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Dissipating energy stored in an inductor

Discussion in 'General Electronics Discussion' started by (*steve*), Aug 12, 2018.

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  1. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I'm toying with designing a device for measuring inductor saturation currents. There's plenty of these on the web, but I have a need to go up to 30A and (at lower currents) up to 1H or thereabouts.

    At moderate currents and inductances, there's not a lot of problems dissipating the energy stored in the inductor, but as currents and inductance rises you have to concern yourself with peak currents for resistors and/or the total energy dissipated.

    As an example, at 30A and with an inductance of 44mH, I hit the limit for a 50W 1Ω resistor of the type I have chosen (20J).

    A simplified circuit is shown below:

    upload_2018-8-12_17-55-52.png

    The idea here is that at a current of around 7A the transistor begins to shunt some current around the resistors.

    I've chosen a TIP35 which has a peak collector current of 50A (for 5ms) and a continuous collector current of 25A.

    I have 2 concerns.
    1. Firstly, the turn on time is 1uS which will cause a high initial peak base current. Would a small capacitor across R1 help? Would it be necessary.
    2. Clearly secondary breakdown is an issue, which would possibly limit the current to 20A, although I could parallel 2 transistors with small emitter resistors (say 10mΩ) to increase this to 30A.
    I can handle the total power dissipation problem by slowing the pulse rate for large inductors, but I don't want to risk an open circuit.

    Another alternative is to trigger a beefy SCR from the resistors...
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I don't think so. Any capacitance across R1 will need to be charged and thus delay the onset of the required Vbe for the transistor to turn on. The capacitor should sit across R2 to be effective at all.
    I like that idea. You'd use the inductor's resistance to dissipate the power and this is likely to be designed to match the currents you will see.
    But why not simply use power diodes?
     
    hevans1944 likes this.
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The main problem is that the lower the total resistance, the longer it will take for the current in the inductor to fall to zero. I guess that simply monitoring the inductor current will tell me when it reaches zero...

    The secondary consideration is that i will certainly be testing inductors near, and very probably exceeding their rated current. Causing them to dissipate less heat is desirable.

    Maybe the answer is as simple as a number of these diodes in series?
     
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