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IGBT current limit

Discussion in 'General Electronics Discussion' started by NMNeil, Jan 19, 2017.

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  1. NMNeil


    Oct 3, 2014
    I still haven't mastered posting circuit diagrams but my circuit is simple and easy to describe.
    The purpose is to limit the current through a 1.58mH ignition coil to 4 amps. The coil is driven by a Fairchild IGBT specifically used for ignition coils so all the spike suppression is already on the IC and it's a 5 volt logic level drive.
    To limit the current I first tried limiting the gate voltage. The IGBT got very very hot, and yes, the magic smoke escaped. Then I thought of using a high power NPN in series, and a low or high side current sense resistor through an op-amp which sort of worked, but now the NPN was getting very hot.
    So I thought of a completely different approach.
    When a normally closed switch is opened it goes through a series of hex Schmitt triggers with a resistor/cap delay to give a low pulse of 100 microseconds duration at the output. This goes to the trigger pin of a 555 set to monostable mode making the output pin go high for 0.75 milliseconds (adjustable) which will, via the IGBT charge the coil to about 4 amps, not precise but good enough for me. After the 0.75 milliseconds the IGBT is turned off and the coil sparks.
    I haven't built the circuit yet but the circuit simulator said it will work. Have I overlooked anything, it just seems too simple a solution? I'm asking before I fry my last IGBT.
  2. Kabelsalat


    Jul 5, 2011
    Armed by a pen in my hand, I find it impossible (without much guessing) to draw any circuit by your description.
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    The way to limit the current through the coil is to limit the time it is turned on.

    Once it hits the max current you turn it off
    BobK likes this.
  4. Chemelec


    Jul 12, 2016
    Do you have a HEAT SINK on that IGBT?
  5. BobK


    Jan 5, 2010
    How much voltage are you driving the coil with? From that, and the equation below you can determine the on time necessary to give you 4A.

    dI /dt = V / L

    Say for instance, that the voltage is 12V.

    dI / dt = 12 / 0.00158 = 7594A / sec

    So you would want to leave it on for 4 / 7594 = 526uSec.

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