Your Via's Loss Isn't in the Barrel

The copper that carries no signal

A via is a short piece of plated barrel. At 20 GHz, 1.6 mm of it costs you almost nothing: run the numbers on a 0.25 mm barrel in FR4 and dielectric loss comes to 0.119 dB and conductor loss to 0.030 dB. Under a sixth of a decibel, for the whole transition.

Then look at what the same via actually does to a channel: 2.27 dB at the top of a 0.1–20 GHz band.

The missing two decibels are not in the barrel. They are in the part of the barrel that is not carrying anything.

Where the stub comes from

A through-hole via is drilled through the entire board. If your signal enters at the top and leaves on an inner layer, the plating below that layer is still there, still connected, and still electrically part of the structure. It carries no signal. It is a stub.

Enter at the top of a 1.6 mm board and exit at 0.4 mm, and 1.2 mm of barrel hangs off the through path as an open-circuited transmission line. Its input admittance is jY0*tan(beta*l), and a shunt admittance Y across a line of impedance Z gives

S21 = 2 / (2 + Y*Z)

When the stub reaches a quarter wavelength, that tangent goes to infinity and S21 goes to zero. A notch. In the example above it lands at 30.48 GHz — above the band, but its skirt is not, and the skirt is where those two decibels come from.

Loss inside the stub is what keeps the notch finite rather than infinite. Without it the expression diverges, which no measurement does.

Backdrilling, quantified

Set the exit depth to the full board thickness — a via that exits at the bottom, or one that has been backdrilled — and the stub disappears along with the resonance. Worst in-band insertion loss drops from 2.27 dB to 0.15 dB.

That is the entire case for backdrilling in one comparison. You are not removing loss from the via. You are removing a resonator.

Whether it is worth the cost depends on where the resonance sits relative to your band. Push the transition deeper and the stub shortens and the resonance climbs; a 0.4 mm stub in the same board resonates near 91 GHz and is irrelevant to anything you are likely to be doing. Moving the layer assignment is free. Backdrilling is not.

The pair is not two single vias

Two barrels driven differentially are one problem, and it is worth decomposing it properly rather than solving two different problems and hoping.

Each barrel has a coaxial capacitance to its own antipad. The two barrels have a mutual capacitance to each other. The even mode puts both at the same potential, so the mutual capacitance carries no charge and drops out entirely — the even-mode impedance is just the isolated coaxial value. The odd mode charges it, so the odd-mode impedance is always lower:

C_self/eps = 2*pi / ln(D/d) C_m/eps = pi / arccosh(p/d) Z_even = eta0 / (sqrt(epsr) * C_self) Z_odd = eta0 / (sqrt(epsr) * (C_self + 2*C_m))

For a 0.25 mm barrel, 0.9 mm antipad, 0.8 mm pitch in FR4: Z_even = 37.48 ohm, Z_odd = 22.05 ohm, so Z_diff = 44.10 ohm and Z_common = 18.74 ohm.

Note the ordering. Even above odd, always, because the mutual term loads only one of them. That falls out of the decomposition rather than being asserted, which is the point of decomposing it — a model that computes the odd mode from a two-wire formula and the even mode from a coaxial one is solving two different geometries and can easily produce the ordering backwards.

44 ohms in a 100 ohm channel

That is the number worth sitting with. A carefully controlled 100 ohm differential channel, held to a few percent along its whole length, passes through a discontinuity of less than half its impedance.

No amount of trace tuning fixes it. What you can change is the antipad: opening it raises the impedance toward the trace value, and it is the knob fabricators are most willing to move. The cost is plane copper, which is a return-path problem of its own, so it is a trade rather than a free win.

What this model is, and is not

It is a lumped screening model. The impedances are the standard coaxial and two-wire results for an idealised geometry; the stub loading follows from S21 of a shunt admittance. What it does not include: pad capacitance, plane-cavity resonance, the launch discontinuity where trace meets via.

A real via transition is three-dimensional and none of those are small. Use this to decide whether a geometry deserves a full-wave simulation — then run one. The FDTD S-parameter tool on this site solves the actual structure.

One boundary is worth naming explicitly. The mutual capacitance term assumes the two barrels share a single clearance opening, which is what a differential via pair is. Space them further apart than the antipad diameter and plane copper comes between them and screens them from each other. The model does not account for that screening, so it overstates the coupling and understates the differential impedance — and the calculator says so rather than pretending otherwise.

What to check, in order

Look at the stub resonance against your band first. If it is far above, the transition may be fine as drawn and backdrilling is money you do not need to spend.

If it is inside or close, try moving the transition to a deeper layer before reaching for a backdrill. Shorter stub, higher resonance, no extra process step.

Then look at the differential impedance against your trace impedance. The mismatch is what the channel sees, not the absolute value.

And if the answer sits anywhere near a limit, simulate it. A screening model that tells you something is marginal has done its job by telling you to stop trusting it.