Thursday, January 31, 2008

The First Question

Hello Friends,

I guess nobody's very much interested in a bigger topic like Cell Phone....so let us start with something very basic....
A question for all of you.....

If the doping density is increased on either p or n or both sides of a p-n junction diode, what is the effect on its barrier potential?? Does it increase or decrease and why so?

Hoping for early replies to the answer....hurry up be the first one to answer,so that we know the genius among us....

All the best....Happy Blogging

4 comments:

Kamaldeep Singh said...

let us say we have a pn junction.
if we dope p region heavily leaving n as it is the depletion width on the p side wud reduce as compared to that at the n region. similar is the case when we dope n region more leaving p intact.so doping one side reduces the depletion width at it's end.
when we dope both the regions, the depletion width reduces to a very small length than it was before doping.
so we wud need a lesser potential to break this reduced depletion region and to start the current flow & bring the device into conduction.
this potential is called "Barrier Potential" which is less in case of heavy doping!!

Louis_rocks said...

if doping is done on either one side, concentration of carrier(hole/e-s)increases leading to increase in diffusion across the junction. electron hole recombination will be more so more concentration of ions, hence greater e-field need to break it down.in my view barrier potential must increase with increase in doping!!!

Anonymous said...

Go back to the basics of semiconductor theory and try refreshing the devices with higher and lower carrier concentration and their potential barrier dependence on concentration for eg. zener diode and a simple p-n junction diode....u'll get the answer

Anonymous said...

leakage current will increase but the barrier potential will remain same....