I want to create a very simple 3-phase inverter that is easy to understand, e.g. without using MOSFET drivers. I decided to use logic level P-channel MOSFETs on the high side and logic level N-channel MOSFETs on the low side.

I tried to draw a diagram, it went so-so. But you get the overall picture: enter image description here

The gates on the high side MOSFET gates are "pulled-up" to 3v. And the low side MOSFET gates are "pulled-down" to 0v. I have understood that this is to put all the gates to a known closed state.

I think I understand it so far. Now, I want to control the gates from a microcontroller, using PWM signals to the gates.

Since the P-channel MOSFETs opens on low signal and the N-channel MOSFETs opens on high signal and in each half-bridge either high side or low side should be open, but not both (it would short circuit) - then can I control both gates by a common PWM signal? like the diagram below:

enter image description here

But what is confusing now for me is that I connect the "pulled-up" and the "pulled-down" gates together. Isn't this a problem?

If that does not work, can I control the high side and low side of the half bridge from two identical PWM signals, originating from two different GPIO pins of the microcontroller instead - that way the "pulled-up" and "pulled-down" gate wouldn't be connected, but controlled by an identical PWM signal.

A diagram of this last alternative: enter image description here

  • 2
    \$\begingroup\$ You haven't drawn anything connected to any of the MOSFET gates and it's unclear what you're asking. \$\endgroup\$
    – Hearth
    Dec 20, 2023 at 0:22
  • 3
    \$\begingroup\$ Some terms you might want to read up on: "shoot-through", "cross conduction", and "dead-time". \$\endgroup\$ Dec 20, 2023 at 0:23
  • \$\begingroup\$ @Hearth yeah, it was kind of hard to find an online service to draw circuit diagrams. It might be easier to just follow the text and omit the images. \$\endgroup\$
    – Jonas
    Dec 20, 2023 at 0:28
  • 1
    \$\begingroup\$ @Jonas There is one built in to this website, if you click the little circuit diagram button while editing your question. Or you can use LTspice or KiCad or whatever CAD software you prefer and just take a screenshot. \$\endgroup\$
    – Hearth
    Dec 20, 2023 at 0:30
  • \$\begingroup\$ @Hearth yep, I did the latter and ended up using circuit-diagram.org/editor but it was fairly difficult. \$\endgroup\$
    – Jonas
    Dec 20, 2023 at 0:32

2 Answers 2


picture 1:

Pull up/down on MOSFET gate is measured relative to the source terminals If your 3.3v on the P channel MOSFETS is relative to the 0V that's not going to work.

picture 2:

You have pull-ups and pull downs with competing goals, you will need to pick a rest state for the mosfets. you will probablt need 7.2V amplitide one the gate wire to reliably turn the upper MOSFET off, and you will need fast switching (or an inductor in series with the supply) to prevent excessive shoot-through during the switching when both the upper and lower MOSFET are briefly turned somewhat on.

Picture 3:

seperate controls allows you to prevent shoot through by turning off before turning on the opposite MOSFET. 3.3v is still the wrong voltage for the upper MOSFET gate.

If you don't want to use mosfet drivers you'll need to find some other way to offset the mosfet gate drive to drive the upper mosfets.

  • \$\begingroup\$ Right, "pull-ups and pull-downs with competing goals" is probably what I was woundering about. And, yeah, I have more to learn about MOSFETs, it is a soo difficult topic. Thanks anyway. \$\endgroup\$
    – Jonas
    Dec 20, 2023 at 2:32

A very common is to use an optocouplers. You can connect its leds to one pin but it doesn’t allow you make a deadtimes. If you connect the leds on separate pins the deadtimes can be done.

enter image description here

Note: Be careful, during Mcu startup without led signals the both mosfets are on. Best is to start Mcu as first and connect the power supply as second. Fuse the power supply also for mosfet safety.

  • \$\begingroup\$ Both LEDs will melt immediately. Forward voltage of a coupler LED typically is below 1.5 V. The gate discharge current will be very low here (around 10 mA) and only 10-20 % of the gate charge current. This tends to produce shoot-through. \$\endgroup\$
    – Jens
    Dec 20, 2023 at 2:51
  • \$\begingroup\$ Thx, I updated the schematic. \$\endgroup\$ Dec 20, 2023 at 3:09
  • \$\begingroup\$ Now I realized during high impedance Mcu pin state the both leds are supplied so both mosfets are happy (off). \$\endgroup\$ Dec 20, 2023 at 3:17

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.