I am developing a 4-layer FR-4 board containing a GNSS receiver (L1 + L2). RF tracks must have an impedance of 50 ohms. The board contains two inner layers of GND and a ground plane around the RF track:

RF track

To decide the thickness of the RF track, I used an online calculator: https://www.changpuak.ch/electronics/Coplanar_Waveguide_Calculator.php Thickness calculation on site 1

I considered "Substrate Thickness H [mm]" as the distance between the top layer (which contains the track) and the closest inner layer (GND), ie the prepreg.

I considered the dielectric constant to be 4.300.

But on this other site, the result is a little different: https://spok.ca/index.php/resources/tools/99-cpwcalc

Thickness calculation on site 2

On other sites, the calculation is even different. Some take into account several other aspects.

My question is, did I calculate correctly? Should I use 0.402 mm or 0.384 mm thickness? My layer configuration (provided by the manufacturer of my PCBs):

Layer configuration provided by my manufacturer

thank you so much

I used trace widths that result in 50 Ohms. On each site, the trace width that results approximately 50 Ohms is different. If you use the same width on different sites, the resulting impedance will be different. If you look, the impedance at both sites is approximately 50 Ohm. But the trace width that results from this is different.I don't understand which of the trace widths actually makes 50 Ohms.

  • \$\begingroup\$ Why are you using different trace widths in your comparison? \$\endgroup\$
    – Andy aka
    Jul 16, 2021 at 13:00
  • \$\begingroup\$ Those calculators differ by 18 microns, which is insignificant compared to manufacturing tolerances. \$\endgroup\$ Jul 16, 2021 at 14:41
  • \$\begingroup\$ FR4 is a grade of material, not a specific material in itself. So the stuff you buy from different manufacturers will have different formulations and thus different ε--so an approximation is the best you can do if you're just using FR4. If you have a demanding application that needs exact impedances, you need to get a specific high-frequency PCB material, like Rogers 4003C or something. \$\endgroup\$
    – Hearth
    Jul 16, 2021 at 15:27

1 Answer 1


There's not a lot of difference (a couple percent between the two results) when the same trace widths are specified. 49.9 vs. 48.7 ohms. The accuracy is given as ~10% on the first calculator and unspecified on the other.

Different calculators may use different formulas, with different accuracy and different ranges of applicability. Some use the formulas in IPC-2141A standard and some use the (apparently significantly more accurate) Wheeler's equations from Wadell's Transmission Line Design Handbook.

As there is a large range in er for FR-4 (and manufacturing tolerance in the trace widths), there's not too much point in calculating things to high accuracy when using low grade substrate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.