Is there capacitance between non-aligned parallel plates?

genxium

Our professor gave us a lecture about sensors tonight, but I'm kind of confused about the principals of capacitive touch screen.

As the attached figure shows, when finger touches the screen , capacitance is formed between finger and the ITO bars, but is it true for all the ITO bars?

It's well known that parallel aligned plates can form a capacitor, and [itex]C=\frac{\epsilon A}{d}[/itex] , [itex]A[/itex] is the so called "effective area", in high school I was told that [itex]A[/itex] is the "aligned projected area", according to this, if my finger has no "projected area" that intersects with some far away ITO bars, can I tell that there's no capacitance formed between my finger and these far away ITO bars?

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berkeman

The capacitive effect varies with distance. The bars that are farther away see less and less of a change in capacitance when the finger touches the screen.

genxium

Yes but is it "relatively small" or "0" capacitance ?

mfb

Relatively small, with 0 as a good approximation for distances much larger than the thickness of the glass.

DragonPetter

Think of a capacitor with any 2 opposite electrode points as having an electric field between them. Capacitance is related to how strong this electric field is. The electric field ideally would be perpendicular between the 2 parallel points. If you have a perpendicular electric field between 2 parallel points, then what would be a resultant perpendicular electric field between 1 of those points and another point 5cm away? By projecting the E-field along the distance between the point and a distant point, the resultant perpendicular field will be ideally negligible, and compounded by this is the distance increase in the denominator of the parallel plate capacitor formula, and so the capacitance must be quite small.

genxium

@DragonPetter, I think it's better to go back to the definition between 2 electrodes, allow me to ask a similar problem, assume I have 2 parallel, infinite area plates in space, but they're both half planes such that in their norm vector direction, they don't have overlap projected area , like the figure in my attachment. and they're apart from another by a distance d, then is there any capacitance between them? If so, how to define/calculate this capacitance?

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DaleSpam

Yes, there is capacitance. To calculate it, place a equal and opposite test charge on each and calculate the field. Then calculate the voltage difference by integrating the field from one to the other. The capacitance is the ratio of the charge and the voltage. For the shape you have proposed that will be a messy calculation.

DragonPetter

There would be some capacitance because of the E-field edge effects between where the 2 plates are separated (and so the plates are not infinite). The infinite area (approximated in proportion to the distance separating the plates) is meant to ensure a uniform and perpendicular E-field in the parallel plate formula. So, if the E-field is not uniform and the plates are not ideal as in the parallel plate formula, that formula breaks down and you have to define it in some other geometrical way. This gets back to my first reply with regards to the capacitive touch screen.

I have no idea how to calculate the capacitance in the example you give nor how to define it. I'd need to work that out myself, but if you have the time to work it out, I think you could at least get a rough estimate. It would obviously be much less than an ideal parallel plate capacitor with infinite areas over each other.

Do you have the intro physics book by Halliday, Resnick, and Walker? It has a lot of examples for calculating capacitances with different geometries. The parallel plate formula is very useful, but not the unique rule for defining capacitance.

mfb

This would probably need a simulation and numeric evaluation of the capacitance.
It is possible (and easy to feed into a computer), but you don't want to do it with pen and paper.

genxium

Yep that's really massy calculation, and thank you for providing the idea( of course also thank everyone who donates there ideas), although I'm still confused.

As my lecture note says, when finger touches the upper glass, there're 4 current probes drawn at 4 corners of the lower plate(the one with ITO bars attached) , then for locating based on detected current values, 4 current values should be proportional(I think it's not necessary, monotonic is enough) to the distance between finger and 4 corners, respectively.

As all suggested, it's hard to calculate the accurate capacitance, then how could they make capacitive touch screen work? I knew that from the Professor, this 4 corner type is only used for ATM or anything that does not need accurate location(otherwise they use grids), but the idea is interesting, so unlucky that I'm not using a computer so that I can't attach images for my question >_<

genxium

That geometry really seems tough, I don't have that book but I have similar ones, in EE major, but most of the examples are conventional, intuitive to understand which wraps the other to form a capacitor, as suggested in the replies i will try both finite element simulation and integral methods, after the final exam -- it's really not a good time for me to get into this right now, but 10 days later will be very nice time for calculation~

mfb

Usually, you are not interested in the individual values. They depend on the finger size and pressure and other parameters you cannot control anyway. You are interested in the central position. To determine this, it might be sufficient to take the weighted average of the individual values. This method can be improved if the weights are not simply the measured capacitance, but some function of it.
Or maybe some function can be fitted to the measurements - but I doubt that this is required.

the_emi_guy

I use a program called FastCap for determining capacitance between arbitrary conductors. It was originally developed at MIT and is quite popular. www.fastfieldsolvers.com has a version that runs on Windows (and it's free!). You do need to understand the concept of self capacitance, and read up on how to read a Maxwell capacitance matrix.