1
\$\begingroup\$

I need to switch a load with a 3.3 volts logic level (pic microcontroller) is this design is correct?

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ You need to reverse drain and source. The diode (which you can see in the symbol) will cause the load to be always on if you wire it like this. Also, to turn it off, you need to let the gate go all the way up to 24V. If it would work for your application, it would be easier to put NMOS between load and GND. Then a 3.3V level on the gate will turn it on. \$\endgroup\$
    – user57037
    Dec 13, 2015 at 9:10
  • \$\begingroup\$ Having an NMOS between load and gnd might have issues as you need to keep gate voltage greater than the source voltage. The gate is ,say 3.3V from the PIC. When gate is 0 the source is pulled down to gnd via the load. But, when gate is made high(3.3V) the source will jump to 24V thereby causing a conflict in this. \$\endgroup\$
    – Board-Man
    Dec 13, 2015 at 9:21
  • 1
    \$\begingroup\$ @Board-Man, not really sure what you are talking about. But if you put NMOS between load and GND, you would connect gate to logic_level, source to GND and drain to bottom of load. As long as it is a low-Vgs type FET it will work fine. Things like that are done all the time. But if it is an external load, or one side of it is earth grounded, it may not be possible to switch the low side. Then it has to be PMOS with another signal FET or BJT to invert the gate control signal. \$\endgroup\$
    – user57037
    Dec 13, 2015 at 9:31
  • \$\begingroup\$ @Sorry, I misunderstood it as the load between the NMOS and GND. My apologies. \$\endgroup\$
    – Board-Man
    Dec 13, 2015 at 9:34
  • \$\begingroup\$ I see. No worries. ;-) \$\endgroup\$
    – user57037
    Dec 13, 2015 at 10:09

1 Answer 1

4
\$\begingroup\$

There are two things wrong with your circuit: -

  • The P channel FET is upside down (will always conduct thru its body diode)
  • If you correctly placed the FET it's a little tricky to directly drive it from a 3v3 logic power supply without damage. You need level translation.

The easiest method is to use an N channel device with the load tied to the 24V rail. Here's an idea that controls a motor: -

enter image description here

If you need the load to be grounded you need an extra transistor: -

enter image description here

Note that there is some attention to protecting the gate from over-voltage - the two resistors and the zener diode limit the gate voltage to below the maximum allowed by the P channel FET.

If your MOSFET was a logic level device that adequately operated at 3V3 gate drive you could use a zener diode directly from the 3V3 logic but it can be a little tricky to set up correctly.

\$\endgroup\$
11
  • \$\begingroup\$ In your second circuit, if "Supply+" is 100 V or has a lot of transients present I can see why you need the zener diode. If it's stable +24 V, and the PMOS is chosen appropriately you could simply remove it and still have a working circuit. \$\endgroup\$
    – The Photon
    Dec 13, 2015 at 17:49
  • \$\begingroup\$ Why do we need to use parallel resistor to zener diode? (could someone please explain). \$\endgroup\$
    – user505160
    Jan 29, 2020 at 9:53
  • \$\begingroup\$ Do you understand why we need a zener? \$\endgroup\$
    – Andy aka
    Jan 29, 2020 at 10:18
  • \$\begingroup\$ Yes (at least I think I do)... so we have P-FET which has gate voltage much lower then supply voltage and to compensate for that we are using zener to bring it operating level of p-fet, correct? -- From practical experience I understand that not heaving resistor makes p-fet turn on while n-fet is turned off. I am missing reason why is that happening? \$\endgroup\$
    – user505160
    Jan 29, 2020 at 14:21
  • \$\begingroup\$ The zener protects the gate from excessive voltages when the overall supply might be greater than 20 volts but, the resistor in parallel is needed to turn off the MOSFET. \$\endgroup\$
    – Andy aka
    Jan 29, 2020 at 16:00

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.