I find a lot of online tools to calculate the capacitance between two traces (or a trace and a plane) on opposite pcb layers. One such tool can be found here: http://chemandy.com/calculators/rectangular-capacitor-calculator.htm

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However, I would like to calculate the capacitance between adjacent traces on the pcb. Something like this:

enter image description here

Has anyone an idea how to do this?

Imagine the following example:

  • Two tracks run parallel for 10 mm.
  • Each track is 0.2 mm wide.
  • The distance between both tracks is also 0.2 mm.
  • Copper thickness is about 20 µm.
  • Board thickness is 1.6 mm.
  • PCB material is standard FR4.
  • One track is signal, the other Gnd.
  • There are no other tracks in the area, nor any (Gnd-) planes.
  • \$\begingroup\$ What is your end goal? Transmission line effects are usually the biggest concern over capacitance. \$\endgroup\$
    – Voltage Spike
    Commented Oct 18, 2016 at 17:03
  • 1
    \$\begingroup\$ The signal freq is rather low (250 kHz). The signal is in fact a pin from a capacitive sensor. So routing it parallel to a Gnd trace will cause some "capacitive offset". I'm trying to estimate that. \$\endgroup\$
    – K.Mulier
    Commented Oct 18, 2016 at 17:14

2 Answers 2

  • This is determined by copper thickness, gap, dielectric constant, d and length and thus there is incremental C and L with resulting impedance , Zo for a coplanar microstrip and thus C can be computed from this. For Zo=50R it is about 3.3pF /" or 1.3pF/cm

  • So compute Zo from a coplanar stripline calculator (complex) then compute C from Zo as shown here. This assumes other conductors are >10x further away. If not then you have a mesh of calculations ;)

  • If you can't find a coplanar stripline tool, try this formula.

\$ C [pf/cm] = 0.12 * \dfrac{t}{w} + 0.09 * (1+d)*log_{10}(1 + \dfrac{2w}{g} + \dfrac{w}{g}^2)\$

  • w = width of track [mm]
  • t = thickness of track [um]
  • d = dielectric constant ~ 4.2 FR4
  • g = gap [mm]

hopefully, I got the units correct...you do the math on a 50R pair.

The second term dominates. Using k=3.5 (kapton), w=10 mm, s = 0.5 mm, t = 2 microns, the capacitance is 1.1 pf/cm

Reference https://www.physicsforums.com/threads/capacitance-of-coplaner-adjacent-plates.130876/

  • 1
    \$\begingroup\$ let's say I have two traces. One is a signal, the other is Gnd. They run parallel to each other for 1 cm. The copper thickness is 20 µm, the distance between the tracks is 0.2 mm. The tracks are also 0.2 mm wide.. \$\endgroup\$
    – K.Mulier
    Commented Oct 18, 2016 at 15:56
  • 1
    \$\begingroup\$ If the result is in pf/cm, how can there be l (length of paired tracks) in the equation. pf/cm suggests that the result is length independent but the equation itself required length, any thoughts on this? \$\endgroup\$ Commented Jan 6, 2019 at 5:13
  • \$\begingroup\$ I checked my ref and the next post corrected it by comment \$\endgroup\$ Commented Jan 6, 2019 at 9:47

I use ATLC for this. You can also draw your own setup (with the correct colours) and have it solve that. http://atlc.sourceforge.net/

  • 1
    \$\begingroup\$ Thank you Matthew. This tool looks very promising indeed. Unfortunately, it seems a bit tough to get started with. I don't have the time to dig deeper into it. Do you know a good formula? \$\endgroup\$
    – K.Mulier
    Commented Oct 19, 2016 at 12:43
  • 1
    \$\begingroup\$ @K.Mulier I can't provide a formula, but I did use the tools to solve your problem - 63pF/m. If you are on Windows try this hdtvprimer.com/kq6qv/atlc2.html \$\endgroup\$
    – Matthew
    Commented Oct 19, 2016 at 14:51
  • \$\begingroup\$ Hi @Matthew, I've been trying to use ATLC2 to emulate your results and I'm getting values above 400pF. Is the value that ATLC converges on actually in pF/m? in the "results" panel, it only says pF... \$\endgroup\$
    – Sean Dever
    Commented Mar 10, 2020 at 22:51

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