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inductor switch protection diode

Discussion in 'Electronic Basics' started by CFoley1064, Aug 4, 2004.

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  1. CFoley1064

    CFoley1064 Guest

    Subject: inductor switch protection diode
    A good rule of thumb is to make sure the current rating of the diode exceeds
    the current going through the inductor when it's on. If you can, choose a
    diode with a Peak Reverse Voltage (PRV) rating twice the voltage across the
    inductive load when it's on. I would guess you're using 12 or 10AWG wire (30
    amp) to the load. A similar size gauge going to the diode would definitely be
    helpful, if you can.

    I'd be a little more concerned about the turn-off for the FET. It looks like
    you've got a driver that will turn on the FET quickly, but turn it off slowly
    with the 10K pulldown. You might want to modify that.

    Good luck
  2. The diode across the inductor will see a peak current almost equal to
    the current at the moment the mosfet turns off. However, the current
    falls with something like an L/R time constant, with L being the
    inductance of the coil and R being its resistance. So the diode does
    not need to be rated for the continuous current in the coil, though it
    should have a surge current rating well above the peak current it will
    pass. A 3 amp rectifier should be fine.

    Note that adding this diode will also lengthen the time that current
    passes through the coil by the same L/R time constant. Is this a
  3. I think so.
  4. andy

    andy Guest

    I'm using this circuit to control a short (1s) pulse of 14A through an
    electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
    should give about 13.5-14A when switched on. I'm putting in a protection
    diode to stop the inductor destroying the mosfet when it's switched off.
    The question is, how do i work out the current rating for the diode and
    the wiring to it? I'm using heavy 30A wire for the wires between the
    battery, magnet, and mosfet, wired through an electrician's terminal
    block, but i'm not sure if i need to do the same with the diode.

    12V ---------o------------o-------
    | |
    |< |
    On --| ----|
    |\ - C|
    | ^ C| Electromagnet
    ----o | C|
    | | -----
    .-. | |
    10k | | | ||-+
    | | | ||<- BUZ10
    '-' o-----||-+
    | |
    created by Andy´s ASCII-Circuit v1.24.140803 Beta
  5. andy

    andy Guest

    P.S. the way i'm thinking of constructing this is to put the BUZ10 in a
    piece of stripboard, with all the terminals soldered just to hold it in
    place, but leave the ends of the terminals sticking out so i can solder
    the high current wire directly to them. I.e. the gate is using the
    tracks on the board, but the source and drain are soldered directly to
    the cable. Is this a good way to do it?
  6. andy

    andy Guest

    no. The time just needs to be just long enough for the coil to reach its
    peak (resistance determined) current, so as not to waste power keeping it
    alive longer than needed.
  7. andy

    andy Guest

    why will it be a slow turn off?

    Would decreasing the resistor to 1k or 200ohm be enough? The transistor is
    being driven from the inverting output of a CMOS monostable.
  8. The switch won't be slow to turn off. The current through the
    inductor will be slow to reach zero, since it will free wheel through
    the diode. The time constant of the current decay is roughly L/R of
    the inductor. If you put a resistor in series with the diode, you
    allow more peak reverse voltage to develop (the peak inductor current
    times the resistance) but the decay time constant then becomes
    L/(Rind+Rseries). And the resistor also has to handle the peak
    current. Other voltage drop elements can also be used, like several
    diodes in series, or a big zener in series with the diode. If a slow
    release of the magnetic field is no problem, then just use one diode.
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