The common mode that has three slopes

The millimetre that becomes common mode measured one way a differential pair stops being differential: a length mismatch between its halves. A millimetre and a half of skew on FR-4 is ten and a half picoseconds, and a tenth of the signal has become common mode by 3.04 GHz while the differential has lost only five thousand parts per million of itself — which is why a pair can pass its own eye diagram and fail an emissions test. The conversion was $\sin(\omega\Delta\tau/2)$, first order in the skew and rising at twenty decibels a decade.

Skew is not the only way the two halves of a pair can differ, and it is not always the largest. The driver’s two outputs may not swing equally, and they may not switch equally fast. Each of those is a different imbalance, and each converts the signal into common mode with a different dependence on frequency. This essay puts the three side by side at the far end of the pair — equal-sized and timed edges with one thing wrong at a time — and finds that the three slopes are not an accident of the arithmetic but a statement about which moment of the common-mode waveform vanishes. It also finds a case where two imbalances hide each other.

Two edges and their difference #

The model is the far end of the pair as two edges. The positive conductor rises through its swing in its rise time, after its delay; the negative conductor falls through its swing in its own rise time, after its own. The differential signal is their difference over two and the common mode is their sum over two, exactly as in the earlier essay. What changes is which of the three parameters of the two edges disagree.

The spectrum has a closed form. An edge is a ramp, whose derivative is a rectangle, whose transform is a sinc; each edge contributes its amplitude times $\mathrm{sinc}(\omega t_r/2)$ times a phase for its delay and half its rise, and the conversion is the magnitude of the common-mode sum over the differential one. The second route samples the two edges every quarter of a picosecond, transforms their derivatives by FFT, and divides: at 100 MHz, 1 GHz and 5 GHz it agrees with the closed form to two per cent for all three imbalances, the difference being the sampled grid’s resolution.

A 2 per cent amplitude imbalance — one conductor swinging 1 per cent more than nominal and the other 1 per cent less — gives a common mode of 1.00 per cent at every frequency. A millimetre of skew, 7.00 ps on FR-4, gives $\sin(\omega\Delta\tau/2)$: 2.20 per cent at 1 GHz, rising at twenty decibels a decade. And edges of 50 and 55 picoseconds with their midpoints aligned — the same timing, the same size, only the slope different — give a common mode rising at forty decibels a decade: 0.043 per cent at 1 GHz and 5.38 at 10 GHz. The three slopes are measured on the spectra between 10 and 100 MHz, and they are 0, 1 and 2 decades per decade to two parts in a thousand.

The three waveforms, and the moments they lack #

The slopes have a reason that is visible in time.

The amplitude imbalance leaves a common-mode step: the two conductors finish at levels that do not cancel, 5 mV apart from zero, and the common mode stays there as long as the signal does. A waveform that ends at a level has a spectrum that runs flat to direct current — the flat line in the hero figure.

The skew leaves a pulse. While one edge is rising and the other has not yet caught up, the two do not cancel; when both have finished, they do. Against 50 ps edges a 7 ps skew makes a trapezoid 35 mV high, lasting the edge, whose area is half a conductor’s swing times the skew: 1.749 mV·ns. A pulse that returns to zero has no level, but it has an area, and a spectrum that starts from zero and rises in proportion to frequency — twenty decibels a decade.

The rise-time mismatch leaves a doublet. The faster edge leads in the first half of the transition and the slower one catches up in the second, so the common mode goes up and then down by the same amount: ±11.4 mV, with an area of zero to the arithmetic’s floor. A waveform with no level and no area has a spectrum that rises as the square of frequency — forty decibels a decade.

So each slope is a moment that vanishes. A flat spectrum is a waveform with a final value; a first-order rise is one with no final value and an area; a second-order rise is one with neither. It is the same bookkeeping the second capacitor, and the tail that changes sign found in two coupling stages, where each capacitor removed a moment from a pulse and the tail had to change sign to meet two conditions. Here the imbalance decides which moments the common mode has, and the spectrum’s slope is read straight off them.

Where the skew overtakes #

A pair usually has all three imbalances at once, and which one dominates depends on the frequency being asked about.

The skew’s common mode is about $\omega\Delta\tau/2$ and the amplitude imbalance’s is $a/2$, so they cross at

$$f_\times = \frac{a}{2\pi\,\Delta\tau},$$

and the bisected crossings follow it to two per cent wherever the edge is slow against the crossing. A picosecond of skew against a 1 per cent amplitude imbalance crosses at 1.59 GHz; ten picoseconds against 2 per cent at 318 MHz; the millimetre of the hero figure against 2 per cent at 455 MHz. Halving the skew doubles the frequency below which it can be ignored; halving the amplitude imbalance halves it.

That turns a question about which imbalance to control into a question about where the emissions limit is written. An emissions test measured to a few hundred megahertz sees the driver’s amplitude match. One measured to several gigahertz sees the routing.

Which one leads at a link’s rate #

A link’s fundamental is half its bit rate, and that is where most of its common-mode energy sits.

For a link with a 1 GHz fundamental the three give −40.0, −44.0 and −67.3 dB: the amplitude imbalance leads. At 2.5 GHz the order changes — −40.0, −36.1 and −51.3 — and 2 ps of skew, a third of a millimetre of routing, is now the largest source. At 5 GHz the skew is −30.1 dB and the rise mismatch −38.9, closing. At 12.5 GHz the rise mismatch, −20.0 dB, has passed the skew’s −22.1. A driver whose two outputs differ in edge rate by ten per cent is, at the speeds of the fastest serial links, a worse source of common mode than the whole routing budget.

That is the practical consequence of three slopes. A design review that checks length matching and nothing else is checking the dominant term in one band of frequencies. Below it, the driver’s balance matters more; above it, the drivers’ edge rates do, and those are set by the silicon and by the loading on each output rather than by anything on the board.

A driver’s balance, as a budget #

The three slopes turn into a specification once a target is named. Take a link with a 5 GHz fundamental — ten gigabits a second — and ask that each source alone keep the common mode 30 dB below the differential, 3.2 per cent.

The amplitude imbalance allows twice that, 6.3 per cent: its common mode is half the imbalance, flat. The skew allows the delay at which $\sin(\pi f\Delta\tau)$ reaches 3.2 per cent, which is 2.0 ps — about 0.3 mm of FR-4, a length most length-matching rules would call matched. The rise-time mismatch allows $t_1^2 - t_2^2$ up to $48 \times 0.032/\omega^2$, which around 50 ps edges is a difference of 15 ps in rise time — thirty per cent.

Read that way the three are not equally hard. The driver’s amplitude and edge-rate balance are generous at this rate and tighten as the rate rises — the edge-rate term by the square of the frequency, so a link twice as fast allows a quarter of the rise mismatch. The routing term is already tight, and tightens in proportion. Halve the fundamental and all three relax, but by different factors: the amplitude by none, the skew by two and the edge rates by four, which is the order in which the three stop mattering as a link slows.

The budget also shows why the common mode is so often underestimated. Each source’s allowance looks comfortable measured against the differential signal — a 6 per cent amplitude error is invisible in an eye diagram — and each adds to the others as a phasor, the skew in quadrature and the other two in line. A pair meeting all three allowances exactly has anything from 3.2 to 7.1 per cent of common mode — the last 23 dB down rather than 30 — depending on whether the amplitude and edge-rate terms reinforce each other or cancel, with the skew’s in quadrature either way. A budget that allots each source the whole target has spent it three times over.

Two imbalances that hide each other #

The three common modes are not simply three sizes; they have phases, and two of them can cancel.

The amplitude imbalance’s common mode is in phase with the edge. The rise-time mismatch’s, for small differences, is $-\omega^2(t_1^2 - t_2^2)/48$ of the swing — also in phase, and with the opposite sign when the larger conductor is also the slower one. The two then cancel where

$$\omega^2 = \frac{24\,a}{t_1^2 - t_2^2},$$

which for a 2 per cent imbalance on edges of 60 and 50 ps is 3.32 GHz. The spectrum’s null is at 3.25 GHz and about 74 dB down, the small difference being the edges’ higher-order terms. A skew’s common mode is in quadrature with both — a delay rotates the phase — so it cannot cancel either, and half a picosecond of it fills the null back to −25 dB.

The null is a measurement hazard rather than a design opportunity. A pair’s common mode measured at one frequency can read tens of decibels lower than the imbalances producing it, and a small change in loading, temperature or skew moves the null and the reading with it. How wide a null is measured the width of a crosstalk null between two lines, and the same caution applies: a cancellation of two imbalances is as narrow as they are precisely matched.

The same arithmetic elsewhere on the site #

The skew’s slope — common mode rising as $\omega\Delta\tau$ — is the product the corner the instrument has no part in found setting an instrumentation amplifier’s rejection corner, where an imbalance between two input time constants converted common mode into difference at twenty decibels a decade. Here the conversion runs the other way, difference into common mode, and the pair has the same three sources that instrument did: an amplitude mismatch in its resistors, a time mismatch in its source impedances, and a bandwidth mismatch in its two inputs. An imbalance in time converts at twenty decibels a decade in both directions.

And the amplitude imbalance’s flatness is the reason a common-mode choke, which the inductor one mode cannot see measured, is the usual cure at low frequencies: nothing about the routing can fix a driver’s amplitude mismatch, but an element that impedes the common mode and not the differential can stop it leaving the board.

Where the common mode goes #

Whatever its slope, the common mode leaves the pair by the same routes. It is not absorbed by a differential termination, which the millimetre that becomes common mode found to be an open circuit to it. It couples to neighbouring tracks more strongly than the differential signal does, because the two conductors’ fields no longer cancel at a distance — the near-end and far-end mechanisms the far end that cancels measured between two single tracks, now driven by a pair acting as one conductor. And it radiates, with the cable or the chassis as its antenna.

Each route filters it differently, and the slope matters for which filter wins. A common-mode choke, whose best attenuation the depth a resonance does not have found to be a matter of how finely it was measured, works best over a band of a decade or two; a flat common mode from an amplitude imbalance needs that band at the low end, a steep one from rise mismatch at the top. And a lossy line, whose dispersion the delay that is not one number measured, slows the edges as they travel — which reduces the rise-mismatch term, whose size depends on the edges’ rates, and leaves the amplitude term untouched.

How the numbers were obtained #

The far end is modelled as two ramps — a positive edge of amplitude, rise time and delay, and a negative one of its own — with the common mode their sum over two and the differential their difference over two. The closed-form spectra transform each edge’s derivative, a rectangle, as an amplitude times a sinc times a delay phase, and divide the common-mode sum by the differential one. The second route samples both edges at 0.25 ps over 32,768 points, differentiates them, transforms by FFT and divides at three frequencies; the two agree to two per cent. The rise-mismatch cases start the slower edge earlier by half the difference so that the two edges’ midpoints coincide and no skew is hidden in them. Crossings are bisected on the closed forms; the time-domain areas are integrated at 0.25 ps. The line between driver and receiver is taken as ideal: every imbalance here is the driver’s or the routing’s, arriving unchanged.

What it leaves out #

The line’s own loss. A lossy pair slows both edges, which shrinks the rise-mismatch term in proportion and leaves the other two, and a longer conductor is also a lossier one, which adds an amplitude imbalance that grows with frequency — the mechanism the earlier essay found filling the skew’s null.

Coupling between the halves. The two conductors are independent here. A tightly coupled pair carries its two modes at different velocities on an inhomogeneous board, which is a skew between the modes rather than between the conductors and converts nothing by itself.

Reflections. Every edge arrives once. The common mode meets a termination that is usually an open circuit to it, and what returns adds to what arrives.

Still open: the skew nobody routed, the lossy pair, and the receiver’s own balance #

A skew in the board rather than the routing. Two conductors of exactly equal length still arrive apart if they run over different materials, and a woven glass board is two materials in a pattern with a pitch comparable to a pair’s. Which conductor sits over glass and which over resin is set by where the pair happens to fall, and the skew accumulates with length along the weave. How much, and what routing at an angle does to it, is a question about integrating a permittivity along a path.

The loss-induced amplitude imbalance. On a lossy pair with skew, the longer conductor’s extra loss is an amplitude imbalance proportional to frequency — a fourth slope between the flat and the skew’s. Where it overtakes the skew’s own conversion is a solve on the dispersive line the earlier essay used.

The receiver’s balance. A receiver that rejects common mode imperfectly converts some of it back into difference, and its own imbalance has the same three slopes. The common mode this essay produced at the far end is the input to that conversion, and the net error at the decision point is the product of the two.