Stats

  • Locked Locked
  • Replies 13
  • Subscribers 127
  • Views 36708
  • 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

make it clearly beta and betaeff

imagesensor123

 Hi,

    these days i am doing some circuit simulation, and i got confused by beta and betaeff? according to the simple hand calculation equation id=1/2uCoxW/L(Vgs-Vth)^2 or id=1/2beta(Vgs-Vth)^2  or the others. and after simultion, i use the result->print->operation point and i found a betaeff parameter, i think it shold be same with beta, but after calculation id, i found the id(betaeff) is just a half of the id(beta), and how do you calculate the drain current id in saturation region?

Parents Reply 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.