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Capacitor question

Discussion in 'Misc Electronics' started by Ken O, Jun 22, 2006.

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  1. Ken O

    Ken O Guest


    I have a rated 525v 30uF Ac capacitor that I want to discharge into a load
    (coil). I am using a battery as a source, using a chopper circuit, a 12vto
    120 v transformer to a full bridge rectifer rated 1000v. everything seems to
    be ok at this point. Althought I am not getting any discharge from the
    capacitor. I let the capacitor charge before dumping it top the coil. I
    guess I don't have enough charge to creat a great enough current. My
    question is, how do I calculate the charge I need (in farads) to I could get
    a discharge ?
    tried fidling around with with v=dQ/dt=I R but I did not find anything I
    could use
    30uF was a a ball park figure, it does take a little while to charge so I
    assumd it would be powerful enough

  2. HKJ

    HKJ Guest

  3. Tom Biasi

    Tom Biasi Guest

    Making a rail gun Ken?
  4. Ken O

    Ken O Guest

    Making a rail gun Ken?

    hahaha something like it :)
  5. Ken O

    Ken O Guest

    I do have a brief discharge from the cap to the coil. But the kind of
    repulsion i am getting seems that I have a 3v battery attached to it. I am
    getting the opposite effect that I am aiming for . 30uF does produce quit a

  6. Ken O

    Ken O Guest

    I guess my questions would be, would getting a 300uF cap increase increase
    the electromagnet strenght ?
    MAybe a different of capacitor ?

  7. Perhaps most of the capacitor's energy is going into making sparks,
    instead of magnetic field.

    How are you connecting the capacitor to the coil?
    Describe the coil dimensions, turns and wire gauge.
  8. Ken O

    Ken O Guest

    The spark comes from me putting a screw driver in there, Ithough tthe cap
    might be dead. Otherwise no sparks.
    the last test I did was to charge the capacitor and immediatly put the
    positive and negative to the coil. The coil is gauge23, about 500 turns woth
    10.6 ohms.
    I assumed if you connected a 30uF 127 v cap (charged) to a coil it would (
    for a very brief time) repulse a magnet (assuming its north to north). I was
    thinking maybe I need a bigger cap 30uF is not enough? A dc cap or a flash
    cap would do a btter job?
    I was hoping to get the same result as if i would hook up a 120v battery
    (just not as long )

  9. What inside diameter?
    Might repel or attract it, or even remagnetize it in the other
    direction. It will repel a closed loop of conductor.

    Have you checked the capacitor voltage with a meter before doing the dump?
    There is more energy in a larger cap, and the pulse will last longer.
    The coil will interact with a magnet, in proportion to the
    instantaneous current. It will interact with a shorted loop of
    conductor, only when the current is changing.
  10. Ken O

    Ken O Guest

    What inside diameter?
    2.2 cm
    outside is about 4.5 cm
    height 7cm

    127v before dumping
  11. Bob Eld

    Bob Eld Guest

    It would be helpful if you told us what you are really doing and what the
    circuit looks like. The energy stored in the cap is 1/2*C*V^2. At best only
    half of that energy will be available in the coil. The rest will dissipate
    in the switch or spark, etc. What is the voltage the capacitor charges to?
    Have you measured it? Your coil is about 10 ohms. The RC time constant of 10
    ohms and 30 uF is 300 us. If you are trying to get something mechanical to
    happen in 300 usec, you can just forget it. Mechanical interia will insure
    that little will be happen in that short time and it will all be over before
    anything apparent occurs. You need to determine how much force over how
    much time (impulse) is required to do what you want. You need to design the
    coil and discharge system to get that impulse. In otherwords, you have to do
    the math and understand the physics otherwise you are just screwing around
    which is ok but don't expect great results. I suspect your problem is
    insufficient energy storage and way too short of a time constant. Also, the
    coil will have inductace which will enter into the calculations. Good luck.
  12. Okay, by the formula, Energy (in watt seconds or joules)= 1/2 * C
    *V^2 the capacitor has about 1/4 watt second in it. Not a big bang.
    But you would get almost all of that transferred to the coil if you
    dumped it with an SCR, instead of touching wires. Of course, if you
    could jack the cap up to 300 volts, the energy rises to 1.35 watt seconds.
    According to this inductance calculator:
    (with the wire gauge altered to give about 500 turns. You must have
    used vinyl insulated wire)
    your inductor should have about 2.8mHy of inductance, so the half
    period of the resonance with 30 uF is about 1 millisecond. If you
    switch the capacitor on to the inductor with no loss, that is how long
    the first pulse will be (excluding ringing, which you can force by
    having a diode in series).
    That is not a long time to be accelerating a magnet. More turns or
    more capacitance would extend the time.
  13. Jamie

    Jamie Guest

    coils are inductive. the energy will not get
    absorb fast enough for the size that your using
    most likely. that is just a guess, could be
    something much simpler.
    your chop rate maybe to fast.
  14. Ken O

    Ken O Guest

    Okay, by the formula, Energy (in watt seconds or joules)= 1/2 * C *V^2
    Ok so bigger coil. Need to buy more magnetic wire then :)
    Actually I was able to bring the voltage up to 280v, but the discharge is
    done no quickly I can barely feel a thing. Also Bob mentionned the time
    constant RC. ok If I added a 1000ohm resistor I,ll have a t=30.3mSec which
    is quit longer, but the resistance then prevented from getting much
    repulsion. Although the energy formula do not take into consideration the

  15. Resistance is not the way to go, because it lowers the peak current.
    The inductance and capacitance forms a resonant circuit. When the
    charged capacitor is connected to the inductor, it produces a damped
    sinusoid. The frequency produced is roughly 1/(2*pi*sqrt(L*C). The
    first half cycle is what you want to deal with, and you can prevent
    there from being more than that, by adding a diode in series with the
    inductor. So increasing the capacitor by 4 times doubles the period
    of the dump pulse, and also quadruples the total energy in it for the
    same starting voltage. Quadrupling the inductance does nothing to
    increase the total energy, but doubles the time of the pulse.
  16. Ken O

    Ken O Guest

    will try that

    its Ac cap
  17. Guest

    use c=q/v

    then find the value of charge ,use resistors to get required current
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