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simple capacitor test circuit

Discussion in 'Electronic Design' started by [email protected], Mar 18, 2013.

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  1. Guest

    Hello, I am new to discussion groups. I have a home built capacitor (it should have a fairly large capacitance) and I would like to find out what its value is as accurately as possible with a standard multimeter. I wonder if it is possible to build a simple circuit with a led light indicator or something that with a little math and good capacitors can tell me ruffly how large it is. Please help. I only need to get a couple of measurements for my project.
     
  2. amdx

    amdx Guest

    and I would like to find out what its value is as accurately as possible
    with a standard multimeter.

    I wonder if it is possible to build a simple circuit with a led light
    indicator or something that with a

    little math and good capacitors can tell me ruffly how large it is.
    Please help. I only need to get a couple of measurements for my project.
    Charge it to a voltage, say 30 volts, then put a resistor across it's
    terminals and see how long it takes to discharge to 10 volts.
    Compare that to known capacitors, add capacitance until the times are
    equal. Or, do a search on RC time constant and you can calculate, using
    the resistor (R) you pick and the time. Discharge to 7.5 volts is 3
    time constants.
    These pages discus RC time constants.

    The discharge time will depend on the capacitance of your capacitor
    and the resistor you choose. If you think your capacitor is not very
    leaky, I would make the time for discharge 20 or 30 seconds to improve
    your measurement accuracy.

    What is the input impedance of your meter, or what kind of meter do you
    have.

    How did you build your capacitor?
    Mikek
     
  3. Guest

    Thanks for all the feedback! I was actually thinking of some kind of simpletiming circuit with a led light indicator but I think I can figure something out with those methods.

    I actually have two capacitors to measure. Each are made with 2 sheets of stainless steel foil with 2mm thick corrugated plastic as a spacer on the ends of the foil (the center is not supported by the spacers). They are woundaround each other on a PVC pipe in a double coil so they have two plates but they have about 3/4 more plate surface area. I hope that makes sense. The dielectric will be tap water so I plan on them being very leaky.. The smaller coil plates (foil) are 12cm by 22cm (264 square cm). the larger one are double the length at 12cm by 44cm (528 square cm).
     
  4. amdx

    amdx Guest

    Hmm, I'm not sure that will be as big as I was envisioning.

    Let us know.

    Mikek
     
  5. Jamie

    Jamie Guest

    Don't have any idea how large is large on your end?

    But, you can use a resistor to charge the cap and time it for
    large types.

    Time_TO_Charge_To_63.2% = R * C;

    So, if you connect lets say a 1 volt supply through a 1k ohm
    R to the cap and have the other end of the cap back to the
    opposite terminal of the supply and then, measure the time it
    takes to get to 0.632 volts. So Time_Calculated / R = C .

    To start the test, you just short the cap and as soon as you release
    the short, start the timer.

    I am just assuming you have a supper cap!...

    Just a thought....

    Jamie
     
  6. tm

    tm Guest

    Hey, it's even worse than you were thinking :).
     
  7. Guest

    I remember seeing that calculation somewhere.. Thanks for reminding me about it. I am not sure what you mean by pulse forming and I am not quite sure what I am intending on. All I know is I will be using a pulsed (I have to figure out an ideal frequency to step charge the capacitor) DC power supply at about 3KV or higher with a bifilar wound coil and one of these as a capacitor and maybe a choke coil in there.. The goal is to continuously resonate the tank circuit.
     
  8. Tim Williams

    Tim Williams Guest

    So like, Tesla coil stuff?

    You may find the capacitor is lossier than the coil, and consuming all the
    power you're trying to push into it means it won't resonate very well.
    Water has a noticable dielectric loss tangent (think of it this way: it's
    a *really bad* capacitor all the way up at microwave frequencies, in fact
    it's an excellent resistor!).

    Tim
     
  9. Greegor

    Greegor Guest

    What made you choose water?
     
  10. Jasen Betts

    Jasen Betts Guest

    perhaps build ac bridge and compare them with a known capacitor.
     
  11. Jasen Betts

    Jasen Betts Guest

    for some examples of that approach google "555 capacitance meter"
    sheets of stainless steel foil with 2mm thick corrugated plastic as a
    spacer on the ends of the foil (the center is not supported by the
    spacers). They are wound around each other on a PVC pipe in a double
    coil so they have two plates but they have about 3/4 more plate
    surface area. I hope that makes sense. The dielectric will be tap
    water so I plan on them being very leaky.. The smaller coil plates
    (foil) are 12cm by 22cm (264 square cm). the larger one are double the
    length at 12cm by 44cm (528 square cm).

    tap water dielectric won't withstand voltages over 1.48V
    so the 555 approach isn't going to work, perhaps you can put them in
    series with a smaller capacitor that's good to 10V or so and measure
    the pair.
     
  12. Phil Allison

    Phil Allison Guest

    <

    I actually have two capacitors to measure. Each are made with 2 sheets of
    stainless steel foil with 2mm thick corrugated plastic as a spacer on the
    ends of the foil (the center is not supported by the spacers). They are
    wound around each other on a PVC pipe in a double coil so they have two
    plates but they have about 3/4 more plate surface area. I hope that makes
    sense. The dielectric will be tap water so I plan on them being very leaky.


    ** Very leaky is a massive understatement.

    A pair of parallel SS plates the size of your palms will quickly boil tap
    water if connected to a 230VAC supply.

    That is about 5 amps of " leakage ".

    Or 45 ohms of resistance.

    The OP mentions using a 3kV DC supply.

    He is having you on.

    Fucking Google Groups gmailers.


    ..... Phil
     
  13. Robert Baer

    Robert Baer Guest

    And...use the meter resistance (most good DVMs are 10 Megohms) for
    the R in series (for charging) or in parallel (for discharging).
     
  14. Robert Baer

    Robert Baer Guest

    He lives in the South Pole...
     
  15. amdx

    amdx Guest

    That's why I ask for info about his meter.
    Hey diablosdemon, you missed one of my questions.

    What is the input impedance of your meter, or what kind of meter do you
    have.
    Mikek
     
  16. John S

    John S Guest

    Just using the 528 mm^2, 2 mm separation, and 80 for a dielectric
    constant, it calculates out to about 187 pF.

    Not very large in capacitance and, with tap water for a shunt
    resistance, you probably won't even be able to measure the capacitance.
     
  17. He said 12 by 44 cm, not mm. So that should be about 19nF. With
    tap water, it would still be more of a resistor than a capacitor.

    To the OP: What are you trying to do with all this?

    Jeroen Belleman
     
  18. Phil Allison

    Phil Allison Guest

    "John S"

    ** At last, someone with a tad of common sense.




    .... Phil
     
  19. Guest

    Something is getting lost in the translation, or the OP does not understand the plates need to be insulated from the water:
    http://www.electronicsarea.com/pic_display.asp?id=7&title=Capacitive water level sensor

    Maybe he thinks the SS passivation is his insulation.
     
  20. John S

    John S Guest


    My mistake. Thanks Jeroen. And, I believe you are correct that it will
    still not be easy to measure (if at all).
     
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