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If a JFET with length L=10µm, a=2µm, W=8µm, Vp=-4V.What is the value of rds at Vgs = 0V?(a) 2KΩ(b) 5.2KΩ(c) 10KΩ(d) 9.8KΩThis question was posed to me by my college director while I was bunking the class.My question is from The JFET Volt-Ampere Characteristics in chapter Field-Effect Transistors of Electronic Devices & Circuits

Answer» CORRECT option is (d) 9.8KΩ

The EXPLANATION: rds=L/(2aqND µnW) = Nd = 2V/qa^2=1.33×10^21 atoms/m^3

µn=0.15m^2/v-sec

On SUBSTITUTING the VALUES, we get rds=9.8KΩ.


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