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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?

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  • imagesensor123

    Hi Andrew,
    thanks, so for calculating the drain current, we can use the equation like: Id=1/2*betaeff*(Vgs-Vth)^2, here, we neglect channel length modulation effect, is the equation right? for analyzing the basic current mirror, i use a vdd=2.5 w=2u and l=0.35u to configure the circuit, and i get the id=655.2uA, but when i calculate it by hand use the above equation id=1/2*662.3*(2.5-0.5484)^2=1.261mA, betaeff and vth comes from the op results, where i am wrong?

    regards,
    zfeng

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  • imagesensor123

    Hi Andrew,
    thanks, so for calculating the drain current, we can use the equation like: Id=1/2*betaeff*(Vgs-Vth)^2, here, we neglect channel length modulation effect, is the equation right? for analyzing the basic current mirror, i use a vdd=2.5 w=2u and l=0.35u to configure the circuit, and i get the id=655.2uA, but when i calculate it by hand use the above equation id=1/2*662.3*(2.5-0.5484)^2=1.261mA, betaeff and vth comes from the op results, where i am wrong?

    regards,
    zfeng

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