I have this dual pmosfet that we were going to use one element of to implement a low-voltage cutoff for a battery input (battery comes in from the left), yanking on the base of the darlington to cut the current off.

For various reasons we've decided to implement this differently, still need the other transistor, and don't want to spin a new board design yet. So we're going to DNP the resistors and darlington.

What I'm wondering is if there is any problem I should be aware of in connecting the mosfet's gate directly to ground to keep keep it always turned on?

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  • \$\begingroup\$ No problem at all in shorting the Gate to ground. \$\endgroup\$ Dec 9, 2016 at 20:30
  • 1
    \$\begingroup\$ What maximum voltage is the battery (including when being charged)? What is the MOSFET - a link would help. \$\endgroup\$
    – Andy aka
    Dec 9, 2016 at 20:31
  • \$\begingroup\$ As @Andyaka points out, the charging voltage could be an issue. If it is in an automotive application it would not survive a standard load dump test - the 20V gate rating wouldn't be nearly enough. \$\endgroup\$
    – Matthew
    Dec 9, 2016 at 20:50
  • \$\begingroup\$ Thanks for the input guys. @Andyaka The max charge voltage is 16V. Datasheet: vishay.com/docs/62859/sia931dj.pdf \$\endgroup\$
    – joshperry
    Dec 9, 2016 at 20:53
  • \$\begingroup\$ If the FET was designed to work when it was being turned on an off, then shorting the Gate to ground has to be within design specs (allowing slight rise for the missing VCE(sat) of about 0.8 -1 V). However if this is an automotive application, it should be designed to tolerate noise pulses up to at least 21 V. That would exceed the VGS(max) of this device. \$\endgroup\$ Dec 9, 2016 at 21:19

1 Answer 1


Should be fine. The only consideration would be Vgs(max). The Darlington array would always drop some voltage across it, whatever its Vce(sat) is, which would lower the voltage across the gate and the source. If you pull the gate directly to ground, Vgs will be larger.

Of course, with a battery, you're unlikely to be anywhere near Vgs(max), but it's worth considering.

  • \$\begingroup\$ Actually -- Vgs(Max) is a mere 8V on most logic level MOSFETs -- you can blow out their gate with an ordinary 9V battery! \$\endgroup\$ Dec 10, 2016 at 5:20

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