I'm trying to implement high side switch using P-MOSFET with NPN driver. I will drive the whole switch with AVR GPIO running at 3.3V (3.3V GPIO label). KT231B9 (it's an English translation from Russian КТ231Б9) NPN have hFE of 110 or greater so feeding 0.1 mA of current should be 10x times enough to turn of Q1 fully thus effectively turning Q2 to saturation too.

I've build the prototype on the breadboard and found that if I'm feeding 3.3V to the R1 then I have full 12V on the drain of Q2 and if I attach a real load (motorized ball valve) instead of multimeter it works fine. However when no input is given to the R1 (hanging wire) then I can read about 5.6V on the drain of Q2 with the multimeter. It have puzzled me and I've attached a 2.2k resistor instead of the multimeter to check whether current flows through it (other side of resistor was connected to the GND). Interesting enough, voltage drop on the 2.2k was 0V so it seems like no current were flowing through it and the voltage measured with multimeter was some kind of sporadic.

I wonder why this is happening and if I should worry about that. Thank you.


  • \$\begingroup\$ Multiply the leakage current (from datasheet) by the DMM's input impedance. If you're measuring less than that voltage, the FET is OK. \$\endgroup\$
    – user16324
    May 10, 2022 at 10:50

1 Answer 1


When Q2 has VGS==0 ('off'), it wil have some small (very temperature-dependent) leakage current. Your multimeter probably has a 10 Mohm input impedance which is quite high, and so just 0.56 uA leakage will allow it to read 5.6 V. for the same reason, using s 2.2 k resistor would read < 1 mV.

There is no problem.

  • \$\begingroup\$ Consider changing the gate source resistor to something n the 2K range to give you a decent turn off time and minimize heating. \$\endgroup\$
    – Gil
    May 10, 2022 at 0:53
  • \$\begingroup\$ There are some complications if your load is inductive and you turn off the FET (switch) quickly. The LOAD pin will tend to fly negative and could over stress the FET -- you can mitigate that by putting a diode (capable of handling the load current) between LOAD and GND. \$\endgroup\$
    – jp314
    May 10, 2022 at 2:41
  • \$\begingroup\$ I've forgot to draw the diode, I already use one of Schottky type to prevent back EMF \$\endgroup\$ May 10, 2022 at 7:42
  • \$\begingroup\$ @Gil I've always thought that it matters for high frequency switching like PWMing external load. will my application (simple switching on/off once in a minute or so) benefit from improved switching off time? \$\endgroup\$ May 10, 2022 at 7:43
  • 1
    \$\begingroup\$ @DrobotViktor Your application likely will not care about slow turn-off unless the turn-off is extremely slow. A 10k resistor is fine here, and may even mean (slightly) higher efficiency since it dissipates less power when Q1 is on.. \$\endgroup\$
    – Hearth
    May 11, 2022 at 14:46

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