I was testing a circuit using this circuit simulator and found that the nMOS transistor had a current of a few nanoamps running through it when it was in the "off" state. This was unexpected because I was taught (and found online) that the current through a MOSFET in the cutoff region was 0.
After a lot of thinking, I've decided that it seems odd that it would ever be exactly 0 since everything about a MOSFET varies in a continuous fashion. In the same way that there is no real point that the MOSFET changes from linear to saturation (as in, it does not suddenly switch to saturation as soon as , it's more of a transitional phase), it seems reasonable to assume that there is no real "off" state of a MOSFET, rather just a very low current linear region. But if that was true, surely I'd be able to find an equation which models the current in the cut-off region. I thought the linear region equation would work, but the results I get from that do not match those of the simulation.
The reason this is an issue for me is that it means the voltage output is dependent on the value of the resistor when in the off state.
Best Answer
It's all about \$I_{DSS}\$ so pick up a data sheet for a MOSFET and look up that term. It'll tell you something like: -
It's the drain current that flows when the gate and source are connected i.e. \$V_{GS}\$ = 0 volts. It's usually specified for a high-ish drain source voltage of maybe 10 volts to hundreds of volts.
Basically it's the leakage current of the channel.