I have probably overcomplicated this resulting in my utter confusion.

I'm trying to simply turn an LED on and off using a 2n7000 MOSFET as the switch. Is my thinking right by saying that a positive voltage on the gate will 'turn on' the MOSFET and connect the Source to Drain? That is, the Source is literally the source for the drain?

Voltage is always relative. A positive voltage on the gate relative to the source turns it on. Simplest is connect the source to ground, and then the drain acts like it is either disconnected or grounded depending on whether the switch is open or closed.

So 2.2V on the gate when the source has 1V will turn it on and allow the 1V to pass through to the drain?

If so, if you wanted to pass 50V @ 200mA from source to drain, you would need to pass more than 50V (little to no current) to the gate?

No, when used as a switch (with a large drain-source voltage) the gate-source voltage difference determines the drain-source current.

Find the 2N7000 datasheet and look at the current-voltage chart. There's a separate curve for each choice of gate-source voltage. It looks like .

If there is only a 1.2V difference between gate and source, the drain-source current will be negligible. If you want 200mA drain-source current, the gate voltage needs to be about 4V above the source. 5V would be better, and that's what the MCU will provide anyway.

Thank you so much for your help so far. I think I may understand now. Let's see if I understand by using some arbitrary numbers:

If I want to pass 2V @ 750mA, I would need 2V at the source and 4V at the gate?

And why does the datasheet show such high amperage when it cannot support anything over a 500mA burst?

These FETS can actually be used in either direction. Source and drain can be swapped. I don't know if it is really good for them or not but I haven't had any trouble yet.

datasheet shows up to 2amps, because 2N7000 is capable of handling iDS at 2A for only a few nanoseconds. You really don't want to go above 400 or so mA otherwise, if you don't have to.

This fet is for signaling. If you want a FET that handles more current, you want a power-FET.

BM

Oh I see... this FET is for signaling. So if I were actually driving something that requires more amperage andor voltage, then I would use this and a relay, or as you said, a power-FET?

If you have 2v at the source (relative to ground) and 4v at the gate, your gate-source voltage (Vgs) is only 2v (the diiference between gate and source). At that level the switch won't turn on (the curve for Vgs of 2v isn't even shown because it's too low).