I tried to measure noise contribution of a MOSFET in the following circuit arrangement. The current source is ideal. I would measure noise current from ADE noise simulation. However the noise current does not seem to be what is expected. The values are way too low. The noise current does show 1/f pattern over 10mHz to 1kHz, however the noise is much lower than the thermal noise value. The thermal noise should be around 2*q*I and at lower frequencies 1/f would dominate. But I see the noise current six orders of magnitude lower than the thermal noise. Shorting out the current source displays expected result however. I measure noise current by attaching a zero volt voltage source at the source of the MOSFET and probe current from it.
The question is what is going wrong here. If I replace the MOSFET with an analog resistor noise current output is zero.
In had calculations we would open up the ideal current source and all the noise current of MOSFET flows through the MOSFET. So I expect to see the same result from simulation.
It's not entirely clear how you're measuring the noise current - but my guess is that because you have an ideal current source, the current through the device is fixed - the noise is not influencing (it may have a small influence about how much current flows into the bulk hence the small numbers).
Showing your real test bench and the analysis setup might clarify that for certain though.
This is my testbench. Let me know if you need anything else.
I agree that fixed current source is not influencing the noise. However, do you agree that hand calculations will show that all of the noise current is flowing through the mosfet?
Because you have a fixed current source (I0), there's no way for any (significant) current additional current to flow through V0. The only way that any noise current generated in the device could flow would be through the bulk-source branch which is going to be limited.
So your hand calculations must be wrong - the current flowing into the drain must essentially be the same as the current coming out of the source. You're probably neglecting the fact that this is an ideal current source.
If you remove the current source and short it (or replace it with a zero-volt source) then you will see output noise current, because the current flowing through the transistor can then vary.
I guess that is why the current output is so low. Anyway, the hand calculation I was talking about is something like in the picture. Even if the current source is ideal the noise current flows through the drain source resistance. But I think I would not be able to see that current by placing the zero volt source at the source.