Low-Side Switch MOSFET Drain Voltage Drop in OFF state?

Question

I'm trying to replace a normally open switch with a low-side switch mosfet. The drain of the mosfet is connected to an input pin of a non-modifiable 3.3V MCU.

Prior to my changes to the setup the HIGH side of the switch measured around 3.3V since it's a LOW = TRUE logic input.

After I added a 5v long-period square wave (~5V high, 0.070V low) to the gate of the mosfet the voltage changed drastically at the drain side of the mosfet.

MOSFET ON - V_ds = 0V with V_gs = 5V as desired.
MOSFET OFF- V_ds = 1.7V with V_gs = 0.070V - HALF of what it should be with the mosfet switch open.

Anyone have any insight as to why this would be? From what I can tell I'm utilizing the 2N7000 well within spec.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

As for the mosfet, when you switch it off (with \$V_{GS}\approx 0\$), in your schematic, the drain is actually floating (unless your MCU gives you the option to enable internal pullup resistors at the input, check that). You may want to do something like this:

^{simulate this circuit – Schematic created using CircuitLab}

When CTRL is 5V, you get \$V_D=0V\$. When CTRL is \$\approx 0\$, the MOSFET is off and the pullup resistor should set the 3.3V(Logic high) on the MOSFET's drain.

Also a resistor (47k works) from the gate to ground would not hurt, though I don't think this is the reason you're having issues, since your are driving the gate low.