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I'm looking to use a MOSFET to turn on a resistive path, precharging a capacitive load. I'm considering the Nexperia BUK9608-55B, but I'm having trouble determining what my operating point is on the SOA plot.

I'll be using this as a high side switch with a 30VDC supply, ~7mOhm R.DSon, 1 Ohm power resistor down stream, then the capacitive load (an inverter with around 14mF of capacitance). With those conditions, my inrush will be around 30A. So where am I on the SOA plot? Am I at A, because we're switching 30A, or am I around B because most of the voltage drop is over the power resistor?

If we're operating around point A, would it make sense to use a load switch IC instead of a discrete MOSFET, and use it's dV/dt limiting capabilities?

SOA plot

It seems like it really depends on turn on speed and how it is driven. I've been planning to use this NMOS with this PV driver (https://docs.broadcom.com/doc/AV02-0259EN). We'll be up at 7V so that should be enough to get sub 10mOhms.

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  • \$\begingroup\$ It merely depends on how fast your MOSFET will reach low Rdson. \$\endgroup\$ Aug 3, 2021 at 11:07
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    \$\begingroup\$ Show a schematics with a gate driving detail, it could be simulated on LTSpice. \$\endgroup\$ Aug 3, 2021 at 11:10

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You're at point B. The FET's 7mOhm Rds_on forms a voltage divider together with the 1 Ohm resistor, which means that almost no voltage will be dropped across the FET.

However, during the switching process, the FET will be at point A for a very short period of time while its on-resistance gradually drops to its final low value. Therefore you have to make sure to switch the FET within 1ms, which is the time that the FET can operate for under these conditions (as you can see in the SOA diagram). 1ms is quite a long time, however. Any semi-decent gate driver can switch that FET in microseconds, not milliseconds.

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  • \$\begingroup\$ Just the answer I was hoping for. I'll be using a 7V PV gate driver, so we should get to the low RDSon, and hopefully rather quickly. \$\endgroup\$
    – matth
    Aug 3, 2021 at 11:17
  • \$\begingroup\$ Photovoltaic gate drivers are pretty much the slowest gate drivers you can get! Be careful, it might not be able to meet that 1ms switching time requirement. Which gate driver do you plan on using exactly? \$\endgroup\$ Aug 3, 2021 at 11:31
  • \$\begingroup\$ I was looking at this one docs.broadcom.com/doc/AV02-0259EN because it is used on another one of our PCBs. And I'll plan to drive it pretty hard, at 20mA, so maybe that'll help it as well. \$\endgroup\$
    – matth
    Aug 3, 2021 at 11:36
  • \$\begingroup\$ @matth That is dead slow gate driver, most likely your MOSFET will burn in the first attempt. \$\endgroup\$ Aug 3, 2021 at 11:39
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    \$\begingroup\$ Some numbers for the AV02-0259EN: Its short-circuit(!) current is 15µA minimum. This means that the FET will need at least 3ms to switch on (~40nC gate charge), most likely even more given that the gate driver's output current will drop off as the gate voltage rises. This is way outside of the FET's SOA and may result in FET death. \$\endgroup\$ Aug 3, 2021 at 11:45
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When you turn on your pre-charge FET, and the cap bank is initially empty, it is essentially a short circuit. If you have 30A flowing through a 1 ohm resistor and your FET, the FET will have 30A running through it at 0.007ohm*30A or 0.21V. This falls well within your SOA.

Just make sure you are driving it with a high enough Vgs to achieve 7mOhm on resistance, and fast enough to avoid any issues with turn on. You will also want some high side driver that allows for 100% duty cycle or the FET will turn back off after the caps have charged to ~Vgs.

I'd be more worried about the 900W you're putting through that 1 ohm resistor.

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  • \$\begingroup\$ Yes, it's some huge power in the resistor, but I at least feel comfortable I know how to deal with that. I'll use PV gate driver which gives us 7V, so we should be well within the low RDSon region. \$\endgroup\$
    – matth
    Aug 3, 2021 at 11:16

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