1
\$\begingroup\$

If a P-channel MOSFET have absolute maximum values:

  • VDSS = -60 V
  • Vgs = +/-20 V

"refering to high switch side":

Does this means we cannot switch the MOSFET by the gate driver that ties the gate to ground if Vd > 20 V?

this means i cant use this gate driver on most of P-mos (most of them +/-20Vgs) if we apply VG(max) on Drain: enter image description here http://ww1.microchip.com/downloads/en/DeviceDoc/21424d.pdf

\$\endgroup\$
0

2 Answers 2

1
\$\begingroup\$

Yes, you will exceed the absolute maximum gate voltage rating if you do that. The MOSFET may not die until 30V or 50V- on the other hand you should stay well away from the absolute maximum ratings for reliability and to consider any transients that may occur on the supply line.

You may be able to add a resistor and a zener diode to limit the gate voltage, for example you might use a 10V Zener. That will tend to slow the switching, which may or may not be an issue depending on what you are doing.

For example:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here


Edit: Okay, you've added some stuff about 100kHz distinguishing this from a simple static switching situation.

You can hang a driver off the 30V supply, with a regulated (negative regulator) voltage if you like. Here is a very simple driver that does level shifting. You can keep the Zener diode in there on the gate to prevent any possibility of exceeding the maximum.

Or just use a negative regulator and a gate driver chip with a level shifter (or fast logic-output optoisolator) to the input. One advantage most gate drivers offer (along with fast, high current drive) is that they include some type of undervoltage lockout, which could save your MOSFET in brownout conditions.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ hello, i was looking for Gate driver to have fast switching. a Zener is used widely but in my opinion is not the right solution for fast switching i think \$\endgroup\$ Commented Oct 16, 2018 at 13:18
  • \$\begingroup\$ See edit regarding additional information. \$\endgroup\$ Commented Oct 16, 2018 at 14:22
  • \$\begingroup\$ does it matter if i used positive regulator from 30v to 15v , and used that 15v rail as ground for the gate driver and the optoisolator ? i mean whats benifit of using negative regulator? \$\endgroup\$ Commented Oct 16, 2018 at 14:52
  • 1
    \$\begingroup\$ The negative regulator will control what you want (Vgs) and will supply sink current from the output (most positive regulators will only source current). You can use a shunt regulator (eg. LM431) hung off the +30V if you like. \$\endgroup\$ Commented Oct 16, 2018 at 16:33
1
\$\begingroup\$

If you go beyond +/-20 between the gate and the source, the MOSFET will die, so to answer your question, no you cannot.

This is why there are isolated gate driver to drive mosfet on high side rails.

Check here for some driver, and you can filter by "High Side Voltage - Max (Bootstrap)"

\$\endgroup\$
3
  • \$\begingroup\$ when you filter by the high side voltage in digikey, all gate types are N-channel which doesnt make sense? \$\endgroup\$ Commented Oct 16, 2018 at 11:58
  • \$\begingroup\$ @Hasanalattar N-channel FETs have better characteristics in general than P-channel ones, and driving them on the high side requires special circuitry, so high-side N-channel drivers are pretty common. \$\endgroup\$
    – Hearth
    Commented Oct 16, 2018 at 13:07
  • \$\begingroup\$ @Felthry so how would some one drive P-channel Mosfets "correct way" when VSupply = 30V and the Vgs = +/- 20V. when most of the gate drivers i know are pulling down to ground. something really doesnt make sense here. \$\endgroup\$ Commented Oct 16, 2018 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.