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Capacitance of non uniform field capacitor

  1. Nov 17, 2005 #1

    Integral

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    I Would like to come up with the capacitance between an infinite sheet and a finite area rectangle as a function of distance. I looked into my old E&M book, it did not consider edge effects I need to include those.

    Here is the situation. I just spent the day working with a capacitive sensor. One plate is a rectangle ~(1X5 cm) the other is a 8" Si Wafer. The sensor measures the distance to the wafer with error less then 100 [itex] \mu m [/itex] . To calibrate this sensor the software collects 5000 data points as the sensor is moved from contact to 8mm away, it is then fit to a 4th order polynomial.

    All very good, but through the day I noticed that the software was rounding and numbers, well not really rounding them, my thought was that they must be using single precision variables. I still not understand why or how 2 would be come 1.999999 ???

    The kicker came when they used Matlab to come up with the "5th" order polynomial. The coefficient of the 3rd order was on the order of 10-14 the fourth was order 10-18 the fifth seemed to be set at 0.

    I do not under stand how software which cannot handle integers can meaningfully compute that polynomial.

    Soooo.... I want to know what theory says..
    Thanks for any help you can provide.
     
    Last edited: Nov 17, 2005
  2. jcsd
  3. Nov 18, 2005 #2
    I'm curious what your E&M book had to say on the subject even without fringe effects. Capacitance is the ratio of a charge +/-Q placed on two conductors to the potential difference of those conductors. How do you place equal magnitudes of charge on an infinite and finite conducting plane?

    We can calculate the capacitance (per unit length, for example) of two infinite coaxial conducting cylinders, but I don't think it's possible to define the capacitance between an infinite plane and a finite plane.

    I could be wrong about all that.

    I don't think what you're asking is possible, but if it is, you may want to look in Jackson. I recall some variational theorems he had on capacitors, but I also recall that you had to find a Greens' function that obeys the same boundary conditions as your problem (ie in your case, "5 volts on an infinite plane, -2 volts on a finite plane floating 1 meter away", or whatever the setup is. You can get bounds on the capacitance, if that's useful.

    What's a 4th order polynomial? You mean your doing a 4th-order spline fit to a numerically computed capacitance?

    Well, in infinite precision, 2 and 1.999999... are the same number. :)

    I seriously doubt any software you're using uses single precision numbers. It really doesn't pay to use floats. A float is allowed to have as few as 6 sigdigs. A double can have no fewer than 10 (although a double with less than 15 sigdigs is rare? non-existant?) And based on alignment issues, doubles are often faster than floats. I'm pretty sure the software isn't using single precision numbers. People get paid a lot of money to figure implementations out. Really smart people.

    Anyone who is used to numerical calculations wouldn't blink twice at a 2 turning into a 1.9999999. It's the nature of the beast! Unless you're using infinite precision packages, like Mathematica. But even there, I've seen hellacious expressions resulting from integrals that were tremendously glorified versions of zero (or "very small"). Terms that you KNOW should've been zero if just a "few" more digits were kept track of! :)


    I don't know anything about Matlab. It can't handle integers? But it isn't such a stretch. It's entirely possible to do all calculations without knowing anything about integers, because the the integers are contained within the reals. That's how the computer knows what you mean when you add a float to an int.

    Sorry if I didn't answer the proper question. I was just trying to say something which might be useful.
     
    Last edited by a moderator: Nov 18, 2005
  4. Nov 18, 2005 #3

    Integral

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    The key question here is the formulation of an expression for the capacitance. If you bring a finite plate near a second plate which has a much larger surface area you have created a capacitor, this capacitor must have a unique capacitance, depending upon the separation and area of the small plate. So a unique solution does exist simply because it is a well defined physical situation.

    As for the numerical puzzles, it has occurred to me that rounding would occur if the were converting to HIGHER precision.
     
  5. Nov 18, 2005 #4
    Right. All I was really saying (which may or may not be of help) is that I don't believe you can define a capacitance between a conductor of infinite extent and a conductor of non-infinite extent. Nothing more.

    Of course big conductor and small conductor form a well defined problem, but I don't think infinite conductor and small conductor is.

    What I was getting at was that I believe you won't be able to get a nice expression for the capacitance by making simplifying approximations like infinite this or zero that. You may have a full-fledged Jacksonesque chapter 3 type potential problem on your hands. Sorry if I was being abstruse.

    Yeah, that could be. Who knows what goes on in those little black boxes? I really wish the open source community would develop something along the lines of mathematica with an SDL graphical interface to make it mega-portable, mega-free, and mega-transparent so we could look at the nitty gritty details of how things are computed.
     
  6. Nov 20, 2005 #5

    pervect

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    There is no problem finding the capacitance between a a conductor and an infinte sheet. The "method of images" is commonly used to do this. If you have a charged conductor of some particluar shape with a charge of +q at a height of +z, and a charged conductor of the same shape with a charge of -q at a height of -z, the potential at z=0 will be 0 by symmetry. Thus, when you solve the first problem for V(x,y,z), you automatically generate a solution for the second.

    Integrating it out, you find that the voltage for the second case is half that of the first case, or the capacitance C = Q/V is doubled.

    The solution of the problem for cylindrical capacitors isn't too bad because of the circular symmetry - the theoretical solution with edge effects in terms of Bessel functions for a circular capacitor with edge effects is given at

    http://scienceworld.wolfram.com/physics/ParallelPlateCapacitor.html

    (but it's rather sketchy)

    http://chemandy.com/calculators/circular_capacitor_calculator.htm

    has a usable standard-units calculator for this formula. Of course you don't have a circular probe, or even a square one, you have a rectangular probe, so it's not exactly what you're looking for.

    The theoretical problem is to solve Laplace's equation in three dimensions for the potential function V and with V zero at z=0, and V= constant over some thin rectangular plate. You then can find the voltage directly, and Q by Gauss's law, giving C = Q/V.

    I have a recollection that conformal mapping can be used to solve the above problem, but the details escape me at the moment. The idea is to perturb the above solution with circular plates into one with the appropriate square (or rectangular) boundary conditions.

    I'm pretty sure I saw a table once in some IEEE publication for computed values for rectangular parallel plate capacitors with a dielectric. (You don't need the dielectric for your problem, but its presence made numerical methods necessary).

    I'm not sure how much this will help, but I hope it will help some.
     
  7. Nov 21, 2005 #6

    Integral

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    Thanks Pervect, that is the kind of info I am looking for. I'll see what I can do with it.
     
  8. Nov 21, 2005 #7

    pervect

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    Unfortunately, on a closer examination, for a large enough value of d the computed capacitance from the formula on Eric's webpage eventually becomes negative. So I don't trust the formula as written anymore :-(
     
    Last edited: Nov 21, 2005
  9. Nov 21, 2005 #8

    pervect

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    I've been doing more digging, and

    http://www.ttc-cmc.net/~fme/captance.html

    (the meat is in a postscript download on that webpage)

    seems to be making sense - their analytic solution actually satisfies Laplace's equation :-). Unfortunatley it's valid only for a single-plate capacitor :-(.

    This simple result is still useful - it gives a finite positive limit for the capacitance as d gets very large, where the other plate (or the ground plane) doesn't matter

    C = 8 pi E0 a

    a being the radius of the plate. The rest of the paper is about numerical methods, with a graph of some results of simulations at the end.

    http://physics.usask.ca/~bradley/pll_plate_edge.pdf

    has a much more concise and readable description of the conformal mapping method, which appear to me to apply only to the field near an infinitely long edge, so it's not all that useful for solving your problem.

    I'm beginning to think my memory is playing tricks on me as far as a simple solution existing for a cylindrical geometry.
     
  10. Nov 22, 2005 #9

    pervect

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    Last edited: Nov 22, 2005
  11. Nov 23, 2005 #10

    Integral

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    Thanks again Perfect.

    I am starting to see that there is not going to be a nice clean functional solution. Perhaps this vendors method of just fitting to a high order polynomial is not a bad solution.

    I have also realized that since the return signal, at about 1cm, is on the order of 20000, there may be significant contributions even with a 10 -18 multiplier.
     
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