The guiding equation for diode current is
Idiode = Ireverse saturation × (ekV/T – 1)
Now, as far as I know, the reverse saturation current is contributed by minority charge carriers. So, if some day it becomes possible to remove all impurities from a diode, this Is is going to be 0. Then how to calculate the diode current in that case?
Best Answer
It is actually possible to get very low impurity Si now. It's just that if you look at it crossed eyed it gets doped.
Your question really comes down to answering the question of what is a minority carrier when there is no doping? There will be carriers of both types present because of the intrinsic concentration but as the doping goes down these become more and more balanced. At the limit, the net polarity of the material is determined by impurity ionization from an applied field.
To form a junction you need to look at the fermi level and do the volume integration to balance the charge through out the semi-conductor. If that doesn't make sense to you, then think about what it means to balance the charge when there is very low charge density. That means that the volume must increase. In simplest terms this means the the depletion volume has to extend to include sufficient # of impurities for the fermi level to establish itself. If you have low impurities the depletion layer moves through the whole volume available.
A case in point is the deep depleted CCD's built on 1e10 /cm^3 substrates with depletion regions that are 500 um or more thick.