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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?

  • Quek

    Hi Zfeng

    Actually there are really too many reasons why your hand calculated results are different from simulation results.

    a. You might have calculated wrongly
    b. Bug with simulator
    c. For advanced process nodes, you might have simplified too many things

    If there is really a need to clarify this, I think it would be good for you to file a service request to your local Cadence support.

    Best regards
    Quek

  • imagesensor123

    Hi Quek,
        thanks, i will check it carefully, and i will generate a request if necessary. any way, thanks a lot for your help.

    regards,
    zfeng

  • EthanRFIC

     I think for nowadays, you may check "Gm/Id" design methodology, which is much MORE accurate for your design, and just get rid of the "squared law equation". I design RFIC circuit using this method, all the design procedure is calculated in Matlab script, even "no need to simulate in spice", I mean, when I click the Matlab, I get the W/L for all the transistors, then simulate it in cadence ----  everything mathched well with my theoretical analysis.

     If you need more information, you can google "Gm/Id" method on IEEE, there are a lot of papers talking about it. Good luck!

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