Calculating capacitance of integral PCB components

I posted this question originally with Electrical Engineering Stack Exchange but nobody seems to know the answer. So I thought a physicist might find the question more approachable.

Formulas for Integral Components:

I am looking for "simple" approximative formulas for integral components. (I.e. devices that are etched into the copper of printed circuit boards.) In particular for:

• A grounded interdigital capacitor (i.e. above a ground plane and with one set of fingers connected to ground), and
• a square pad on an infinite ground plane (to estimate SMT pad capacitance). I believe that due to the method of images this is equivalent to a finite parallel plate capacitor with twice the separation but I am not 100% sure.

Alternatively: Is there a free software that can simulate this and can be used to answer the question without investing more than a few hours to learn how to use it?

Sanity-Check Request:

Maybe somebody is also able to sanity-check the following reasoning.

In the absence of a formula for a grounded interdigital capacitor, would the following provide a reasonable estimate:

1. Calculate the capacitance between the two sets of fingers (with the available formulas)
2. Add the capacitance of a microstrip of equivalent length to the total length for microstrip contained in one set of fingers

Example using the formula provided here:

• $\epsilon_r$ = 4.2
• t = 70 µm (copper thickness)
• h = 1.6 mm
• W' = 9.9 mm
• l = 8.4 mm
• W=S=l'=s' = 0.3 mm

A1 = 3.63*10^-3 pF/mm, A2 = 8.27*10^-3 pF/mm, N = W'/(W+S)+1 = 17

$$C_{interdigital} = 2.58 pF$$

Length of microstrip in one set of fingers (the one with 9 fingers):

$$L_{microstrip} = 9.9 mm + 9\times(8.4+0.3) mm = 88.2 mm$$

$$C_{microstrip} = 88.2 mm \times 0.0431 pF/mm = 3.80 pF$$

It follows that the interdigital capacitor presents an approximate capacitance to ground of about:

$$C_{total} = C_{interdigital} + C_{microstrip} = 6 pF$$

Does this calculation make any sense at all?

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