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I am trying to make a kicker solenoid. My circuit is:

  • 12 V battery connected to DC step-up booster 300 V

  • diode A10A. The diode is rated for 10 A and 1000 V.

  • thyristor BTW69-1200 with external circuit (battery 1.5 V - resistor 1/4 W, 39 Ω). The thyristor is rated for 50A and 1200 V.

  • capacitor 1000 μF - 450 V

  • solenoid coil (3.6 Ω)

enter image description here

Since the capacitor will send about 60 A I thought the diode would burn out when tested, but surprisingly, it did not.

How it did not burn out while the current is 60A and this diode is rated for 10 A?

Is the back EMF current low and does it not follow Ohm's law?

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    \$\begingroup\$ What is the DC resistance of the solenoid and what is the ESR of the 1000 uF capacitor. Both will determine what the peak current seen by the diode is. What is the inductance of the solenoid and how long is the thyristor activated for. These will also dictate the forward flowing current. You calculate 60 amps; I'm not convinced (yet). \$\endgroup\$
    – Andy aka
    Nov 27, 2022 at 21:04
  • \$\begingroup\$ @Andyaka The dc resistance of coil is 3.6 ohm, unfortunately the datasheet of capacitor is not existed so can not determine its ESR, but i stimulated the circuit here and it is around 60A circuitlab.com/editor/#?id=j7847thgtv42 \$\endgroup\$
    – Tito
    Nov 27, 2022 at 21:37
  • \$\begingroup\$ If you didn’t simulate the ESR and the inductance then it’s unlikely your 60 amp figure is reliable. \$\endgroup\$
    – Andy aka
    Nov 27, 2022 at 21:58
  • \$\begingroup\$ see my answer; I think you'll realize that the diode current will be about 1 amp maximum. \$\endgroup\$
    – Andy aka
    Nov 28, 2022 at 10:53

2 Answers 2

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Your diode is rated for 10 A average, but its peak surge current rating is for 600 A. That means that for a very brief surge of current, like the freewheeling diode sees here, it's fine with currents of as much as six hundred amps, just as long as they're not too long (where "too long" is defined by the datasheet).

The length of the pulse that the freewheeling diode sees is related to the diode's forward voltage and the inductance of the solenoid coil (which we can't know since you just gave a resistance). It's probably less energy than the 16.7 ms half-sine specified in the datasheet, however. If needed, you can artificially increase the diode's forward voltage by putting a resistor in series with it, or an anti-series zener; both are common.

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  • \$\begingroup\$ the time of pulse is about 12 ms, thanks hearth i got that \$\endgroup\$
    – Tito
    Nov 27, 2022 at 21:39
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How it did not burnt out while the current is 60A and this diode rated for 10A

Once activated, the thyristor/SCR will only deactivate when the holding current reduces to about an amp hence, when it commutates off, the peak current seen by the reverse diode will be about 1 amp. This current then flows into the diode so, it won't even see a peak current of 10 amps let alone 60 amps.

enter image description here

This is how thyristors/SCRs work even though it might be tempting to believe it switches to an open circuit when the switch opens. It doesn't!

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    \$\begingroup\$ Oh dammit, you're right! I completely missed that they were using a thyristor. \$\endgroup\$
    – Hearth
    Nov 27, 2022 at 23:58

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