I try to develop an application where I need to switch 55V at 20A in both directions using MOSFETs.

Because of the high current, I decided to use N-MOSFETs (because they are available with lower rdson) back to back and using a photvoltaic MOSFET driver to switch the gates.

My problem now is that the load I try to switch has a lot of capacitance (about 20 mF) which kills my MOSFETs every time because of the high power dissipation during the switching transition.

I already tried to to use an RC Lowpass to drive the gates more slowly through the threshold but the power dissipation is still too high, peaking at 200W (see simulation)

enter image description here

Is there any way I can limit the inrush current using this back to back topology?

  • 1
    \$\begingroup\$ Should we guess at the node names and where they are on the circuit? \$\endgroup\$
    – Andy aka
    Sep 22 '20 at 16:46
  • \$\begingroup\$ you should be switching them faster, not slower. \$\endgroup\$
    – dandavis
    Sep 22 '20 at 21:06
  • \$\begingroup\$ Back to Back, to my knowlege only has one advantage: if closed no Current flows from eigher side on the other. It does not hel with switching. Why a photvoltaic MOSFET driver and not one with a charge pump? Have you thought about a fully integrated High Side Switch? \$\endgroup\$
    – schnedan
    Sep 22 '20 at 21:23
  • \$\begingroup\$ @dandavis switching them faster just leads to a shorter, but even higher power dissapation (1.2 kW according to simulation) which the die of the FET can't handler either. \$\endgroup\$
    – pockined
    Sep 23 '20 at 19:30
  • \$\begingroup\$ @pockined the faster the switching, the less dissipation. If a fet can't take Ids, it can't take Ids... \$\endgroup\$
    – dandavis
    Sep 23 '20 at 19:32

@20 A, I'd have been looking at relays already, but you likely have a reason for Mosfets.

If opening an alternate path to alleviate the load on the mosfets (routing it through a high wattage resistor) is an option, you could try that. Essentially you:

  1. Allow a path through a high wattage resistor (this reduces the current that your mosfets would have to halt)
  2. Halt flow through your mosfets
  3. Halt flow through high wattage resistor

General idea is that by splitting it up over 2) and 3) you need to halt smaller currents at each step. The resistor gives another outlet for the 20mF to discharge while the main path was shut.

That all said, something is bugging me about your mosfet use. I feel like using all N mosfets (or even all P) and "both directions" is setting off a big red flag somewhere, the kind that tells me not to power a circuit up when those words are together.

  • \$\begingroup\$ the reason to use MOSFETs was because of the high costs of DC relays which can handles this much current (the MOSFET solution only costs ~ 5$). I had the idea with the resistor already but since I would have to use an additional back-to-back config just for this inrush switch, I would basically double the number of FETs which I would like to avoid. Using this configuration in general should be fine and is basically just a high side load switch. \$\endgroup\$
    – pockined
    Sep 23 '20 at 19:36

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