NJFET and breakdown avalanche

Thread Starter


Joined Aug 1, 2005
Hey every one,

here is my first post, and it seems very dumb I know, but I got that question, and could not answer it completly, so the teacher asked me to think about it, and to come back to tell him what I found.

Anyway I attached a file with the circuit. he asked me to draw him the voltage Vd and the current iL... so for me at the beginning when the NJFET is off the current is null thus Vd = 12V and iL=0 A.

when we turn the transistor ON, the current starts to grow through DS... for me iL = iDS and since VL = L diL/dt, it increases proportionnaly until it reaches the Isat or a constant value of Ids, then VL = 0, and thus iL is constant.
concerning Vd, it decreases since it equals 12v-VL, after what it goes back to 12V since VL --> 0V.... well it seemed that my problem was there cause for him Vd is 12V when the NJFET is OFF, and directly after we turn it ON Vd = 0V ???

Anyway after that I think I got the avalanche breakdown right, and the a way to avoid it was to put reduce the iD by using a Resistor.

thanks guys for your help