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I designed a PCB to drive mosfets with a ESP32 micro-controller. The ESP32 will be connected to the PCB with wires from a separate PCB. When there is no micro-controller connected to the input of the Mosfet PCB all Mosfets should be turned on. So I connected a pull-up resistor to their gate. The only onboard voltage is 12V so the pull-up is connected to the 12V as well. When nothing is connected the voltage at the input pins will be 12V because of the ESP32. this doesn't seem a safe scenario to me for connecting a micro-controller. I worked out a few possible solutions.

  1. Default scenario
  2. Using a voltage divider
  3. Using a 3V zener diode at the input
  4. Creating a separate 3V source with a zener diode, and use this source to connect the pull-up.
  5. (UPDATE) instead of option 4 use a voltage regulator to connect a pull-up to

For each scenario I've written my thoughts next to it. I'm not sure which one I should pick. What is the best practice in this scenario? Opto-couplers might be but they slow down my switching time to much.

Thanks in advance

enter image description here

Link to simulation in falstad

UPDATE This is my actual schematic I'm using at the moment. R7 Needs to be changed to a pull-up that's the one I was talking about.

A little bit more info on my circuit.

  • Q4 is the power-mosfet that will get a voltage of 0V or 12V at its gate. From Mosfet driver U2.
  • U2 is a mostfet driver with the internal schematic of a totem pole transistor configuration. To get around 12V at it's output we need 12V at it's input.
  • To to make a 12V to 0V input voltage at U2 controller by only a 0-3.3V control voltage I use Q3 a N-CH Mosfet with a pull-up resistor.

Actual schematic

The reason I'm using this mosfet driver and I'm not controlling it directly is because I need a very high switching speed. At the moment when I measured this I only have a 33ns delay between input and output. And I'm afraid that adding an extra transistor or mosfet will slow down the propagation time even more.

If a transistor is faster than a mosfet which I guess it is I would gladly change Q3 for a transistor. But then I'm in the same situation as before with the pull-up at the gate/base.

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2 Answers 2

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A suggested way to properly drive your FET is using a circuit like this when you just have a 3.3V GPIO.

enter image description here

Update:

This answer was based upon the information that you originally supplied. You should have provided that schematic from the outset. Based upon that schematic I'll tell you that you need to change approach and use a gate driver like a DGD0215 or DGD0216 instead of that driver you have now. This chip as a 3.3V logic in compatibility with no additional supply rail required. You can then get rid of that 2N7002 completely.

Mouser price listing for the suggested part is the same as that ZXGD3002E6 that you have in your schematic now.

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  • \$\begingroup\$ Please see my update, this is indeed what I'm doing already. But I still need a way to keep Q1 switched on when there is no input connected. In order to keep M1 switched off. Thanks for the fast reply! \$\endgroup\$
    – Bruce
    Apr 3, 2020 at 9:01
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    \$\begingroup\$ This answer was based upon the information that you originally supplied. You should have provided that schematic from the outset. Based upon that schematic I'll tell you that you need to change approach and use a gate driver like a DGD0125 or DGD0216 instead of that driver you have now. This chip as a 3.3V logic in compatibility with no additional supply rail required. You can then get rid of that 2N7002 completely. \$\endgroup\$
    – Michael Karas
    Apr 3, 2020 at 9:16
  • \$\begingroup\$ Wow these "DGD0125 or DGD0216" look very promising. Do you know of similar drivers that are faster? shorter propagation times? Thanks for the suggestion. \$\endgroup\$
    – Bruce
    Apr 3, 2020 at 10:13
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For all circuits you proposed, there is a slight risk of damaging the MCU due to back feeding. Why don't you isolate the control circuit from the power one, by using a simple bipolar transistor?

circuit

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  • \$\begingroup\$ Please see my update, this is indeed what I'm doing already. But I still need a way to keep Q1 switched on when there is no input connected. In order to keep M1 switched off. Thanks for the fast reply! \$\endgroup\$
    – Bruce
    Apr 3, 2020 at 9:00

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