simulate this circuit – Schematic created using CircuitLab

I have a general question about thyristors. I know that the di/dt limitation of thyristors is due to high currents flowing through regions of the thyristor that have not yet fully reached a state of full on-state conduction/uniform current density. This can cause localized overheating.

However, I am wondering about this case: What if we preemptively turn-on the thyristor before any anode->cathode current flows. For example, consider that we turn-on the thyristor under zero voltage and zero load current. Then, after waiting a millisecond or so, we ramp up the current at a very high slope, such as at 20,000\$\frac{\text{A}}{\mu\text{s}}\$. In this case, will there be any such localized overheating/damage?

My gut says there shouldn't be, since other transistors can handle very high di/dt with no ill-effects. Please let me know your thoughts!

EDIT: For example, consider this circuit. First, we turn-on the SCR1 thyristor. Then we wait. Then, we turn on the MOSFET. Will this cause deleterious effects?

  • \$\begingroup\$ You cannot turn on thyristor at zero/zero, they have a minimum trip current (for gate) and turn on voltage/current. ST Micro has a series of application notes on applying Thyristors and triacs . \$\endgroup\$
    – Barleyman
    Feb 19, 2023 at 17:28
  • \$\begingroup\$ I am assuming that we preemptively applying a gate->cathode voltage, that forward-biases the PN junction. Then, we reduce the thyristor to the same structure as a PN diode, right? Which can handle high di/dt? I updated the question to show a circuit of what I mean. \$\endgroup\$ Feb 19, 2023 at 17:30
  • \$\begingroup\$ Turn-on at low sine voltage portion does of course limit the current. But then you're stuck at (near) 100% power factor. I'd recommend those STM appnotes for basics. \$\endgroup\$
    – Barleyman
    Feb 19, 2023 at 17:36
  • \$\begingroup\$ What about just the figure shown, with a DC source. You can assume that there is some inductance so the di/dt isn't infinite, but very high. I'm not concerned with turn-off for this example. Would the thyristor be damaged if a gate voltage is preemptively applied before the MOSFET is turned on? \$\endgroup\$ Feb 19, 2023 at 17:39
  • \$\begingroup\$ How are you delivering "simmer" current to the thyristor? The gate alone will not reproduce the charge distribution that MT current has. If it's through another transistor... what's the thyristor doing at all, just use that? \$\endgroup\$ Feb 19, 2023 at 19:52

1 Answer 1


I strongly suspect this is not allowed. The "base current" in a thyristor is primarily derived from the load current. The gate only turns a small region of the die 'on', and conduction spreads regeneratively out from that region. Without load current, there is no conduction to spread.

However there are many different constructions of thyristors and some are undoubtedly better than others.

  • \$\begingroup\$ Not allowed vs doesn't work.. With a decent zero crossing detection you can time the turn on at dozen volts or even less, though. \$\endgroup\$
    – Barleyman
    Feb 19, 2023 at 17:38
  • \$\begingroup\$ 'Not allowed' as in damage may occur to the thyristor due to excessive di/dt. Outside of absolute maximum ratings. \$\endgroup\$ Feb 19, 2023 at 17:41
  • \$\begingroup\$ Thanks a lot, @SpehroPefhany. Then, does the same limitation apply for diodes? It seems that once a gate voltage is applied, and allowed to propagate through the thyristor, the thyristor is functionally equivalent to a diode? \$\endgroup\$ Feb 19, 2023 at 17:43
  • \$\begingroup\$ No, diodes and BJTs are not regenerative. Darlington transistors probably would be similar to thyristors. \$\endgroup\$ Feb 19, 2023 at 17:44
  • \$\begingroup\$ @SpehroPefhany Thanks again for the answer. My understanding of a thyristor is that they are essentially two stacked diodes, P-N-P-N. With the second "P-N" forward-biased, does it not behave identical to a diode? Also, it was my understanding that diodes are also regenerative, since they have a reverse recovery? If it appears I am being argumentative, my apologies. I truly am just trying to fully understand. \$\endgroup\$ Feb 19, 2023 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.