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'Rolled' homemade capacitors

  1. Sep 25, 2012 #1
    By rolled, I mean this kind of basic homemade capacitors:
    http://www.ehow.com/how-does_4852927_building-a-capacitor.html

    Kind sirs, I have few questions regarding this kind of capacitor:
    1.) Would I still be able to get the 'theoretical free-space' capacitor of the same dimensions using [itex]C_o = \frac{A \epsilon _o}{d} [/itex]? I'm guessing A should be the area of the paper dielectric.
    2.) For those who didn't bother looking up the link, the parallel plate capacitor is made by sandwiching aluminum foils and papers; in this case it's: foil, paper, foil, paper, in that order. My question is, this is just one capacitor right? The last paper layer is there just so the two conductors doesn't touch and cause a short circuit, which also doubles as a solid dielectric.
    3.) What would be a good substitute (and cheap) dielectric if we are still to use the same set-up besides paper?

    Thanks in advance.
     
    Last edited: Sep 25, 2012
  2. jcsd
  3. Sep 25, 2012 #2

    Integral

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    It is a single cap.

    Try using saran (plastic) wrap.
     
  4. Sep 25, 2012 #3
    Is that plastic polyethylene? Sorry, I'm not familiar with materials. I'm asking because I'd also like to look up the dielectric constant of the materials I'm using.
     
  5. Sep 25, 2012 #4
    Oops, it's polyvinylidene chloride I could've looked that up earlier. Anyways, anyone willing to answer 1?
     
  6. Sep 25, 2012 #5

    Integral

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    A is the area of your plates.
     
  7. Sep 26, 2012 #6
    I tried doing this, and I tried to read the potential difference across the plates after I charged it using a voltage source. I did get some reading, but it diminishes steadily over time, am I doing it right?
     
    Last edited: Sep 26, 2012
  8. Sep 26, 2012 #7

    sophiecentaur

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    If the volts are dropping slowly then it must be working (well done) and it is discharging slowly through the meter. You can find an approximate value for the Capacitor you have made as follows.
    Charge it and find how long (seconds) it takes to discharge to 1/e of the start value (1/2.7); this is called the time constant. Assume the input resistance of the meter is about 10MΩ.
    The Capacity will be approximately 1/(10e7 X time constant) in Farads.

    Try applying the meter after a short while. If the volts have dropped without the meter there, there is another 'leakage path'. Not unlikely but make sure you are not causing a path through your fingers - they can easily cause some leakage. Also, some damp can be to blame.
     
  9. Sep 26, 2012 #8
    By discharge, do you mean the actual charge of the capacitor or its voltage? Also, by start value, is it the applied voltage or the initial charge?

    Also, you are talking about [itex]\tau = RC[/itex] where tau is the time constant, R is the resistance, and C is the capacitance. Right?
     
    Last edited: Sep 26, 2012
  10. Sep 26, 2012 #9

    sophiecentaur

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    The voltage is proportional to the charge in it. (Q = CV)
    The discharge follows an exponential law. The volts drop by the same proportion in the same interval, whatever voltage you start with. You could choose any time and read the meter, then wait till the volts drop to the 1/2.7 value. But it would probably be easier to measure the voltage with the battery (?) connected and calculate the 1/2.7 value before you start - then disconnect the battery and watch the volts fall. How long, approximately are we talking about, for it to go to half its initial value? A stopwatch (an app on your smartphone, perhaps?) would be handy.
     
  11. Sep 26, 2012 #10
    Ok I get it. I have a digital multitester here do you think it's safe to assume 10MΩ as the deflecting resistance of the voltmeter?
     
  12. Sep 26, 2012 #11

    sophiecentaur

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    Mega Ohms!!!!
     
  13. Sep 26, 2012 #12
    Sorry, I edited it just now lol. I'm quite groggy at this time already. Anyway, your posts were really helpful, thank you.

    Edit: I really just want to make things clear, I can use the resistance of the voltmeter which is 10 Mega Ohms as the resistance for the time constant equation right? (My knowledge of how these meters works is still shaky.)
     
    Last edited: Sep 26, 2012
  14. Sep 26, 2012 #13

    CWatters

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    Deleted
     
  15. Sep 26, 2012 #14

    sophiecentaur

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    Yep.
    But check that there is no decay when the meter is disconnected. Dab the meter on after a while to see. Then you can be sure that the dominant (lowest) resistance in the RC time constant is the R of the meter.

    Be careful with your decimal points and expect a 'low value' then let us know how well you have done.
     
  16. Sep 28, 2012 #15
    The voltage drops really quick the last time I tried it, I don't think I can measure the time with a stopwatch. Would having a larger capacitor help?
     
  17. Sep 28, 2012 #16

    sophiecentaur

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    Of course, double the Capacitance, double the time.
    To get a noticeable difference, though, you would probably need ten times the value of C.
    But if you can see any delay at all, compared with the meter without a capacitor, your time constant will be several tens of ms. You could do your calculation on that assumption.

    Digital displays are not too good for this sort of exercise. Ideally, you'd use an oscilloscope which could give you a nice trace or an electronic meter with a needle display (an old valve voltmeter would do the job fine!)

    I have a DMM which happens to have a switch position with capacitance measurements - that would be good, of course. I used to use it when I had students doing the same exercise as yours. Could you borrow one, perhaps? C measurement is quite common on the more expensive meters you can get.
     
  18. Sep 28, 2012 #17
    I also have a valve voltmeter though, but I don't think I could barrow that kind of DMM you're talking about. I'm quite interested how could we do this: "But if you can see any delay at all, compared with the meter without a capacitor, your time constant will be several tens of ms." assumption? I'd be really really thankful if you could expound on that.
     
  19. Sep 28, 2012 #18
    If you have video camera or photo camera with video mode, then you can record at least 15 frames/second. However, not sure if the volt meter is fast enough...
     
  20. Sep 28, 2012 #19

    sophiecentaur

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    I just meant that, if it was too fast for you to think of using a stopwatch yet it was a 'visible' delay, then that would fit in with that sort of time constant. Any quicker and you'd not spot it at all. 10ms is the period of flashing of a mains filament bulb (UK @ 50Hz) and you can only just see flicker. I would have thought that the eye could pick up on several times that time period. Perhaps 100ms would be a more likely value for the threshold of perceiving the decay. But it's only a ball park figure and could give you an idea of just how good your capacitor is. You could also work out the probable capacitance if you could measure / estimate the thickness of the film (measure many thickenesses and divide by n, rather than trying just one thickness) Look up the formula for a parallel plate capacitor and can calculate the area of foil (double it because it's a rolled sandwich) you can also find the relative permittivity of the PVC. See what agreement you get. It won't be very close because the spacing will be far from uniform inside the roll - but don't worry. Even a vague agreement will be satisfying (as in calculating the Mass of the Universe lol).
     
  21. Sep 28, 2012 #20
    Well, the problem is, what I'd like to do is compute for the dielectric constant experimentally so It'd be good if I could get an accurate result for calculating the resulting capacitance with a dielectric. Also, the last time I tried it with a 3V potential difference, the voltage drops straight to about 0.90~80V which is quite lesser than the 1/e of the initial value.

    I guess I'd try doing it with the valve meter now and look at what figures I can get.
     
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