1
\$\begingroup\$

Can one someone please guide me with the current flow direction and amount of current (Through Gate) (how shall I calculate it) for a P Channel Mosfet? For example if I connect the Gate to something that can sink a very low amount of current, then what happens?

\$\endgroup\$
  • 2
    \$\begingroup\$ Then the FET will turn on very slowly, which may or may not be a problem. A FET does not have a permanent gate current (except for leakage), it is voltage controlled, but it happens to have a parasitic capacitor so it does take current to reach a given gate voltage. \$\endgroup\$ – Wesley Lee Apr 9 '18 at 21:02
  • \$\begingroup\$ So, if I need a very quick turn on, do I need to sink more current? Can you please elaborate @WesleyLee? \$\endgroup\$ – Rakesh Mehta Apr 9 '18 at 21:04
  • \$\begingroup\$ You need to sink a lot of current for a brief time, essentially you are driving a capacitor. Its a bit hard to elaborate without knowing which FET you want to turn on at which speeds and voltages with which devices. If only turn on time is important you can drive the gate with an NFET or with a transistor or with a dedicated PFET driver. If turn off time is also important then you have to sink and source current. \$\endgroup\$ – Wesley Lee Apr 9 '18 at 21:07
  • \$\begingroup\$ See this answer.... \$\endgroup\$ – Trevor_G Apr 9 '18 at 21:08
  • \$\begingroup\$ Thank you @Trevor_G for the link. My head is spinning though :) \$\endgroup\$ – Rakesh Mehta Apr 9 '18 at 21:18
1
\$\begingroup\$

The gate of a MOSFET looks like a capacitor to the driving circuit. It therefore takes current to change the gate voltage, but no current to hold it at a particular voltage.

If you connect the gate of a P channel MOSFET to something that can sink only a small amount of current, the gate voltage will go low slowly. Whether that's a problem or not depends on the rest of your circuit.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ So for a quick charge and discharge I have to source and sink more current right? Actually I raised this question while I was trying to build a watchdog circuit to disconnect and reconnect the power supply to reset the MCU. The watchdog IC can source really a little bit of current like 20mA. So I believe I need to put a transistor to quickly turn on / off the FET. \$\endgroup\$ – Rakesh Mehta Apr 9 '18 at 21:21
  • \$\begingroup\$ @Rak: It doesn't sound like you need speed. There is probably a significant delay before the watchdog decides the processor is wedged anyway. Do a few 10s of us additional time really matter? \$\endgroup\$ – Olin Lathrop Apr 9 '18 at 22:07
  • \$\begingroup\$ Hi @Olin, let me put a schematic, you will be able to help better.. I am doing the schematic now. \$\endgroup\$ – Rakesh Mehta Apr 10 '18 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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