Considerthefollowingabruptjunctiondiode(m=0.5):Cj0=3x10^-3(F/(m^2)),AD=1μm^2,andФ0=0.64V.Ifweapplyareversebiasof-2.5V,what’sthevalueofthetotaljunctioncapacitance?(selectthenearestvalue)
Solution: c)Cj=ADCj0(1-VD/Ф0)^(-0.5)=1x10^(-12)x3x10^(-3)x(1+2.5/0.64)^(-0.5)=1.35fF