Hello,

I'm trying to familiarize myself w/ some basics of MOSFET operation + using SPICE to do some easy simulations. (Please see attached image of circuit I am trying to construct)

I ran into the following problem. First, I biased a MOSFET by diode connecting it, drain tied to gate. I then set the source to ground. This means that the MOSFET is operating in the active region by the following: VDS > VGS - VT. I see a drain current, which is what I expect to see. So everything is ok.

Next, I tried to make some slightly different connections to confirm that the MOSFET would also operate in the cutoff region. However, when I tied the Source to the Gate, I still see appreciable current flow. In this new case, Gate = Source, so VGS = 0, and the MOSFET should have no drain current. In theory, there should be no conductive channel that forms from source to drain, right? In simulation, I am seeing a current, and a voltage drop across the drain resistor. I'm trying to figure out what is going on here. Am I missing some key points?

Thanks for the help!

 

All MOSFET has a body diode connect between source and drain.
And this diode very often isn't shown in the diagram.
http://en.wikipedia.org/wiki/Power_MOSFET#Body_diode
http://www.datasheetcatalog.org/datasheets/70/283449_DS.pdf

 

In the case with gate connected to source, you need to swap +5V and ground.

 

In the case with gate connected to source, you need to swap +5V and ground.

I know that this transistor is hooked up incorrectly. My question is when I simulate it, why does it show such a large drain current? Shouldn't I see close to 0 A since VGS = 0, and the NMOS is in cutoff?

 

I know that this transistor is hooked up incorrectly. My question is when I simulate it, why does it show such a large drain current? Shouldn't I see close to 0 A since VGS = 0, and the NMOS is in cutoff?

Jony130 explained it. The body diode is between source and drain, anode to source. You are seeing current through that diode, not through the channel, which is indeed cut off.