Connect with us

Which Diode?

Discussion in 'Electronic Basics' started by Mike, Nov 9, 2004.

Scroll to continue with content
  1. Mike

    Mike Guest

    I'm making a very simple coilgun. This basically works by charging up a
    capacitor and discharging it through a coil. I understand that I should
    place a diode across the coil to stop any current going back into the
    capacitor. Which properties of a diode should I be looking at when I make
    my selection? In this example let's say that the capacitor has a voltage
    rating of 100V and a capacitance of 200uF. When I look at the available
    diodes the properties of each is meaningless to me.

    TIA
     
  2. Reverse voltage capability is cheap, so there is no good reason to tey
    to cur this close. A 200 to 400 volt unit will cost almost the same
    as a 50 volt one. The biggest concern will be peak forward current
    capability (surge current rating), since this is what you are pushing
    through the coils. 3 amp continuous rated lead mounted diodes tend to
    have oversized junctions to let them get rid of the normal heat
    through their leads. This tends to give them high surge current
    ratings. Even more so for 6 amp units.

    For instance:
    1N5404, 400 volt, 3 amp, has a surge current rating of 200 amps.
    http://www.diodes.com/datasheets/ds28007.pdf
    6A4-T, 400 volt, 6 amp, has a surge current rating of 400 amps.
    http://www.diodes.com/datasheets/ds28008.pdf
     
  3. Mike

    Mike Guest

    Okay, that helps a bit, thanks. How do I calculate the peak forward
    current? Do I calculate the resistance of the coil (plus the ESR of the
    capacitor) and then just apply Ohm's law (using the maximum voltage of the
    capacitor)?

    Thanks again.
     
  4. You would do better simulating the circuit with spice. The inductance
    of the coil should be a big part of the current pulse shaping, with
    the wire resistance and the capacitor internal inductance and
    resistance minor players. If not, there will be a lot of energy
    dumped into places you do not want it to go. Of course, you will also
    have to get hold of the formulas that allow you to predict the
    inductance of various coil forms. E.G.
    http://home.earthlink.net/~jimlux/hv/wheeler.htm

    You can download a free spice simulator from Linear Technology:
    http://www.linear.com/designtools/softwareRegistration.jsp
     
  5. rayjking

    rayjking Guest

    Mike,

    You can determine which diode to use by multiplying the cap size times the
    voltage stored, this gives you Jules. Jules = C*V = I * T so if 100 volts *
    200uf = 0.02jules ( V*C ) then a diode with a rating of 30 amps surge for
    60hz ( really only 8.33ms sees current ) ( I * T ) = 0.25 Jules is more than
    enough. You should not go beyond 25% of the surge rating for continuous
    applications. The surge ratings are for fuse considerations and these surges
    are for less than 10 times for the life of the diode The one amp diode with
    a 30 amp surge rating should not be used beyond 7 amps for 8.33ms or 0.058
    Jules.

    Ray
     
  6. Neil Preston

    Neil Preston Guest

    Before answering the question, I would like to know how he intends to
    connect the diode.....

    My first impression was that he was told to put a quench diode in parallel
    with the coil to clamp any oscillatory transients. In this case, I believe
    the diode need not be very heavily rated.

    Most of the responses I've seen, though, appear to assume that the diode
    will be in series with the coil, in which case the coil current will rise
    slowly according to the time constant formula to a maximum of approximately
    Ec/Rcoil (depending on the relative values of C and L.)

    Shouldn't be much more complicated than that, should it?

    Neil
     

  7. Errr... Better double check that energy storage formula. The energy stored
    in a capacitor is E=0.5*C*(V^2), where E is energy is Joules, C is
    capacitance in Farads, and V is the voltage on the capacitor in Volts.

    So for a 200uF capacitor fully charged to 100V, it should store one Joule of
    energy.

    I apologize for disagreeing with you, but one must nip misinformation in the
    bud before it spreads and gets out of hand.


    I'm not sure I agree with these ideas either. Limiting yourself to a peak
    current of 7A (for 8.33ms) on a typical 1A rated diode such as the 1N400X
    series devices is probably more conservative than really necessary in many
    cases (emphasis on the words "many cases", not "all cases"). Do you have
    some references that you can point us to that support your ideas and
    methods?
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-