Low voltage controlled open drain circuit

Question

I have a power regulator with an output enable that can be used to shut the regulator off without actually unhooking it. I am trying to figure out how to control that using a GPIO port of a Jetson TX2 (1.8V logic).

The manufacturer of the regulator says that I will not be able to connect the output enable wire directly to a 1.8V GPIO pin because the output enable is internally pulled up to the regulator Vin (24V). They said to drive it properly I will need an open-drain circuit, which will allow your GPIO to switch the output enable pin between float and ground.

Using https://www.microcontrollertips.com/what-is-an-open-drain/ as a source I looked on Digikey to find an N-channel MOSFET that would work and I think one of these:

https://www.digikey.com/product-detail/en/toshiba-semiconductor-and-storage/SSM3K329RLF/SSM3K329RLFCT-ND/3522426

https://www.digikey.com/product-detail/en/nexperia-usa-inc/PMV40UN2R/1727-1900-1-ND/5056345

would. Is creating the open drain circuit a matter of connecting the output enable wire of the regulator to the mosfet drain, the Jetson GPIO to the gate and the mosfet source to ground like this? Or do I need a current limiting resistor between the GPIO and the gate?

^{simulate this circuit – Schematic created using CircuitLab}

Thanks for your help.

Answer 1

Either the MOSFET or the BJT will work.

You might find the transistor (leaded type) easier to work with and it has a bit more margin on the voltage (45V vs. 30V on a 24V circuit).

Personally I would add a resistor with the MOSFET gate (something like 10K) to make it a little less likely to damage that $500 computer board. It will slightly slow it down (microseconds). It's not necessary for the MOSFET. It is necessary for the BC547 BJT.