The pad the stub comes out of

The via that is a piece of line found a via of half a picofarad of pad and a nanohenry of barrel that leaves no area under its reflection on a line of $\sqrt{L/C}$ = 44.72 Ω, from any edge. It had not vanished. It delayed the edge going past it by 22.4 ps, exactly $\sqrt{LC}$, and it still reflected: a doublet whose largest excursion fell as the square of the edge’s rise time where a lone capacitance’s falls as the first power. A balanced via is a short piece of line, and a short piece of line reflects only through its length.

That essay named the complication every real board adds. A via that carries a signal from the top layer to a middle one has barrel left over below the middle layer — a stub, open at its far end, hanging off the line at the point the signal leaves. It looks like extra capacitance at low frequency and resonates at its quarter wave, and on a via that was balanced it unbalances the balance. The question it left was whether a smaller pad can absorb the stub, or only for edges slower than its resonance. The answer depends on which pad, and on how long the stub is.

The via it starts from #

The via is three elements in a row: half its pad capacitance, the barrel’s inductance, the other half. That arrangement — a shunt capacitance, a series inductance, a shunt capacitance — is the first section of a lumped model of a transmission line, the section a ladder is not a line chains together, and on a line whose impedance is $\sqrt{L/C}$ it is a short piece of that line. The first-order reflection, which the area measures, is the difference between the capacitive and inductive terms, $-Z_0C/2 + L/(2Z_0)$, and at $\sqrt{L/C}$ the two cancel.

The stub, untrimmed #

The stub is a piece of open line of its own, hanging from the node below the barrel. It is modelled here as exactly that: a shunt admittance $j\tan(\omega\tau)/Z_s$, with τ the stub’s one-way delay and $Z_s$ its impedance, taken equal to the via’s own balance of 44.72 Ω. At low frequency $\tan(\omega\tau) \approx \omega\tau$ and the stub is a capacitance $\tau/Z_s$; at the quarter wave, $\omega\tau = \pi/2$, its admittance is infinite and the open end, reflected back up the stub, arrives as a short circuit across the line — the same inversion a quarter wave, and the path the current takes back uses deliberately to transform one impedance into another.

A 10 ps stub is 0.224 pF of capacitance at low frequency. On a via whose pads were 0.5 pF between them, that is enough to unbalance it thoroughly.

Untrimmed, the via’s reflection grows from 24.4 mV to 90.5 mV and has an area again, 5.0 ps — which is $Z_0C_{\text{stub}}/2$ for the stub’s 0.224 pF, exactly as a lone capacitance of that size would give. The doublet is still there underneath, but the first-order term it had removed is back.

Two trims that balance the area #

The obvious repair is to take the stub’s capacitance out of the pads, so that the via’s total shunt capacitance is what it was. There are two ways to do it, and they are indistinguishable by the quantity the balance was defined by. Taking 0.112 pF out of each pad and taking 0.224 pF out of the bottom pad alone both leave the via’s total capacitance at 0.5 pF, and both put its area back to zero to three decimal places.

They do not leave the same via. Trimmed from both pads, the reflection’s largest excursion is 56.1 mV — area zero, and still more than twice the stubless via’s. Trimmed from the bottom pad, it is 24.6 mV, within one per cent of the via with no stub at all.

The reason is the one the earlier essay’s title gives. A balanced via is not merely a total capacitance and an inductance that cancel in the area; it is a piece of line, and a piece of line is symmetric — half its capacitance at each end of its inductance. The stub hangs at the bottom. Trim the bottom pad by what the stub adds and the capacitance at each end of the barrel is what it was, so the structure is again a symmetric section, C/2, L, C/2, with part of its bottom capacitance supplied by the stub. Trim both pads and the total is right but the distribution is not: less than C/2 at the top, more at the bottom. The area cannot see distribution. The doublet can.

The size of the leftover can be read off the arrangement. With the capacitance moved from the top of the barrel to the bottom, the via’s capacitance has its centre below the centre of its inductance, displaced by a fraction of the barrel’s own delay. A displaced centre is not a first-order term — the totals still cancel — but it is a term in the next order, the one the doublet itself lives in, and it has the same dependence on the edge: it falls as the square of the rise time. So the both-pads trim’s error does not go away at slow edges. It stays in fixed proportion to the doublet, which is what the factor of two the figure below finds says. An error at the same order as the thing being balanced cannot be made small by making the edge slow, because both shrink together.

At every edge #

Swept against the edge, the three stubbed vias separate into the two laws the earlier essays found. Untrimmed, the reflection falls as the −1.09 power of the rise time, the law of a lone capacitance, because the stub’s uncancelled area dominates at every edge slow enough to be measured by its area. Trimmed from both pads, the reflection falls faster but stays above the stubless via by a factor of 2.3 or more at every edge from 59 ps up: the asymmetry is a second-order defect of its own, falling with the square of the edge like the doublet it sits on, and never catching up.

Trimmed from the bottom pad, the stubbed via lies on the stubless one within 4.0 per cent at every edge drawn, down to 5.9 ps. That includes edges whose own spectrum reaches well past a few gigahertz, and the reason the stub’s resonance does not show is the second half of the answer.

At an edge of 11.8 ps the picture has changed in one respect. The untrimmed stub now reflects slightly less than the stubless via, 265 mV against 273, because at this speed every via reflects largely through its own delay and the stub’s extra capacitance is a smaller part of a larger event. The bottom-trimmed via is 284 mV, 4 per cent above; the both-trimmed one is 377. The ordering of the trims does not change with the edge. What changes is how much any of this matters against a reflection that is now a quarter of the step.

The pad that has to have room #

The trim takes the stub’s capacitance out of the bottom pad, and that pad has only so much to give: 0.25 pF here. A stub’s capacitance is its delay over its impedance, so the bottom pad can absorb a stub of at most

$$\tau_{\max} = Z_s \cdot \frac{C}{2} = 44.72\ \Omega \times 0.25\ \text{pF} = 11.2\ \text{ps}.$$

With the stub’s impedance equal to the via’s own balance, $Z_s = \sqrt{L/C}$, that limit is $\sqrt{LC}/2$: half the via’s own delay, the 22.4 ps the earlier essay measured it adding to a passing edge. A stub can be absorbed if it is shorter, in time, than half the via it hangs from. A longer stub has more capacitance than there is pad to remove. Taking the whole pad away still leaves the via with extra capacitance at its bottom, an area that no trim of that pad can cancel, and a reflection that grows with the stub.

Below 11.2 ps every stub is absorbed, to within 5.1 per cent of the stubless via at every edge from 12 to 295 ps. Above it the trim runs out of pad, and the worst excursion climbs steeply: 1.2 times the stubless via at 12 ps, where only 0.41 ps of area is left over, and 30.2 times at 30 ps. There is no gradual loss of the fix; there is a length at which the pad is used up, and beyond it the stub is simply a capacitance the via cannot pay for.

The limit has a frequency attached, and it is the answer to the question the earlier essay asked. A stub whose delay is under 11.2 ps has its quarter wave above $1/(4 \times 11.2\ \text{ps}) =$ 22.4 GHz. So the stubs a smaller pad can absorb are exactly the stubs whose resonance is above that frequency. The condition “only for edges slower than its resonance” turns out to be automatically satisfied by the stubs that can be absorbed at all, for every edge slower than a few picoseconds — which is why the 4 per cent figure held down to the fastest edge drawn. The binding limit is the pad’s capacitance, not the stub’s resonance.

The length of a stub is a fact about the board, and so is whether it fits. A via’s barrel runs the full thickness of the board; a signal that leaves it on a layer a depth $d$ from the top leaves the rest as stub. At roughly 6.8 ps a millimetre in an ordinary glass-epoxy laminate, the 11.2 ps limit here is about 1.6 mm of stub. A signal changing from the top layer to one a few tenths of a millimetre above the bottom of a 1.6 mm board leaves a stub of a fraction of that, well inside the limit. The same change on a 3 mm board, or to a layer near the top of a thick backplane, leaves two or three millimetres — past the limit, where no pad can absorb it. The limit also scales with the via: since it is half the via’s own delay, a via with more pad capacitance and more barrel inductance — a longer piece of line — can absorb a longer stub, and a small, fast via can absorb only a short one.

What passes the via #

The reflection is half the story; the other half is what arrives at the far side, the half the staircase in time follows down a line from the start. With no stub, the via passes everything up to the tens of gigahertz, where its own lumped section stops behaving like a line. With a 10 ps stub trimmed from the bottom pad the transmission is the same until the stub’s quarter wave at 25.0 GHz, where the open end, a quarter wave down the stub, comes back as a short across the line and the transmission drops into a notch.

A 40 ps stub — about six millimetres of barrel in an ordinary laminate — cannot be absorbed, and its notch is at 6.25 GHz. That is inside the band of any fast serial link, and no pad size moves it: the notch is set by the stub’s length alone. It is the reason back-drilling exists. Drilling out the unused barrel shortens the stub until its capacitance fits in a pad and its resonance leaves the band, which on this reckoning are the same condition.

What the reflection and the transmission say together #

A stub has two effects and they happen at different edges. In reflection it is a capacitance, and it is repaired by taking capacitance out of the pad it hangs from, provided the pad has enough to give. In transmission it is a resonator, and it is harmless below its quarter wave and a short circuit at it. The two are joined by one number, the stub’s delay: its capacitance is delay over impedance, and its resonance is one over four times the delay.

So the design rule a board needs is short. A stub whose delay is less than its via’s impedance times half its pad capacitance can be absorbed completely by trimming the pad beside it, and its resonance will be above a frequency the same numbers fix. A stub longer than that cannot be repaired by any geometry above it, and has to be removed.

Stated in the units the earlier essays used, the rule becomes a comparison of two delays. The via adds $\sqrt{LC}$ to an edge passing it; the stub’s round trip is twice its one-way delay; and the stub can be absorbed exactly when that round trip is shorter than the via’s own delay. A stub whose echo returns before the via has finished passing the edge can be hidden inside the via. One whose echo returns later cannot.

There is also a warning here for anyone checking a board with a reflectometer. An edge slow enough that the via is small against it measures the via by its area, and the area of a stubbed via trimmed from the wrong pad is zero: the reflectometer reports a balanced via. Only an edge fast enough to resolve the doublet — a few tens of picoseconds for the vias here, where the both-pads trim is twice the stubless via’s excursion — can tell the right trim from the wrong one. A balance verified at a slow edge has verified the total, not the arrangement.

The via two lines can weigh measured a via’s capacitance and inductance separately from its areas on two lines. The same measurement on a stubbed via would read the stub as extra pad capacitance, correctly at low frequency, and would say nothing about where the capacitance sits — which is the one thing that decided the trim here. The dip whose area is fixed established that the area is additive and blind to arrangement. This page is a case where the arrangement is the whole answer.

How the numbers were obtained #

Each via is a cascade of two-port matrices on the reference line: shunt capacitances, a series inductance, and for the stub a shunt admittance $j\tan(\omega\tau)/Z_s$, the exact input admittance of an open lossless line. The cascade’s reflection and transmission at each frequency multiply the spectrum of a raised-cosine pulse, and the inverse transform gives the waveforms, on 16,384 samples, or 65,536 for edges under 30 ps. The area is the integral of the reflected waveform over the half of the record holding the rising edge’s reflection. The largest excursion is the largest magnitude in that half. Each notch is located on a fine grid around its own first quarter wave.

What it leaves out #

Loss in the stub, which would fill the notch and lower the resonance’s effect on the transmission; a real barrel in lossy laminate has a notch tens of decibels deep rather than infinitely deep, for the reasons the delay that is not one number measures on a line’s own loss.

The stub’s own impedance being different from the via’s. It is taken equal here; a barrel with a larger anti-pad has a higher impedance and so less capacitance per picosecond, which moves the absorbable length up in proportion.

And the pads’ own shape. A pad is a small capacitance treated here as a lumped element; a large pad on a fast edge has a delay of its own, and the symmetry argument then applies to its distribution too.

Still open: the stub’s loss, the pair of vias, and the loss in front #

A lossy stub. Resistance and dielectric loss in the stub turn its notch from a zero into a finite dip, and whether a lossy stub that cannot be absorbed is still a problem at the notch depths a real board gives is a transmission measurement on the same cascade with a lossy stub in it.

Two vias close together. A signal that changes layers and changes back crosses two vias, which are two short pieces of line with a stretch of track between them. Their doublets add with a delay between them, and the spacing that makes a pair quieter or louder than one depends on the edge’s shape as much as on the vias.

The loss in front of the via. A doublet reads the edge’s curvature, and a lossy line in front of the via changes the edge’s shape as well as its rise time. Whether a balanced via behind a length of lossy line is suppressed as its arriving rise time says, or less, is the measurement a real channel makes.