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

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

• So if you remove the mosfet, what's the voltage at that pin? – Big6 Apr 21 '18 at 3:00
• I hope M1 GND and 555 GND are actually connected (unlike the schematic) – Brian Drummond Apr 21 '18 at 10:14
• @sixcab With mosfet removed voltage is at a steady 3.3v. – ItsAPhaseLocked Apr 21 '18 at 23:14
• @Brian Drummond lol yes the grounds are connected. – ItsAPhaseLocked Apr 21 '18 at 23:17
• My next move would be to try a different mosfet – Big6 Apr 21 '18 at 23:22

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:
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.