Alright so I have a circuit designed to switch very high currents, it reliably switches 1500 A using these: https://au.mouser.com/ProductDetail/Vishay-Siliconix/SQJQ140E-T1_GE3?qs=TuK3vfAjtkXA5wP331tO2g%3D%3D

I have designed the circuit such that there are 6 MOSFETs driven by 3 totem poll drivers, so they are in pairs. The gate capacitance is 288 nC and they are switched by 1.5 A each, 3 A per driver. This results in very fast switching and low losses. My main question is because I've been keeping the current below 1800 A because the max current through the MOSFET at <300 us is 1820 A. Because of tolerances one MOSFET is bound to turn on before the others, meaning a current higher than the individual max could go through before the rest turn on. Even with a very worst case tolerance of 10% and the worst case combinations the maximum time for one to be on alone is roughly 40 ns.

I initially designed the circuit to work in pairs to attempt to make them work as one unit, allowing for currents over the max individual, but I am apprehensive to actually test this in case it doesn't work. Earlier today I accidentally shorted the output in a way that probably should have broken them, unless they are working on pairs as I hoped. This may have been luck but I am wondering if anyone with greater MOSFETs experience than me has made something like this and knows if my pairs idea works for currents greater than individual maximums. Thank you.

  • \$\begingroup\$ What are you switching? \$\endgroup\$
    – TQQQ
    Commented Oct 21, 2023 at 5:21
  • \$\begingroup\$ My understanding is that in a lot of cases 40nsec might not matter, however the a problem may appear if same mosfet is always first. In a similar case I am thinking about intentionally turning on one before the rest, every time another one. \$\endgroup\$
    – TQQQ
    Commented Oct 21, 2023 at 5:23
  • 1
    \$\begingroup\$ What is the actual rise time of that current? Even a tiny inductance of the wiring will prevent you from reaching this current in such a short time. A slope of 45 kA/µs is quite a lot. \$\endgroup\$
    – asdfex
    Commented Oct 21, 2023 at 9:35
  • \$\begingroup\$ I agree with @asdfex. The MOSFET is rated for 40V. At that low a Voltage, even 1nH of series inductance prevents the current from rising above 1600A within that 40ns period. \$\endgroup\$
    – Rainer P.
    Commented Oct 21, 2023 at 15:34

3 Answers 3


Can you exceed max mosfet current for a few nanoseconds?

Maybe. But if you violate the absolute maximum rating on the datasheet, and your MOSFET fails, who are you going to blame? If you made the circuit for a customer, who is the customer going to blame? If you made this circuit for your employer, who is your employer going to blame?

  • \$\begingroup\$ it's for personal use so blame myself ig? \$\endgroup\$ Commented Oct 21, 2023 at 3:10
  • \$\begingroup\$ Avoiding blame can only get you so far. \$\endgroup\$
    – Drew
    Commented Oct 21, 2023 at 3:15
  • \$\begingroup\$ I just wanna know if it's possible :'( \$\endgroup\$ Commented Oct 21, 2023 at 4:01

The current rating on a MOSFET datasheet is useless. It's an imaginary number and not reflective of real world use. Max current for a MOSFET is a rather meaningless term. You need to calculate your load, pulse waveform, duty cycle against the thermal resistance of your actual board implementation.

MOSFETs have pulse ratings which are ultimately limited by the imperfections in the material. As pulse durations get shorter, heat gets produced faster than it can spread out and equilibriate. That's why you can't keep decreasing the pulse length and get near infinite current for near zero time without blowing up the MOSFET. There is also the fact that the channel can saturate and cannot pass any more current.

enter image description here

  • \$\begingroup\$ I am not going to lie to you I have no idea what that graph means in relation to the question. Sorry. \$\endgroup\$ Commented Oct 21, 2023 at 5:22
  • \$\begingroup\$ @leasthollow First, learn how to calculate the temperature rise of the MOSFET under some continuous current. That requires power dissipated and the thermal impedance of the junction to ambient of your package on your board setup with or without a heatsink. Obviously the temperature rise + ambient temperature must remain below the max listed. \$\endgroup\$
    – DKNguyen
    Commented Oct 21, 2023 at 8:22
  • \$\begingroup\$ @leasthollow Once you learn how to do that then it is trivial to go one step farther and use the graphs above to figure out the heating for non-continuous currents. The graphs indicate that as the current pulse gets shorter, the thermal resistance is effectively less. That means you can get away with more current for the same temperature rise. \$\endgroup\$
    – DKNguyen
    Commented Oct 21, 2023 at 8:24

I suppose your best option is to estimate the mass of the die and calculate the temperature rise during that time.

  • \$\begingroup\$ bond wires might still act as fuses though \$\endgroup\$
    – Wesley Lee
    Commented Oct 21, 2023 at 4:20
  • \$\begingroup\$ do you mean the wires connected to the mosfet drain source ? nah they can handle it \$\endgroup\$ Commented Oct 21, 2023 at 4:43
  • \$\begingroup\$ @leasthollow Wesley means the wires inside the package, connecting the source of the MOSFET die itself to the lead that your wires are soldered to. \$\endgroup\$
    – Hearth
    Commented Oct 21, 2023 at 13:33
  • \$\begingroup\$ oh i see, maybe it's best to just stick too 1600 amps. \$\endgroup\$ Commented Oct 21, 2023 at 23:54

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