I use a Schottky diode as reverse polarity protection on earlier versions but have decided to change to something more appropriate and energy efficient.

The voltage input on the left is a 12-14.8 V vehicle battery and current will not be more than about 500 mA @ 12 V.

I want to use an AO3401 N-MOSFET as it is a basic part with JLCPCB for manufacture and it's small. However, I'm not sure it will work as Vgs is about 12 V but Vds is 30 V.

Does that mean it is not suitable for this use? I also know I need a Zener diode between the source and gate of the FET.


  • \$\begingroup\$ why do you think that Vgs_max being 12V and Vds_max being 30V would be a problem ? \$\endgroup\$
    – Rahmany
    Sep 14, 2022 at 7:44

1 Answer 1


If the vehicle battery is discharged so that the MOSFET is partially on then the MOSFET will overheat and likely be destroyed due to the diminutive package of the AO3401.

I don't see any significant disadvantage in using a Schottky diode if that is your actual circuit- the drop out will be increased by a few hundred mV but it won't save any energy at all, just move some mW from the linear regulator to the diode.

As you know, vehicle electrical systems can have large and destructive transients (load dump and so the MOSFET ratings may also be insufficient depending on what protection you've added. As you say, the Zener is definitely required, but 30V Vds rating does not leave much margin for the suppression circuitry to work with.

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  • \$\begingroup\$ Ill bin my plan and just use the schotkey in that case. I dont want to use a physically larger component really as space on the board is minimal. \$\endgroup\$
    – robbrown92
    Sep 14, 2022 at 9:25

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