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I would like to add reverse polarity protection to my robot. Supply is 320 VDC, 20 A max. The load is mainly inductive (DC brushless motors, some DC/DC converters).

A simple solution would be using a high-voltage power diode in series. The loss in efficiency wouldn't be much of an issue at such high voltages, but the thermal management will still make integration far more difficult.

For low voltages a P-MOS would provide the needed reverse polarity protection, but with a very small voltage drop (so low power dissipation). But as there are no PMOS which allow a 320 V gate voltage (as far as I know), this won't work.

I found ideal diode drivers (mainly designed for power ORing) that drive those MOSFETs even for higher voltages, but so far the highest I found is the LTC4357, but it is only specified for 80 V (100 V abs. max.).

What is the best way for reverse polarity protection for rather high voltages (320 VDC) and rather high currents (20 A)?

Note that both blocking reverse current or rectifying the reversed current are acceptable (I would prefer blocking), and that solutions involving blowing a fuse or other easily replaceable components are acceptable (reverse polarity is unlikely, but damage is very high if the robot gets destroyed). Cost is no big issue.

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  • \$\begingroup\$ What is the budget? A >400V, <50mΩ MOSFET is not the cheapest, plus the drive circuitry to run it (as you say, PMOS are not readily available in this range). How much space is available? Maybe the simplicity and cost of the diode is worth the dissipation after all. \$\endgroup\$
    – Tim Williams
    Jan 31 at 9:55
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    \$\begingroup\$ Maybe a simple voltage divider can be used to generate a gate voltage that is only +/- 30V instead of +/- 320V \$\endgroup\$
    – user253751
    Jan 31 at 9:57
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    \$\begingroup\$ Why a PMOS instead of an N-MOSFET? Can you switch the negative rail? \$\endgroup\$
    – greybeard
    Jan 31 at 10:03
  • \$\begingroup\$ @TimWilliams for budget, 100$ should still be OK (so I don't expect it to be the limiting factor). For space, as it is for a new iteration, there is again no big constraint, the smaller the better. Heat dissipation IS currently an issue in some conditions, so dissipate 10-20W just for reverse current protection is probably not an option. \$\endgroup\$
    – Sandro
    Jan 31 at 10:09
  • \$\begingroup\$ @user253751 : it's worse thinking about it. First intuition is hat it should work, excepted maybe some issues that the resistors prevent the gate capacitance to to charge/discharge fat enough. \$\endgroup\$
    – Sandro
    Jan 31 at 10:12

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Couple of options:

1- Diode and fuse: This is the simplest one. If the power is applied in reverse polarity then the current will flow through the diode and the fuse. The fuse will blow, and the diode will quite possibly be destroyed as well. The fuse should be selected according to your load needs as the whole current will flow through it during normal operation.

schematic

simulate this circuit – Schematic created using CircuitLab

2- Series MOSFET: This can be implemented with either PMOS or NMOS. Low RDS-on (for low losses) and high (at least 400V but >500V for a safe margin) VBRDSS are needed. NMOSs are quite common and cheap, but the load won't be grounded (if grounding is needed). So if you want the load side to be always grounded then use PMOS but the problem is, PMOSs with the required specs are expensive compared to NMOSs, and are not common (i.e. might be difficult to find one that suits).

schematic

simulate this circuit

3- Mechanical: This involves extra circuitry to detect polarity, and a routing device such as a relay or a contactor to power the rest of the circuit. So it brings extra complexity, but is safe to implement:

schematic

simulate this circuit

Series diode is okay for the low side as the required power will be lower.

NOTE: I used a relay for simplicity but can be anything that suits your needs.

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  • \$\begingroup\$ I find it odd that I don't find applications using an N-channel MOSFET in the positive rail, high voltage or not. Can it be that hard to drive the gate? \$\endgroup\$
    – greybeard
    Jan 31 at 10:26
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    \$\begingroup\$ @greybeard Because you need a gate voltage higher than the supply voltage to turn it on, so you have to generate it somehow - usually with a charge pump of some sort. \$\endgroup\$
    – Finbarr
    Jan 31 at 11:00
  • \$\begingroup\$ @Rohat Kılıç : for option 2, how do you deal with the fact that the gates only support about 30V? Just using a voltage divider as user253751 suggested in comments, or something else? The power supply is isolated, to cutting either voltage rail should be fine \$\endgroup\$
    – Sandro
    Jan 31 at 11:31
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    \$\begingroup\$ @Sandro see the edit. I've shown zener but can be a divider as well. The problem with divider is that you can put as bigger as you want to decrease the dissipation but the turn-on will get longer. \$\endgroup\$
    – Rohat Kılıç
    Jan 31 at 11:34
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    \$\begingroup\$ Indeed, the DC-DC conv. of the relay version could similarly power an isolated gate drive. Which I suppose doesn't need to be very much, an oscillator into a transformer, rectifier and load resistor, and the gate itself can provide the filter capacitance. It wouldn't switch quickly, but might do well enough for purposes. The downside of putting a semiconductor here is handling inrush and fault currents. \$\endgroup\$
    – Tim Williams
    Jan 31 at 14:56

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