# Doubt about 50 ohm impedance calculation

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:

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

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

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):

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.

• Why are you using different trace widths in your comparison? Jul 16, 2021 at 13:00
• Those calculators differ by 18 microns, which is insignificant compared to manufacturing tolerances. Jul 16, 2021 at 14:41
• 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. Jul 16, 2021 at 15:27