# PCB trace caps (a.k.a. finger / interdigital caps) formula

I am trying to calculate a two-fingered, both on the same layer, capacitor. I have run into a few papers on it but they are taking a ground plane on the other side of the PCB into consideration and they seem to be showing a meandered capacitor like this one

which is not exactly what I was looking for. I am after a simpler design:

I know that the capacitance formula is where A = t * L; t being the thickness of the copper pour and L being the length on which the two traces juxtapose, thus forming the plates of our capacitor. Should I apply it as such (no ground plane beneath) ? Am I also right to be using these values ?

• epsilon_r = 1 for air
• t = 35 microns for 1 oz copper
• $\epsilon_R$ is 1 for air, but some of the field will be in the PCB material rather than in air. Aug 11, 2017 at 22:35
• @ThePhoton I was afraid taht would be the case :S Aug 11, 2017 at 22:35
• I found one formula for calculating Er in a coplanar waveguide. You could try using that. The formula is kind of involved. You can find it in this PDF: jlab.org/accel/eecad/pdf/050rfdesign.pdf. Make sure you use the formula for the ungrounded coplanar waveguide. Aug 12, 2017 at 0:58
• Also don't forget about solder-mask. Either remove it or account for its Er of 3 to 5 (depending on the material, there's also outliers of 2 and 7 for speciality purposes) Aug 12, 2017 at 2:50
• Yeah, I am not following the terminology. But from a later example, it appears that the effective Er is around 2.44 or 2.45. Maybe just try working out your problem using that, and see how it goes? Aug 12, 2017 at 19:49