I mounted a Resistor (SMD, Size: 0603, Resistance 500MOhm) on a PCB. What is the value of its parasitic capacitance? How would it change if it were 0402 in size? Does it depend a lot on the individual component or is it solely dependent on the distance/size of the soldering pads?

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    \$\begingroup\$ Not enough information.. you need to define what its capacitively coupling to. If its a ground plane, you need to know the distance . I.e. how thick us your prepreg? \$\endgroup\$
    – Kyle B
    Commented Apr 22, 2020 at 7:39
  • \$\begingroup\$ no ground plane \$\endgroup\$
    – luis
    Commented Apr 22, 2020 at 7:48
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    \$\begingroup\$ Read the data sheet for the resistor - it should tell you its inherent capacitance then add 0.1 pF for good measure. \$\endgroup\$
    – Andy aka
    Commented Apr 22, 2020 at 9:13
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    \$\begingroup\$ The datasheets I checked did not provide this value. What is the typical range? \$\endgroup\$
    – luis
    Commented Apr 22, 2020 at 11:17
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    \$\begingroup\$ Ahh, you're asking about parasitic capacitance of the resistor to itself. Not it's capacitance to other things. Andy's right, check a datasheet. If the ones you're using don't have this value, check some other manufacturers, they should all be similar. \$\endgroup\$
    – Kyle B
    Commented Apr 22, 2020 at 15:10

1 Answer 1


To approximate such parasitics, I just use the parallel-plate capacitor formula:

C = E0 * Er * Area/Distance

where Eo = 8.98e-12 farad/meter and your FR-4 PCB Er ~~~ 5 (maybe 4.7, but who cares)

This formula becomes about 45picoFarad/meter * Area/distance

If you have a 1/16" thick FR-4 PCB (1.5 milliMeters) and 3mm by 3mm solder pads, then the parasitic capacitance (ignoring the fringing) will be

C = 45picoFarad/meter * 3mm * 3mm/1.5mm * 1meter/1000mm

C = (45pF * 3 * 3 / 1.5 )/ 1,000 ==== (45*6/1,000) picoFarads

C = 0.27pF

Now about that fringing ... given you have fringing on all 4 sides, and the Distance is approximately the size of the solder-pad (yes, I know 1.5mm is not 3mm), I'd just double what we computed, and have 0.54000 pF as our estimate.

And this doubles, because you have TWO solder pads for the capacitor.

You can use a Finite Element Model, using SPICE resistor grids, and get much more accurate.


We've just computed/estimated the capacitance from each solder pad to Ground.

For capacitance across the resistor, in parallel with the resistance element, you need a **distributed R+C" model, which with the embedded tiny pieces of resistance will be lossy.

Or you can ignore the distributed model and just model the endcaps where you use solder. For that, you can approximate with "wire to wire" models, which use "hyperbolic-cosine" in the differential-equation of the electric fields. If your resistor is anywhere near other pieces of metal, some/most of the efield flux can be sketched as being attracted to those metal bits.

Since PCBs install these capacitors over FR-4, the cap is in a mixed air/FR-4 dielectric, and 80% of the field passes thru the FR-4 because its relative dielectric is 5x that of air. In this case (the very common case), any underlying Ground planes will disrupt the field model, and invalidate the pseudo-precision of hyperbolic-cosine coefficients.

Long time ago, I heard you should assume 0.5 picofarad between terminals on SMT components. Your mileage will vary.

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    \$\begingroup\$ Except the pads aren't parallel ???? They're co-planar. \$\endgroup\$
    – Kyle B
    Commented Apr 22, 2020 at 17:51

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