law the seeded drives were held against. β = 0.5, 100 ns of headroom in summed imbalance. For bursts of 100, 1000, 10000 periods, the largest skew for which the probability of saturating within the burst stays below one in a thousand, against the threshold noise. With no noise the answer is the skew whose ramp reaches the headroom in exactly the burst: 1499.5 ps, 150.0 ps, 15.0 ps. Noise eats into it and ends it: no skew at all is safe above 7.62 mV, 2.41 mV, 0.762 mV, respectively. — Fig. 6 The largest comparator skew for which no more than one drive in a thousand saturates a core with 100 ns of headroom within a burst, against the threshold noise, for bursts of a hundred, a thousand and ten thousand periods. Each curve falls to zero at the noise that saturates one drive in a thousand with no skew at all.

With no noise the answer is arithmetic: the skew whose ramp reaches the headroom in exactly the burst, 1,500 picoseconds for a hundred periods, 150 for a thousand and 15 for ten thousand. Noise eats into it. For the thousand-period burst, a millivolt of noise takes the allowance from 150 picoseconds to 93.5, two millivolts to 34.2, and at 2.41 millivolts no skew at all is safe, because the noise alone saturates one drive in a thousand within the burst. For a hundred-period burst the allowance falls from 1,500 to 1,324 picoseconds at a millivolt and lasts until 7.62; for ten thousand periods it is already only 5.9 picoseconds at half a millivolt and is gone by 0.762.

The ends of the curves move as they should. The quiet-end allowance scales inversely with the burst, because a ramp reaches the headroom in a count proportional to it. The noise at which the allowance vanishes scales as one over the square root of the burst, because a walk does: 7.62, 2.41 and 0.762 millivolts are in the ratio of $1 : 10^{-1/2} : 10^{-1}$ to three figures. Between the ends the contour is the two-barrier law solved for the skew, and it bends — at a millivolt and a thousand periods the noise has already cost 38% of the quiet allowance, though the noise alone would saturate about one drive in ten quadrillion within the burst.

That bend is the interaction the comparison of sizes could not show. Noise that is harmless on its own is not harmless beside a skew, because the skew has already carried the walk most of the way to the barrier and the noise only has to cover the rest. Near the quiet end, the skew’s ramp reaches the headroom just as the burst ends, so a draw that lengthens the last few hundred periods’ imbalance by a fraction of a standard deviation is enough; the early tail is the ramp’s arrival time, smeared by the walk’s spread at that moment, and the walk’s spread after a thousand periods at a millivolt is about twelve nanoseconds, an eighth of the headroom.

Read as a design, the figure turns a specification argument into a lookup. A designer with a thousand-period burst, a core with a hundred nanoseconds of imbalance to spare, and a comparator quoted at a millivolt of input noise can allow about 90 picoseconds of skew, which is a demanding figure for a discrete comparator and an easy one for a matched pair inside one package. The same designer with two millivolts of noise has a third of that, and at two and a half has nothing: the core must be larger, the burst shorter, or the noise lower, and the skew is no longer the variable that matters. Which of those is cheapest is a judgement the figure does not make, but it says where the judgement starts.

What the law is, and is not, being asked

Three bounds on what these numbers say.

The core is ideal. Its flux is the running sum of volt-seconds, so every core saturates eventually and the question is purely one of when. Any resistance in the magnetising branch changes that — it bounds the walk and the ramp alike — and the spread essay drew the boundary in that case. A core that runs long enough to reach its stationary spread needs that analysis instead; a core that runs for much less than its L/R time constant needs this one, and the crossover between them is the leak time constant itself.

The noise is a held draw, one normal sample per threshold, as in the decision taken where the ramp is slowest. The walk’s spread per root period, s, is the only property of the noise the first-passage law uses once the count is long, so a different noise model would change these numbers through s alone — provided its consecutive draws telescope the way held ones do.

The headroom is a hard edge. Saturation is taken to happen at a fixed imbalance; a real core’s permeability falls gradually, its magnetising current rises sharply rather than instantly, and a boundary in volt-seconds is where that edge is drawn in the units the drive is specified in. The first-passage count is to whichever level a designer calls the edge.

Still open: a burst’s start, a skew that wanders, and two failures that are one

Where a burst starts. Every drive here starts with a demagnetised core. A real burst starts wherever the previous one left the core, less whatever its reset removed, so the starting point is itself distributed and is correlated with how long the previous burst ran. The probability of saturating within the burst is then a first passage from a random start, and a reset that removes a fixed fraction of the offset sets up a renewal process whose stationary distribution of starting points is the quantity a burst-mode design actually needs.

A skew that drifts. A comparator’s skew depends on its temperature and its overdrive, so over a long burst it is not a constant. A drift that changes slowly turns the ramp into a curve and the first-passage law into one with a time-varying drift, which has no closed form; whether a skew that wanders by a tenth of its value moves the early tail more or less than a tenth of the noise does is a march with the skew as a second slow state.

Two failure modes that are one event. The core saturating and the oscillator’s period exceeding a tolerance are driven by the same draws — the timing walk and the imbalance walk share every threshold sample — so a burst that fails one way is more likely than average to fail the other. The joint first passage of the two walks, which are correlated by exactly the coefficients $A + B$ and $B - A$ , would say how much margin is double-counted when the two are specified separately.