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 at most 0.011 dB. Under a sixth of a decibel, for the whole transition.

Then look at what the same via actually does to a channel: 2.25 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 $jY_0\tan(\beta l)$, and a shunt admittance $Y$ across a line of impedance $Z$ gives

When the stub reaches a quarter wavelength, that tangent goes to infinity and $S_{21}$ goes to zero. A notch. In the example above it lands at 30.48 GHz if the planes are ignored — 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 2.25 dB to 0.13 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.

Why the calculator does not give the via's impedance

The calculator used to report the pair's differential impedance from a closed-form model: a coaxial capacitance from each barrel to its antipad, plus a two-wire capacitance between the barrels. A field solution showed that model reading about half the true value.

Solved in three dimensions with the planes the via passes through in place, and cross-checked full-wave, the same 0.25 mm pair in a 0.9 mm antipad is 85.6 Ω when the planes are 0.1 mm apart and 96.0 Ω when they are 1.0 mm apart. The planes are transverse to the barrels: they add capacitance at every hole edge but barely change the inductance, so the impedance depends on how densely they are stacked, and the plane pitch is not one of the via's dimensions.

Across three reference geometries the solved impedance spans a ratio of up to 1.31 over that range of plane pitch, so no formula in the barrel, antipad and pitch alone can land within 10 % of every case. The calculator therefore reports no impedance. For the number, put your actual stackup into the full-wave tool.

The stub is slower than it looks

The same field solution says something about the stub. Because the planes load the barrel capacitively, the wave travels slower than $c/\sqrt{\varepsilon_r}$: its effective permittivity is 1.17–1.86 times $\varepsilon_r$. The quarter-wave resonance $c/(4L\sqrt{\varepsilon_r})$ is therefore an upper bound. For this via the solved stub resonates near 26.3 GHz rather than 30.48 GHz, and the worst in-band loss rises to roughly 3.4–4.1 dB.

So read the calculator's resonance as the highest the notch can be, and its worst-case loss as taken at that unloaded resonance. If the bound already sits inside your band, the real notch sits deeper inside it.

What this model is, and is not

It is a lumped screening model. The stub loading follows from $S_{21}$ of a shunt admittance, and the conductor loss is bounded at the dense-plane limit, with each barrel inside its own coaxial wall: $Z_{dense} = \frac{\eta_0}{\pi\sqrt{\varepsilon_r}}\ln(D/d)$, 74.95 Ω here and below every solved impedance. What it does not include: the radial currents in the planes, 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. Whether the two antipads merge into one opening or stay separate, with plane copper between the barrels, changes how the barrels couple. That is part of why the impedance needs your stackup and a field solution rather than a formula; the stub resonance and the loss do not depend on it.

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 get the via's differential impedance from a full-wave simulation of your stackup and compare it with 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.