Can a GPIO drive MOSFET gates in parallel?

Question

I'm trying to make a circuit that tests if there is any connection in the little box between the transistors. If there is connection on all the boxes, I'll be able to read something out on the ADC.

Switching time is not an issue. It does not matter how long it takes for the transistors to turn on, just that they do.

Lets say I want to connect something like 7-8 transistors in series like this, and activate them with the same GPIO signal from Raspberry Pi. Would it work, or do I need a gate driver to make sure the transistors will turn on?

I'm using transistors with Vth 0.4V-1V and Ron <100mOhm.

Answer 1

Assuming the transistors' \$V_{GS(TH)} << (3.3V - 160mV)\$, then yes this will work, but the GPIO might struggle with the current that will flow momentarily, on each high/low transition.

The combined capacitance of all those gates could be in the nanofarads, depending on the MOSFET model and the number of them. I don't think a GPIO would have a problem with 10s or even 100s of picofarads, but nanofarads is pushing your luck.

A simple solution is to place a resistor directly at the GPIO, and connect all the gates to it. Not just any old resistor; let's calculate a suitable value. The Raspberry Pi outputs are very weak, and if I recall correctly, can only sink or source about 2mA. We shouldn't exceed that. By Ohm's law:

$$ R = \frac{V}{I} = \frac{3.3V}{2mA} \approx 1.8k\Omega $$

The circuit would look like this:

^{simulate this circuit – Schematic created using CircuitLab}

That limited output current is certainly going to slow down the switching on and off of the transistors, but I don't think your application requires speed, so unless there's something you're not telling us, a driver won't be necessary.

If the total gate capacitance was 1nF, switching fully on/off will take approximately:

$$ \tau = R_1 \times C_{GATES} = 1.8k\Omega \times 1nF \approx 2\mu s $$

If that's fast enough for you, then job done. If not, then think about using a gate driver.