The millimetre that becomes common mode

Every line drawn in this field so far has been one conductor and a return. A source at one end, a load at the other, a staircase of reflections between them if the two do not match.

A differential pair is two conductors and a return, and it does not carry two signals. It carries two modes: the odd mode, in which the conductors are driven in opposition and the return current is in the other conductor, and the even mode, in which they are driven together and the return is in the plane. The signal is entirely in the first. Nothing in a differential system measures the second.

This essay is about the routing error that puts energy into it.

Two modes, two impedances #

With a coupling of a quarter between the conductors, the odd mode sees 35.36 Ω and the even mode 55.90 Ω. Those are properties of the geometry: the odd mode’s mutual capacitance carries twice the voltage so its impedance falls, and its mutual inductance subtracts; the even mode’s mutual capacitance carries nothing and its mutual inductance adds.

The numbers a data sheet quotes are those doubled and halved — a differential impedance of 70.7 Ω and a common-mode impedance of 27.95 — which is a statement about how they are measured rather than about the line. It matters because the two are not related by anything simple, and a pair specified as “100 Ω differential” has a common-mode impedance that depends on the coupling and is not 50.

Where the two modes come from, in a netlist #

It is worth saying how the mode impedances are arrived at, because the derivation is the reason the even mode is a mode rather than a leakage.

Two coupled conductors over a return have a per-metre inductance matrix with L on the diagonal and a mutual Lm off it, and a per-metre capacitance matrix with C + Cm on the diagonal and −Cm off it. The odd and even excitations are the eigenvectors of that pair of matrices: drive the conductors in opposition and the mutual capacitance carries twice the differential voltage while the mutual inductance subtracts, giving √((L − Lm)/(C + 2Cm)); drive them together and the mutual capacitance carries nothing while the mutual inductance adds, giving √((L + Lm)/C).

Both are ordinary transmission lines with ordinary impedances, and each obeys everything this field has established about a single line — the same reflection arithmetic, the same lumped-model boundary, the same loss and dispersion. What makes a pair a different subject is not that either mode is unusual: it is that a circuit driving one and measuring one has no instrument pointed at the other, so an error that would be obvious on a single line is invisible here until it leaves the board.

Coupling appears nowhere in this essay’s conversion arithmetic, and that is worth noticing. The skew’s conversion depends on Δτ alone. What coupling decides is the impedances the two modes see, and therefore what the terminations have to be and how badly the converted energy resonates once it exists. A tightly coupled pair converts exactly as much as a loosely coupled one and is harder to terminate for it.

What a length mismatch does #

Route the two halves of a pair to different lengths — around an obstacle, through a via field, past a connector — and the signal arrives at the two conductors at different times.

The arithmetic is two lines and one subtraction. A differential drive puts +V on one conductor and −V on the other; they arrive with delays τ ± Δτ/2; the differential half of what arrives is their sum over two and the common half is their difference over two:

$$|S_{cd}| = |\sin(\omega\Delta\tau/2)| \qquad |S_{dd}| = |\cos(\omega\Delta\tau/2)|$$

A millimetre and a half on FR-4 is 10.50 picoseconds. A tenth of the launched amplitude is common mode at 3.04 GHz, and the closed form puts that edge at 3.04 as well — which is the check that the solved pair and the expression are about the same thing.

Two things fall out of the pair of expressions and both matter more than the numbers.

The conversion is first order in the skew and the loss is second order. At the frequency where a tenth has converted, the differential signal is down by 5,011 parts per million — half a per cent. At one per cent converted it is down by fifty parts per million. The energy that leaves the differential mode is not missing from it to any accuracy an eye diagram can see.

That is the whole practical problem. A pair converting one per cent of its signal to common mode looks perfect on the measurement its designer makes and radiates like an antenna, because the common mode has no return path in the other conductor and finds one wherever it can — the mechanism the crosstalk essay measures between two lines, arriving inside one. The signal-integrity measurement and the emissions measurement are looking at quantities that differ in order.

And the product is ωΔτ. That is the same product that appears in the instrument whose rejection corner belongs to its source, where an imbalance between two input time constants sets a corner at 1/(ωΔτ) and the rejection falls at twenty decibels a decade above it. Here an imbalance in time converts; there an imbalance in time constants fails to reject. Two fields, two circuits, and one dimensionless group — and in both cases the cure is to balance the times rather than the components.

The null that the loss fills in #

The closed form has a null. At ωΔτ = 2π the two conductors’ delays differ by a whole period, the signals arrive back in phase, and the conversion returns to zero — 95.3 GHz for this skew.

The solved pair does not do that.

At the null frequency the lossless calculation gives 3.2×10⁻⁴ and the lossy one gives 3.6×10⁻², a floor at −28.9 decibels rather than a null. The reason is that the longer conductor is also the lossier one: an extra millimetre and a half of copper and dielectric attenuates a little more, and an amplitude imbalance has no null in it at any frequency.

So the periodic nulls a skew model predicts are an artefact of assuming a lossless line, and a real pair’s conversion rises, ripples and never comes back. The dispersion essay found the same thing about a different quantity — a lossy line’s delay is not one number — and the two findings have the same root: the loss makes the two conductors differ in magnitude as well as in phase, and every cancellation that depended on phase alone stops being exact.

In time, it is a pulse of half the amplitude #

The frequency-domain answer says how much and says nothing about the shape. The shape is the more alarming half.

The common mode is not a small residue spread over the waveform. It is a pulse, present only during the transition and gone in between, reaching 9.3 per cent of the amplitude and 34 picoseconds wide for this skew and this edge.

Its peak depends on the edge rate — a faster edge gives a higher, narrower pulse — but its area does not. The common mode is the difference between one edge and the same edge Δτ later, so its integral is Δτ/2 times the amplitude whatever the edge rate: measured at 5.39 ps against Δτ/2 = 5.25, the same number for a 10 ps launched edge, a 30 ps one and a 100 ps one.

That invariant is the useful one, because the quantity a radiated-emissions measurement responds to is the area rather than the peak. A designer who slows the edges to fix an emissions problem has lowered the peak and moved the spectrum down, and has not reduced the charge at all.

There is a second observation in the same figure. The 10 ps and the 30 ps launched edges arrive as 122.5 and 123.7 picoseconds after 150 mm of FR-4 — the line has forgotten the difference between them, which is the dispersion essay’s finding arriving as a practical constraint: past a certain length, the edge rate at the receiver is the line’s property rather than the driver’s, and slowing the driver changes nothing.

The termination that is an open circuit to the error #

The mode a skew creates is the mode nothing absorbs, and that is not an accident of any particular design.

A single resistor of 70.7 Ω across the pair is an exact match to the odd mode. To the even mode it is an exact open circuit, with a reflection coefficient of one, and this is true whatever its value: no even-mode current can flow through a resistor whose two ends are at the same potential. So the most common differential termination is, by construction, transparent to common mode, and the converted energy runs back and forth along the pair until something else absorbs it.

Two resistors to ground of half that value each are better and are not right: they match the odd mode and leave 22.5 per cent of the even mode reflected, because the even-mode impedance is not the odd one.

The arrangement that matches both is three resistors — two of $Z_{\text{odd}}$ to a common node and one of $(Z_{\text{even}} - Z_{\text{odd}})/2$ from that node to ground, 10.27 Ω here. It is exact rather than a compromise: the odd mode makes the common node a virtual ground and never sees the third resistor, while the even mode drives both halves together and sees $Z_{\text{odd}}$ in series with twice the third, which is $Z_{\text{even}}$. And the third resistor’s value is the difference between the two mode impedances, so it is zero for an uncoupled pair — which is the check that the whole arrangement collapses to the obvious one when there is nothing to fix.

In practice the third element is usually a resistor and a capacitor in series, so that the common-mode path is terminated at signal frequencies and open at direct current. That is a refinement of the same idea and does not change the value.

What a skew budget should be written against #

The three results above give a budget with a different shape from the usual one.

The usual instruction is a length-matching tolerance in millimetres or in mils, applied uniformly. The measurements here say that the quantity is Δτ, that the conversion is ωΔτ/2, and therefore that the tolerance is a time which has to be set against the highest frequency with meaningful energy in it — not against the bit rate, and not against the edge rate the driver was specified at, because the line has already changed that.

For one per cent conversion — a fifty-parts-per-million differential loss, which is negligible, and a −40 dB common mode, which is usually the specification — the requirement is ωΔτ/2 = 0.01, or Δτ = 3.2 ps at 1 GHz and 0.32 ps at 10. On FR-4 that is 0.45 mm and 0.045 mm of copper. The second of those is not a routing tolerance anybody achieves by matching lengths on a layout; it is achieved by making the structure symmetric, which is a different instruction.

And the budget has to include the things that are not routing. A via pair whose two vias have different stub lengths is a skew. A connector whose two pins have different lengths is a skew. Fibre weave in the laminate makes the two conductors of a pair see different local permittivities and is a skew that varies along the board and from board to board. All of them add in time, none of them is visible on a length report, and the sum is what the expression above takes.

What is not modelled #

The two modes do not interact here. The pair is symmetric apart from its lengths, so the odd and even modes propagate independently and the only coupling between them is the skew at the ends. A genuinely asymmetric pair — one conductor closer to a plane, or wider — converts along its whole length as well, and that is a different calculation with a different frequency dependence.

The velocities are equal. In a microstrip they are not: the even mode has more of its field in the air and travels faster, so a perfectly length-matched pair still has two arrival times. That is a skew nobody routed and it is worth a rung of its own; it does not convert on a symmetric pair, but it does spoil the odd mode’s own eye once the two modes are recombined at a receiver that is not perfectly balanced.

Nothing here reflects. Both ends are matched to each conductor’s own impedance at every frequency, so the conversion drawn is the pair’s and not a standing wave’s. A real pair terminated by the bridged resistor above has an even-mode reflection of exactly one, so the converted energy makes round trips, and what a probe at the far end sees is the conversion multiplied by whatever the resonance of those round trips does at that frequency. That resonance is the same object the branch essay measures and it can be several times worse than the number here at particular frequencies.

And the common mode is not followed anywhere. This essay says how much is made and stops. Where it goes, what it couples to, and what it radiates are the questions the return-path essay begins and neither answers.

What the gate checks #

The two routes are asserted against each other rather than one being drawn over the other. On a lossless pair the solved conversion and sin(ωΔτ/2) must agree at every frequency drawn, and the frequency at which a tenth has converted is asserted against the closed form’s own edge to two per cent.

The differential loss at that edge is asserted to be parts per thousand rather than per cent, which is the first-order-against-second-order claim stated as a number, and is what would fail if the two halves had been assigned to the wrong modes — a slip that gives a conversion of very nearly one at every frequency and was caught exactly that way.

The null is asserted to be filled in: the lossy pair’s conversion at ωΔτ = 2π must be at least five times the lossless pair’s, which is the amplitude imbalance being real rather than a rounding error.

In time, the area under the common-mode excursion is asserted against Δτ/2 to six per cent and to be the same across a factor of ten in edge rate, which is two assertions because the invariance is the finding and the value alone would not establish it.

And the terminations carry three claims: that the bridged resistor’s even-mode reflection is exactly one, that its odd-mode reflection is exactly zero, and that the three-resistor arrangement’s is zero for both — with the third resistor’s value checked against the difference of the mode impedances rather than against a number, so the claim holds at every coupling on the slider.

The product ωΔτ, and where else it appears #

The answer here is a product of a frequency and a time mismatch, and that combination is the same one the instruments field’s rejection corner is the reciprocal of — which is worth drawing out, because the two circuits look nothing alike and are doing the identical arithmetic.

The corner the instrument has no part in puts an instrumentation amplifier’s rejection corner at 290 Hz for a kilohm of source imbalance against ten picofarads, and identifies the mechanism as the difference of two time constants. A time-constant difference and a length-derived skew are the same quantity in different units: both are a Δτ between two paths that were supposed to be identical, and in both cases the fraction of common mode converted to differential — or differential to common — is $\omega\Delta\tau$ to first order.

That the conversion is first order is what makes both specifications the shape they are. A picosecond budget for a pair and a matched-source requirement for an instrument are both budgets on Δτ, and both are linear, so halving the mismatch halves the conversion at every frequency. What is not linear is what a system does with the result, which is why this essay’s five thousand parts per million lost from the differential signal reads as negligible and its tenth of the signal appearing in common mode does not.

The inductor one mode cannot see is the component fitted downstream to deal with the second of those, and its ratio of two thousand between the two modes is what a Δτ this small requires — which is the honest measure of how expensive a millimetre and a half of skew is: it is removed by a component whose whole design problem is presenting two inductances four orders apart.