The walk the core sees

The decision taken where the ramp is slowest put noise on the thresholds of a relaxation oscillator and measured what it does to the period. Each decision is a fresh draw, and each draw does two things: it moves the crossing it was armed for, with a sensitivity $B = RC/V(1-\beta)$ set by the slow slope the ramp has at its threshold, and it moves the level the next half cycle starts from, with a sensitivity $A = RC/V(1+\beta)$ set by the fast slope the ramp has as it leaves. A period is two half cycles and three draws, and the one in the middle — shared between the halves — enters with $A + B$, the largest coefficient of the three. So the period’s spread is twenty per cent above the two-crossing estimate at β = 0.5, consecutive periods are positively correlated, and the jitter accumulates at 3.771 units per root period.

It is the fourth look at one circuit. Two thresholds because there is a floor added the hysteresis to stop a comparator chattering, and the period a delay lengthens closed the loop into an oscillator. The noise essay was about a clock. An oscillator of this kind is as often a drive: its square wave switches a transformer, and the delay that is two delays showed what a transformer asks of it. The core does not care when an edge arrives. It integrates the voltage, so it cares about the difference between the time spent high and the time spent low, because that difference, multiplied by the rail voltage, is the volt-seconds left behind each cycle. That essay found ten nanoseconds of comparator skew saturating a hundred-turn core in 231 cycles.

The same draws are in the difference. They enter it with different signs, and that changes nearly every conclusion about the noise.

Adding and subtracting the same draw #

Write a period’s high half and low half in terms of the three draws that decide them. The first draw set the level the high half started from; the second ended the high half and started the low one; the third ended the low half. With every draw a threshold noise of standard deviation σ, a draw that moves the upper threshold up lengthens the high half that is climbing towards it, by $B$ per volt, and also lengthens the low half that starts from it, by $A$ per volt, because the ramp has further to fall.

So the shared draw enters the high half with $B$ and the low half with $A$, both positive. The period is their sum, where the shared draw enters with $A + B$. The imbalance — high less low — is their difference, where it enters with $B - A$.

Everything else follows from that one subtraction. The per-period spreads are

$$\sigma_T = \sigma\sqrt{A^2 + (A+B)^2 + B^2}, \qquad \sigma_D = \sigma\sqrt{A^2 + (B-A)^2 + B^2},$$

and at β = 0.5, with $A$ and $B$ at 0.667 and 2.000 in units of $RC/V$, these are 3.40 and 2.49 nanoseconds for the oscillator the essays have been measuring — a microsecond RC, five-volt rails, five millivolts of noise. The event map, run for forty thousand periods, measures 3.39 and 2.49.

The duty error is quieter than the period, but not by as much as the shared draw’s coefficients suggest. A period’s middle coefficient is 2.667 and the imbalance’s is 1.333, a factor of two. The spreads differ by 1.36, because the draws at the two ends of the period — $A$ at the start and $B$ at the end — enter both quantities alike. The end draw $B$ is the largest single term in the imbalance and second only to the shared draw in the period, so the two quantities share their biggest contribution and differ only in the one they do not.

The anticorrelation #

Consecutive periods share a draw with the same sign in both — $B$ at the end of one and $A$ at the start of the next — so their covariance is $AB$ and they are positively correlated, 0.115 at β = 0.5 in closed form.

Consecutive imbalances share a draw too, and the signs are not the same. The draw that ends one period’s low half enters that period’s imbalance with $-B$, because it lengthens the half being subtracted, and it enters the next period’s imbalance with $+A$, because it lengthens the high half that starts from it. The covariance is $-AB$, and consecutive duty errors are anticorrelated:

• at $\beta = 0.2$, −0.432 measured against −0.429 in closed form;
• at $\beta = 0.5$, −0.217 against −0.214;
• at $\beta = 0.8$, −0.064 against −0.062.

The anticorrelation is strongest where the hysteresis is smallest, because there $A$ and $B$ are close and the shared term is most of the variance. At β = 0.2 a period whose high half ran long is followed, more often than not, by one whose high half runs short. The duty error is partly a dither that undoes itself.

That has a direct consequence for anything that averages the duty cycle — a filter, a core, a thermal mass. An average over N periods of a quantity whose consecutive values are negatively correlated has a smaller spread than the per-period figure divided by $\sqrt{N}$ would suggest, because neighbouring errors cancel rather than merely failing to add. The period’s positive correlation made the clock’s accumulated jitter worse than the per-period figure predicts; the imbalance’s negative correlation makes the core’s accumulated error better.

The walk, and why it is β #

How much better is exact, and it is the reason the per-period figure is the wrong number to hand a transformer designer.

Sum the imbalance over N consecutive periods. Every interior draw appears in two consecutive half cycles — once ending a half and once starting the next — and the two appearances are a $B$ and an $A$ with opposite signs in the difference, so each interior draw contributes $B - A$. There are $2N - 1$ of them. The first and last draws of the stretch appear once each and do not cancel. The spread of the summed imbalance is

$$\sigma\sqrt{A^2 + B^2 + (2N - 1)(B - A)^2},$$

and for large N that is $\sigma\sqrt{2N}\,|B - A|$. The summed period has the same form with $A + B$ in place of $B - A$, which gives the 3.771 per root period the clock essay found. The ratio of the two walks is

$$\frac{B - A}{B + A} = \beta,$$

exactly: $B - A$ is $(RC/V)\cdot 2\beta/(1-\beta^2)$ and $B + A$ is $(RC/V)\cdot 2/(1-\beta^2)$. The volt-second walk a core sees is β times the timing walk a clock sees, and at β = 0.5 that is 1.89 nanoseconds per root period against 3.77.

Six hundred independent seeded runs of the event map, each summed over a hundred periods, give a spread of 18.3 nanoseconds in the imbalance and 35.9 in the timing — a ratio of 0.511 against β = 0.5 — and the two curves sit on the exact finite-N expressions to within the six per cent that six hundred runs buys. At β = 0.2 the hundred-period figures are 5.86 and 28.0 nanoseconds, a ratio of 0.209; at β = 0.8, 60.9 and 74.7, a ratio of 0.815.

The per-period duty error and the walk therefore disagree about the hysteresis entirely. As β goes to zero the per-period duty error does not vanish — at β = 0.2 it is still 1.56 nanoseconds, and it tends to $\sigma\sqrt{2}\cdot RC/V$ as the two end draws take over — while the walk goes to zero with β. What a single period’s imbalance mostly contains is the two draws at its ends, and those are exactly the contributions a sum over many periods telescopes away.

So a measurement of cycle-to-cycle duty jitter overstates what a core will integrate, by a factor that grows without limit as the hysteresis shrinks. At β = 0.2 the per-period figure would predict a spread of 15.6 nanoseconds after a hundred periods; the core sees 5.86.

What an instrument that measures duty jitter is reporting #

The disagreement between the per-period figure and the walk is not a curiosity of the model; it decides what a measurement means. An oscilloscope’s pulse-width statistics, or a time-interval analyser set to measure high time less low time, reports the per-period spread — 2.49 nanoseconds at β = 0.5 — and a designer extrapolating to a core will multiply it by $\sqrt{N}$.

The exact expression says how wrong that is at each count, and the error arrives immediately. Over two periods the imbalance’s spread is $\sigma\sqrt{A^2 + B^2 + 3(B-A)^2}$, which is 3.13 nanoseconds against the 3.53 that independence gives. Over ten it is 6.18 against 7.89. Over a hundred it is 18.9 against 24.9, and the six hundred runs measured 18.3. The ratio heads towards $|B - A|\sqrt{2}/\sqrt{A^2 + (B-A)^2 + B^2}$, which is 0.756 at β = 0.5 and 0.378 at β = 0.2: the extrapolation is a third high at the first ratio and more than two and a half times high at the second.

The measurement that does predict the core is not a pulse-width histogram at all. It is the spread of the summed imbalance over a record as long as the core integrates for, which is what a volt-second integrator on the drive would read directly — and it is the same quantity the six hundred seeded runs measure. Read the other way, a pulse-width histogram that looks alarming on a low-hysteresis oscillator may be almost entirely the two end draws of each period, which a core never sees.

A skew is a ramp, and the noise is a walk #

The delay essay’s comparator had a skew — one transition slower than the other — and no noise. A skew adds the same imbalance to every period, so the volt-seconds ramp linearly and a core saturates after a number of cycles proportional to its headroom. The flux that walks measured that ramp directly: a one per cent imbalance saturates its core in 64 cycles, and halving the drive only doubles the count. Noise adds a walk instead, whose spread grows as $\sqrt{N}$, so it reaches a given headroom in a number of periods proportional to the square of it.

A real comparator has both, and which one a core feels depends on the count. The ramp’s mean after N periods is N times the per-period imbalance; the walk’s standard deviation is $\sigma\sqrt{2N}\,|B - A|$. They are equal after

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

periods, with the ramp measured on the noiseless event map for each skew. Before $N^*$ the core wanders at random about a drift too small to see; after it, the drift dominates and the wandering is a ripple on it.

At β = 0.5 and five millivolts of noise, a skew of a nanosecond is overtaken after 7.99 periods, a tenth of a nanosecond after 800, and ten nanoseconds after less than a tenth of a period. At β = 0.2 the same skews give 312, 3.11 and 0.03 periods; at β = 0.8, 5,000, 50.0 and 0.5.

Two readings follow from the numbers, and both are cleaner than the expression suggests.

The dependence is an inverse square, visible down every column: a tenfold smaller skew is outrun by the noise for a hundredfold longer. That is the ramp against the square root, and it is why trimming a skew has diminishing value — past a point, the core’s worst case is set by the noise however well the skew is trimmed, and a designer could stop trimming there.

And the β-dependence collapses to one quantity. At a nanosecond of skew the three counts are 3.11, 7.99 and 50.0, which are $2/(1-\beta)^2$ at β = 0.2, 0.5 and 0.8 to within half a per cent. Since $B\sigma = \sigma RC/V(1-\beta)$ is exactly one nanosecond divided by $(1-\beta)$ for this oscillator, the counts are $N^* = 2(\sigma B/d)^2$ for a skew $d$: the skew is outrun by the walk until it exceeds $\sqrt{2/N}$ times the timing error of a single slow crossing. The fast-slope sensitivity $A$ drops out, because the skew’s own imbalance carries the same factor of β as the walk’s.

That answers the question the clock essay left about whether a twenty-nanosecond skew is visible above the noise. On this oscillator it is visible within a fiftieth of one period. Skew only becomes invisible once it is comparable to the single-crossing timing error, and on this oscillator that is a nanosecond or two.

At a fixed frequency, the core and the clock disagree #

Everything so far is at a fixed RC, where more hysteresis also means a longer period, so comparisons across β mix two effects. The clock essay removed that by rescaling RC at every divider ratio to hold the frequency, and found the accumulated timing jitter, as a fraction of the period, smallest at β = 0.648, where $\beta\ln\frac{1+\beta}{1-\beta} = 1$.

Do the same for the two duty quantities and they disagree with it and with each other.

The per-period duty error is smallest at β = 0.533, found by golden section on the closed form, where it is 1.131 units of σT/V. It has an interior minimum for the same reason the timing jitter does: small hysteresis makes the period short, so a fixed timing error is a large fraction of it, and large hysteresis makes the crossings slow.

The imbalance walk has no interior minimum at all. It is β times the timing walk, and the timing walk rises as $1/\beta$ at small β because the period shrinks as β, so their product tends to a floor of $1/\sqrt{2}$ as the hysteresis vanishes and rises at every step from there: 0.886 at β = 0.53, 1.023 at 0.65 and 1.430 at 0.8. At a fixed frequency, less hysteresis is always better for the core, down to a floor that no divider ratio can go below.

So a relaxation oscillator used as a drive and one used as a clock want different hysteresis, and the difference is not small. At the clock’s optimum of 0.648 the core’s walk is 45 per cent above its floor; at 0.2 it is within three per cent of it, where the clock’s accumulated jitter is more than twice its own optimum. A designer choosing the divider ratio has to know which of the two the oscillator is for.

What the noise model leaves out #

Two things bound these conclusions, and both are worth stating before anyone sizes a core from them.

The noise model is a held draw. Each threshold carries one normal draw, armed when the threshold is armed and held until it is crossed — noise with a correlation time long against a crossing and short against a half cycle. The telescoping that makes the walk small depends on it: a draw appears in two consecutive half cycles precisely because it is held across the flip. A wideband floor, of the kind the bandwidth noise sees prices, would decorrelate the end of one half from the start of the next, and the cancellation in $B - A$ would weaken. How much is the first of the questions below.

And the core is ideal. A winding with no resistance integrates the imbalance for ever, so a walk and a ramp both grow without limit and the question is only when they reach saturation. The walk that stops showed that resistance in the winding gives a ramp a fixed point: the magnetising current settles at the drive’s direct component over the resistance. A random imbalance under the same resistance should settle too, to a stationary spread rather than a growing one.

Still open #

A wideband floor in place of a held draw. With noise whose correlation time is short against the crossing, the timing error at each threshold is set by the noise bandwidth against the slope, and the draw that ends one half is no longer the draw that starts the next. The cancellation that makes the interior coefficient $B - A$ then weakens, and the walk’s ratio to the timing walk should move away from β by an amount that depends on the comparator’s bandwidth. Measuring the ratio’s dependence on β is what would say which model a real comparator obeys.

The stationary spread under winding resistance. An RL magnetising branch leaks the walk away with a time constant L/R, so the flux should settle at a spread of the order of the walk’s rate times the square root of that time constant, in periods. Whether the anticorrelation between consecutive imbalances reduces it further — it does at short counts, and the leak makes every count short — is a march with a seeded drive, and it would turn this essay’s inverse square into a boundary in millivolts and ohms.

And the flux as a boundary rather than a count. A boundary in volt-seconds drew the core’s limit in the units the drive is specified in. With a skew and a walk together, the quantity the core sees after N periods is a ramp plus a spread, and the probability of reaching a given headroom within a given number of periods is a first-passage problem with drift, whose answer is a contour in skew and noise rather than a single count.