Every tooth the same height

The millivolts in the wire put a return conductor in the netlist — five milliohms and ten nanohenries for ten millimetres of ordinary copper — and measured what somebody else’s hundred milliamps does to a ten-millivolt reading sharing it. Half a millivolt at low frequency, five per cent of the signal; a corner at 79.6 kilohertz where the inductance takes over; and above that an error rising a decade per decade with no ceiling, reaching the whole signal at 1.59 megahertz.

Every number there is for a sinusoid. The essay’s own list of things it does not claim ends with the admission that the interfering current does not have to be a current anybody meant to draw, and names two shapes from elsewhere in this collection that are the wrong shape: a rectifier drawing its average as a pulse thirteen times larger, and a switch-on transient twenty-six times larger still.

A converter’s current is neither of those. It is a trapezoid, repeating, and what it puts into a reading is a comb. The interesting thing about the comb is that it is flat.

The cancellation #

A trapezoid’s harmonic amplitudes are the product of two sincs — one for the duty, one for the edge — and what matters here is only their envelopes. Between the first corner and the second the amplitude of the $n$-th harmonic falls as $1/n$. Above the edge’s corner, $1/(\pi t_r)$, it falls as $1/n^2$.

The conductor’s impedance does the opposite. Below $R_g/2\pi L_g$ it is a constant five milliohms; above it, it is $2\pi f L_g$, which rises as $n$.

So in the band where the current falls as $1/n$ and the conductor rises as $n$, the product is a constant, and not approximately:

Flat to a tenth of a per cent across every harmonic from three times the conductor’s corner to a third of the edge’s, at every setting of the slider. A switching load puts the same voltage into a shared reading at every one of its harmonics, which is not what a spectrum of a square wave looks like and is not what the impedance curve looks like either. It is what their product looks like.

The two exponents that cancel are worth naming, because neither is a coincidence and neither is adjustable. The $1/n$ is the Fourier series of anything with a step in it — a property of the waveform having a discontinuity, and the same $1/n$ that makes a square wave’s harmonics audible. The $n$ is the impedance of an inductance, and the inductance is there because the conductor has a length. Both are first-order and they are exactly opposite.

What that does to the total #

A comb whose teeth are all the same height has a total decided by its length, and the length is set by the edge.

Harmonics of one periodic current sit at different frequencies, so they are orthogonal over any whole number of periods and their powers add exactly — two solves that add is the general statement, and this is its one case where the addition is a theorem rather than a mistake. So the total is the root sum of squares of the teeth, which for $N$ equal teeth is $\sqrt N$ times one of them, and $N$ is the number of harmonics before the edge’s corner: $1/(\pi t_r f_0)$.

$$v_\mathrm{total} \approx v_\mathrm{tooth}\sqrt{\frac{1}{2\pi t_r f_0}}$$

Fitted over those five, the total goes as the edge time to the power −0.500. A faster edge raises no harmonic at all; it adds teeth to the comb. Ten times the edge rate is ten times the harmonics and $\sqrt{10}$ times the error, which is 3.16 rather than 10 — better than a linear dependence and worse than none, and neither of the two guesses anybody makes.

That is the number a design can act on, and it inverts the usual advice in a specific way. Slowing an edge is the standard remedy for radiated emissions and it is worth a factor of ten there, because emission is about the amplitude at a frequency. Here it is worth $\sqrt{10}$, because the mechanism is a conducted sum rather than a peak — so the same series resistor that buys twenty decibels of emission buys ten of this.

The direct-current term, which is a different quantity #

The comb starts at the fundamental. There is a term below it and a slow measurement sees only that one.

The zeroth harmonic is the average current, and it crosses the conductor’s resistance with no inductance in it because it has no frequency: a hundred milliamps through five milliohms, 500 microvolts, which is exactly the number the earlier measurement reports and is unchanged by every edge rate on the slider.

So a measurement’s bandwidth decides which of two quite different errors it sees.

A slow measurement — a six-digit meter integrating over twenty milliseconds, or any reading averaged in firmware — keeps the average and rejects the whole comb, because the comb’s lowest tooth is at a hundred kilohertz. It reads 500 microvolts of error, five per cent of the signal, and the edge rate is irrelevant to it.

A fast measurement keeps the comb and reads 12.6 millivolts of error on a ten-millivolt signal, which is 126 per cent. The signal is not corrupted, it is buried, and the edge rate is the only thing that moves it.

Between them is a third case and it is the one that catches people. A measurement that samples rather than integrating folds the comb down into its own band, and the window the square-root law has is the arithmetic: a front end wider than a fraction of half the sample rate brings everything above it into the reading at full amplitude. A twenty-millisecond average of samples taken at thirty kilohertz, behind a front end at fifteen, has the 100 kHz fundamental and the 300 kHz third harmonic and everything after them folded somewhere inside its 25 hertz of noise bandwidth — not attenuated, and no longer at a frequency that identifies them.

The crest factor is not the quantity #

There is an older way of describing a current that is not a sinusoid, and it is worth saying why it does not answer this question.

A crest factor is the ratio of a waveform’s peak to its root-mean-square value, and is used here it twice: the direct voltage that is a sawtooth finds a rectifier drawing its 157 milliamps of average as a 2.098-ampere pulse over 28.8 degrees of each half cycle, a crest factor of 13.4 that gets worse as the reservoir capacitor is made larger. That number is exactly the right one for the question it answers, which is how much a conductor heats: heating is $I_\mathrm{rms}^2 R$ and a crest factor is the bridge between a peak somebody can see and an root-mean-square value they cannot.

It is the wrong number here, and the reason is that a crest factor has no frequency in it. Two currents with identical crest factors — one switching at ten kilohertz with a slow edge, one at ten kilohertz with a fast one — put entirely different totals into a shared reading, because the conductor’s impedance weights their harmonics and a crest factor has already summed over them.

The arithmetic makes that precise. The heating a conductor does is $\sum I_n^2 R$, with $R$ a constant, so it is decided by $\sum I_n^2$ — which is the root-mean-square value, and is dominated by the low harmonics because they are the large ones. The error a conductor adds to a reading is $\sqrt{\sum (I_n Z_n)^2}$, with $Z_n$ rising, so it is spread evenly over every harmonic up to the edge’s corner and is dominated by nothing. The two sums weight the same current in opposite ways.

So a switching current is characterised by two numbers that do not determine each other. Its root-mean-square value says what it heats, and its edge rate says what it corrupts, and the second is absent from every specification that quotes the first. That is a distinction this field keeps arriving at from different directions — the ammeter that is a resistor is where the same copper is the measurement rather than the interference, and the quantity it needs is again not the one a data sheet prints.

Why the usual remedies do what they do #

Three things a designer does about a shared return, priced against this comb rather than against a sinusoid.

Shortening the run divides both $R_g$ and $L_g$ by the same factor, so it divides the conductor’s impedance at every frequency and every tooth with it. The comb’s height falls in proportion and its length does not change, so the total falls in proportion. Halving the shared length halves the error, as the earlier measurement found for a sinusoid, and the corner stays at 79.6 kilohertz because it is a ratio of two quantities that both scaled.

Slowing the edge shortens the comb and leaves its height alone, so the total falls as the square root. This is the remedy the sinusoid picture cannot price at all, because a sinusoid has no edge.

Lowering the switching frequency does something neither of those does. The teeth get closer together — $N$ is $1/(\pi t_r f_0)$, so halving $f_0$ doubles the number of teeth — while each tooth’s height is unchanged, because the height is the product of a current falling as $1/n$ and an impedance rising as $n$, and $n$ is a harmonic number rather than a frequency. So halving the switching frequency makes the total worse by $\sqrt2$, which is the opposite of what anybody expects and is a real consequence of the cancellation: a slower converter draws its current for longer in each cycle, and the shared conductor does not care how long.

What a difference amplifier does to a comb #

The neighbouring essay’s repair is a difference amplifier across the shared conductor, and the rejection four resistors decide prices it: 54 decibels for tenth-per-cent resistors, set by one plus the gain over four times the tolerance, with the amplifier having nothing to do with it. Applied to a comb that is a flat 0.828 millivolts a tooth, 54 decibels leaves 1.65 microvolts a tooth and 25 microvolts of total at a ten-nanosecond edge — back under the signal, which is the arrangement working.

It works less well than that, and the reason is in the neighbouring field. The corner the instrument has no part in finds a 95-decibel instrument falling twenty decibels a decade above 290 hertz when its source has a kilohm of imbalance and ten picofarads at each input — a mechanism that is a difference of two time constants rather than anything about the part. Every tooth of this comb is above a hundred kilohertz, which is two and a half decades past that corner, so the rejection there is not 54 decibels but something in the twenties.

Which gives the honest arithmetic for the arrangement: a flat comb of 0.828 millivolts a tooth, rejected by twenty-odd decibels rather than fifty-four, is about 80 microvolts a tooth and a millivolt of total — ten per cent of the signal, from interference the same amplifier removes to a part in five hundred at direct current. A difference amplifier is a direct-current repair applied to a megahertz problem, and the comb is entirely in the band where it has stopped being one.

What it looks like on a bench, and the two tests that identify it #

The essay before it gives a signature for the direct-current version of this fault: the reading changes when a different circuit changes, it scales with the interfering current rather than with the signal, and two one-minute experiments identify it uniquely. The comb has a signature of its own and it is a different one.

It is at the switching frequency and its harmonics, and it is flat. That is unusual enough to be diagnostic on its own. Almost everything else that puts a switching frequency into a measurement puts it in with a falling spectrum — capacitive coupling gives a comb rising with frequency, magnetic coupling gives one rising too, and the load current’s own spectrum falls. A flat comb over two decades is the signature of a $1/n$ current through a rising impedance, and nothing else on a board produces it.

Its amplitude does not move when the load current’s amplitude is halved at constant edge. Halving the current halves every tooth, so that test is not the discriminating one. The discriminating test is the edge: put a resistor in the gate drive, slow the edge by three, and watch the total fall by 1.73 while every individual tooth stays where it was. A coupling mechanism that depends on $dV/dt$ or $dI/dt$ falls by three; this falls by the square root of three.

And it does not care about the loop area. The essay before it separates two faults that are both called ground loops — a shared impedance and an enclosed area — and notes that a fix for either does nothing for the other. The comb is entirely the first, so folding the circuit flat, twisting the pair, or adding a ground plane over the top changes nothing at all until the plane is used as the return. Where the current comes back is where the plane’s own return path is computed, and moving the return onto it is what actually shortens the shared conductor.

The three tests together are worth having as a set because each of them fails on the others’ mechanisms. A flat comb, insensitive to geometry, falling as the square root of the edge rate, is a shared return carrying a switching current — and none of the three readings requires the interfering circuit to be turned off, which is the one experiment a working system often cannot offer.

Where this model stops #

The conductor’s resistance is a constant. Above a few megahertz the current crowds towards the surface and the resistance rises as the square root of frequency. The essay before it argues that this arrives in the term that has already stopped mattering — at 3.5 megahertz the inductive part is 220 milliohms against the resistive five, so a doubled resistance moves the total by a per cent — and that argument holds here for the same reason, because every tooth above the corner is inductive.

The trapezoid is symmetric and ideal. A real converter’s current has a reverse-recovery spike, a ringing tail, and rise and fall times that differ. The first of those is a much shorter edge than the nominal one and, by this essay’s arithmetic, adds a great many teeth: the diode that conducts backwards is where the spike’s own duration is measured, and a ten-nanosecond recovery inside a hundred-nanosecond edge sets the comb’s length rather than the edge does.

And the load current is independent of the conductor. It is not: the shared conductor is in the load’s own loop, so a rising impedance in it slows the load’s own edges. The effect is second order here — five milliohms against a fifty-ohm load — and it is the direction that helps, which is worth stating because it is the only thing in this essay that does.

Still open: the spike inside the edge, and the comb a real converter draws #

The recovery spike, as a second edge. Everything above takes one edge rate and finds the comb’s length from it. A real switching current has two: the nominal edge and the much faster recovery transient inside it, and this essay’s own arithmetic says the shorter one wins — the comb runs to $1/(\pi t_\mathrm{rr})$ rather than to $1/(\pi t_r)$. Measuring a current with both features would say which term sets the total, and the answer decides whether slowing a gate drive helps at all.

The comb at the sampler. The third case above is described and not solved. A sensor read by a converter behind a stated front end has this comb folded into its baseband, and the folded total is computable from the two essays’ results together: every tooth’s frequency maps to a baseband frequency by the sampler, and the front end’s response at the tooth’s own frequency decides how much of it arrives. That would turn “not attenuated, and no longer at a frequency that identifies them” into a spectrum a bench could be shown.

And the switching frequency, swept. The claim that halving the switching frequency makes the total worse by $\sqrt2$ follows from the cancellation and is not drawn. It is the most surprising consequence in the essay and the easiest to check, and it is the one a converter designer could act on — because the switching frequency is usually chosen for magnetics and efficiency, with no term in the decision for what it does to a measurement sharing its return.

What is checked #

The flatness is stated as a spread over the harmonics that exist, between three times the conductor’s corner and a third of the edge’s, and required to hold to twelve per cent at every setting of the slider. The teeth that do not exist are excluded on a relative threshold rather than on being non-zero: at a duty of a half the even harmonics are identically zero, which in a double is $10^{-17}$ rather than nothing, so the first version’s flatness ran over teeth of $10^{-22}$ volts and reported $8\times10^{19}$ per cent.

The square-root law is stated as a fitted exponent across five edge rates, against −0.5 to six hundredths. The sum has to reach past the edge’s corner for that to mean anything, and at a one-nanosecond edge the corner is the 3,183rd harmonic: fitted over a sum truncated at two thousand terms the exponent came out −0.41, which reads as a departure from the law rather than as a missing tail. The number of terms now follows the edge.

And the direct-current term is held against its own closed form — the average current times the resistance, with no inductance and no edge in it — and required to be the smaller of the two at this switching frequency. The first version of that check looked for harmonics below the conductor’s corner and averaged them; at a hundred kilohertz of switching against a corner of 79.6 there are none, so the set was empty and the requirement reported a NaN rather than a failure.