Stats

  • Locked Locked
  • Replies 5
  • Subscribers 64
  • Views 19179
  • Members are here 0
This discussion has been locked.
You can no longer post new replies to this discussion. If you have a question you can start a new discussion

Noise simulation shows wrong result when ideal current sources are present

mhasan14

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.

Parents

  • 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.

    Regards,

    Andrew.

Reply

  • 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.

    Regards,

    Andrew.

Children
No Data

Community Guidelines

The Cadence Design Communities support Cadence users and technologists interacting to exchange ideas, news, technical information, and best practices to solve problems and get the most from Cadence technology. The community is open to everyone, and to provide the most value, we require participants to follow our Community Guidelines that facilitate a quality exchange of ideas and information. By accessing, contributing, using or downloading any materials from the site, you agree to be bound by the full Community Guidelines.