Pad Capacitance Calculation

Hi All,

I'm interested in finding out the real life capacitance of two pads which I have shown in my attached picture. Essentially this is a capacitively sensed switch. Also in my attached picture is the metal foil which, when pressed, comes down and hovers just over the two pads on the left side of the picture. So, you start out with the switch unpressed and you get the model on the left. This is just the capacitance of the two metal half moon pads along with trace capacitance because there is an excitation signal(square wave at about 1Mhz) on one of the pads coming from the Micro. When you press down the metal foil to hover over the two metal pads this creates two capacitors in series ultimately decreasing the capacitance. This change in capacitance is sense by electronics which I probably don't need to get into at this point.

What the best way to approach this to find the capacitance of first the pads? I'd like to actually calculate than say have an excitation signal and experimentally find it. I would also like to find the equivalent capacitance of adding the metal foil into the circuit.

Any thoughts, theories, or web pages with suggested reading would help. Thanks!

I think this is more complex than it looks. In the un-pressed photo the two pads have a track running between them, which will significantly reduce the capacitance, which would be very small even without it since the plates of the capacitor are formed by the edges of the pads. Next you have to consider whether this is a double or single sided board. If there is copper on the opposite side it will form a much larger capacitor than the one you are interested in. Next, I think (and please correct me if I am wrong) that you will find the opposite effect to what you expect, since the two capacitors in series you are creating by pressing the pad will still have a much larger value than the single capacitor which exists with just the pads.

Well that's about the limit of my knowledge here, hope it helps.

Hope these links are useful.
**broken link removed**
http://www.daycounter.com/Calculators/Plate-Capacitor-Calculator.phtml
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html
http://en.wikipedia.org/wiki/Capacitance

Have a look at the attachments for my thoughts on this.

When the circuit board pads are on their own without the switch disc anywhere near to them, there will be three components to the capacitance between the two pads.
C1, the capacitance due to the area of copper on the ends of the two pads.
C2, the capacitance due to the fringing electric field passing through the circuit board material.
C3, the capacitance due to the fringing electric field passing through the air.

C2 will be greater than C3 due to the dielectric constant of the circuit board material.

The capacitance between the two pads is then Ctotal = C1 + C2 +C3


When the pads have a switch disc in close proximity to them, I think that there will be six components to the capacitance between the two pads.
C1, the capacitance due to the area of copper on the ends of the two pads. (As before)
C2, the capacitance due to the fringing electric field passing through the circuit board material. (As before).
C4 and C5, the capacitance due to the fringing electric field passing through the air at the edges of the pads/switch disc.
C6 and C7, the capacitance between each pad and the switch disc.
I think that the C3 capacitance may dissapear in this configuration, or at least be very much reduced.

The capacitance between the two pads now becomes Ctotal = C1 + C2 + ((C4 + C6) series (C5 +C7))

The capacitance between two parallel plates is easily calculated using the standard formula C = E0.Er.A/d
How to calculate the fringing capacitance, I don't have the faintest idea.

I hope this helps.

JimB

Wow. Let me digest. Thanks!!!!!