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Capacitor measurement and understanding

  1. May 1, 2007 #1
    Hello everyone,

    I'm trying to find more information to get a better understanding on capacitors. I made a small parallel plate capacitors basically to use it as a sensor and for data collection/analysis to see if there is any correlation with my ongoing research.

    The capacitor formula C= (epsilon*A)d does not tell anything about the conductivity of plates (material used), nor talks about any side plates. After gathering some information, and comparing it with my calculation, I'm more confused than ever. I'm posting in physics because some of the answers might just boil down to electrons, e-field, maxwell etc

    Questions:
    1. The capacitance goes down if you put a conductor inside the plates or move your hands near it. I suppose this is causing the electrons drift elsewhere creating less e-field?

    2. I compared readings between having side plates/top plates and with the parallel plates (basically having a cube of 6 sides) with just 2 side plates. The readings differ from 15pF (with just 2 plates) to 60pF (cube with 6 sides).
    Why is that? Is it because with the side and top plates it is containing the e-field thus reducing stray capacitance? Again the formula does not say anything about side plates but most books show capacitors built with just the 2 parallel plates and i suppose they are assuming d is so small. My d is approx 2inches

    3. The inductance of my parallel plate capacitor decrease when i increase the frequency. Also I read in the manual it says capacitance measurement accuracy increase when increasing frequency. Why? I thought with a higher frequency the capacitor acts like a short. Isn't capacitance a measure of how many e- can be stored with some potential applied? The higher the capacitance, the more energy or e- are stored.
    Inductance suppose to resist change so it should go up (unless I'm contradicting myself here between impedance of capacitor and inductor and would just need to fix Zc, so if Zc goes down and Zl = Zc then L goes down?)


    Please move this if its not in the right place. Just wanted some ideas, explanation on some of my questions. If would be helpful if could answer some of my questions or point me to the right information.
    Thanks for your time
     
  2. jcsd
  3. May 2, 2007 #2
    I think that you need to learn what capacity is. You can begin with wikipedia. It is not very good but it is better than nothing.
    The formula that you use is an approximation valid only if the plate's dimensions are far greater than de distance between the plates.

    As for your questions:
    1- Capacitance does not go down when you put a conductor between the plates. If you measure so, it is your measure method that is wrong.
    2- As I said before the formula only applies to the given geometry.
    3- Inductance is fairly insensitive to frequency. The precision of the capacitance measurement may improve at higher frequencies (not too high). But it depends on the measurement method, the measuring devices and the measuring ability of the operator.

    You are right that impedance of a capacitor diminishes with frequency and impedance of an inductor increases with frequency.
     
  4. May 2, 2007 #3
    Ok here's what I've come up with. Someone correct me if I'm wrong. Need further help and resources

    - Capacitance is a measure of amount of charge,Q appearing on each plate. More voltage means more charge can accumulate.

    - Capacitance is a function of the physical size of the plates plus dielectric. Increasing voltage does not increase capacitance, but increase the amount of charge/energy stored. If the voltage goes to high beyond the breakdown voltage of the dielectric, it shorts out and becomes useless.

    - Putting a conductor between 2 plates reduces the capacitance. If the conductor touches both the plates, it is a short so no energy stored. If there is gap between the conductor inserted and the plates, it is essentially 2 capacitors in series. Hence, if C1 and C2 are in series, the product of their capacitance is less (1/Ceq = 1/C1 + 1/C2). If the dielectric is partially in maybe 1/2 and the other 1/2 is air, the it is in parallel. The capacitance would simply be the C = kx + x with k being the dielectric constant and x being the capacitance of air (k=1).

    - The overall E-field inside the capacitance is reduced. The Q+ and Q- creates E-field in one direction for the plates, while the Q+ and Q- of the dielectric gets attracted the other way. Product of both e-field is less then. This means more e- can crowd onto the plates increasing the charge and energy stored. Does this even make sense? Higher e-field means more voltage, which also means more current flow and energy stored. With a dielectric inserted, capacitance increase but energy stored becomes less (due to less e-field) thus you need more voltage to get more energy stored.

    - I still cannot find a solution for solving my parallel plate capacitance theoretically for my research. d = 2", A is approx 3"x3". If the formula applies for A>>>d, how can i derive for mine then? Somebody have a equation/method that i could start with like deriving from .......?

    - In DC, the charge accumulates onto the plates such that eventually there is no more current left to flow. Voltage stays constant, since there is no voltage change, current flow is zero. In AC, the charge,Q goes back and forth between plates creating movement of e- in the E-field. This allows current to flow...??

    - Anything else i left out? Suggestions or questions to make me probe further is appreciated

    Thanks.
     
    Last edited: May 2, 2007
  5. May 2, 2007 #4
    Uhm, say what? Last time I checked, putting a conductor between the plates always reduces the capacitance.

    This is a non-trivial problem. I think you're going to end up with all sorts of non-closed-form functions; you won't get some nice general formula you can write down. The right thing to do is get a computer program that simulates arbitrary capacitor geometries, and just make plots of capacitance versus whatever characteristic you want. I don't know what programs do this (but you could always write one in MATLAB/octave or something).

    I think everything you said up there is correct. If you want to learn everything in ridiculous detail, get a good book on electrostatics. Or a good book on electromagnetics and just read the chapter on electrostatics. Learn what E and D mean.
     
  6. May 3, 2007 #5
    Was the conductor isolated or you hold it with your bare fingers?
    Imagine a plane plate capacitor, plate's dimensions larger than separation, etc. Put a plate (very thin for now) in the middle of the two other plates. Now you have two capacitors in series, and each one has a capacitance double of the original. The two in series give a capacitance identical to the original. You can derive the formula for any position of the intermediate plate.
    If the intermediate plate is thick, the capacitance of each capacitor will be higher than the double of the original one, and the global capacity greater than the original one.
     
  7. May 3, 2007 #6
    I see what you're saying. Capacitance is proportion to 1/d, and if i made the d twice smaller by putting a really thin conductor between both plates, assuming i had 1pF with 2 plates, now each thin plate would have 2pF. Resulting Caps in series would be 1pF then. Likewise for a thicker plate means smaller d. So i suppose this is how some capacitor manufacturer create large Caps, by putting multiple plates in series or making it in parallel to add up capacitance.

    Between, yes i was holding momentarily with my hand. But i also took a small piece of conductor plate and put it in the middle and stayed far away and it still fell down little bit. This is using a calibrated capacitance meter (and i did zero the offset in the meter before using it to compensate for stray capacitance). I know my A > d just not a whole lot larger, maybe there's something else i'm missing
     
  8. May 3, 2007 #7
    Yes, to manufacture high capacity capacitors, you must diminish "d" and have the biggest "A" possible. You put capacitors in parallel to increase capacity.

    If putting an isolated metallic part in between, capacity went down, you modified some other thing inadvertently. Do not touch anything other. Do not recalibrate the offset between the two measures.
     
  9. May 4, 2007 #8
    lpfr: Oops, I was thinking about something different. My mistake!
     
  10. May 7, 2007 #9
    ughh what were you thinking? I'm going to look into some D and E i guess. lpfr, i calibrated the meter before performing the test, inserted a conductor in the middle and it fell. Maybe there's additional variables affecting the equation in my case since A !>>> D. Anyone else out there to help me out here with ideas?
     
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