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honestly I'm drawing a blank. I am trying to turn on and off a load with a P-CH MOSFET. I used generic resistor values just to be clear.

enter image description here

Now, in this example the VGS will clearly never go to 0. It just can't, unless I try to drive the gate with the 24V source.

Is there any way to turn off this PFET without using a second transistor? Specifically, I was using an N-CH MOSFET to "pull up" to 24V like in this second example.

enter image description here

Is there any advantage to doing it this way? Am I forgetting something about P-CH FETS? Is there a better way to do this? More specifically, is there an easier way to do a load switch that is connected to ground, and the switch connects power?

Thanks :) I appreciate any advice, app note links, etc.

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    \$\begingroup\$ You answered your own question : " VGS will clearly never go to 0. It just can't, unless I try to drive the gate with the 24V source.". So that's what you do. OFF = 24V; ON = somewhere around 12-14V is good. \$\endgroup\$ – user16324 Nov 18 '20 at 15:19
  • \$\begingroup\$ Look at the datasheet for your chosen PFET. That will tell you the maximum -Vg that can be applied safely. Zeners can work for this. Oh Andy beat me to it. :) \$\endgroup\$ – rdtsc Nov 18 '20 at 15:29
  • \$\begingroup\$ @rdtsc thanks for pointing it out anyways :). I normally would just do a resistive voltage divider on the N-CH fet drain side, but was just trying to throw together some stuff for screenshots to support my question haha. \$\endgroup\$ – Punchki Nov 18 '20 at 15:45
  • \$\begingroup\$ I would be cautious about simple resistor dividers and stick to something more robust. You want to ensure the gate voltage can never exceed the +-Vg rating, else the PFET goes poof. \$\endgroup\$ – rdtsc Nov 18 '20 at 16:51
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Even your 2nd circuit is going to suffer problems because you will apply the full 24 volts between gate and source of the PFET when turning it on and that is highly likely to exceed the maximum rating. Most MOSFETs have a gate source maximum voltage of \$\pm\$20 volts and 24 volts is too much. An added resistor and a zener can fix this of course: -

enter image description here

Is there any advantage to doing it this way?

Yes, it's a good way to do it but....

Am I forgetting something about P-CH FETS?

Yes, you are forgetting the maximum VGS voltage

Is there a better way to do this?

The way shown above is better because it prevents the PFET being destroyed by over voltage between source and gate.

More specifically, is there an easier way to do a load switch

If by easy you mean "once done, you can sit back and relax and not worry about things" then use the circuit above. Is this easier to build? No.

Is there any way to turn off this PFET without using a second transistor?

You can use a level shifter made from a zener diode and a couple of resistors but it's a bit fiddly and will only be effective if the supply rails are maintained at the same value. A 2nd transistor is a more general solution.

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  • \$\begingroup\$ Thanks for the answer @Andy aka I should have added the resistor to have a voltage divider on the gate of the pfet. Good catch. That being said, could you explain a little more what that 12V zener is doing in your above drawing? \$\endgroup\$ – Punchki Nov 18 '20 at 15:19
  • \$\begingroup\$ It prevents the gate source voltage exceeding 12 volts. Of course, the zener can be 15 volts (or even 18 volts) but most MOSFETs are happy with the gate source voltage being limited to 12 volts and will produce lowest ON resistance at this value. \$\endgroup\$ – Andy aka Nov 18 '20 at 15:25

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