Question about depletion layer

[majormajor]

I am trying to understand some aspects of the depletion layer in a semiconductor diode and I have come across some questions that none of the textbooks I have seem to explain. Perhaps some of you could help? (I must add I am not a physisist but an electronics engineer, although I have been taught the usual basic semiconductor theory at uni.)

(1) I understand that when a p and n type materials form a junction, electrons diffuse from the n type material to the p type and holes diffuse from the p type to the n type. This is kind of intuitive enough and I understand how this leaves a positive charge layer in the n-type material that is bound to the crystal lattice and vice versa, a negative charge layer in the p type material due to the holes that have diffused into the n-type material. But what the textbooks don't seem to explain is this: let's just look at the n-type material. OK, so electrons have diffused across the junction, leaving positively charged ions behind. But what happens to the electrons that have left the n-type material? Are they just sitting somewhere in the p-type material (let's assume there is no external electrical field on the junction)? Where are they? Clearly they must form another (mobile) space charge layer somewhere on the p-type side?

(2) What is even more baffling for me is how minority carries can pass through the depletion layer (like in a reverse charged pn junction or the base-collector junction of a BJT). The depletion layer means that near the juction on the n-type side there is a positively charged layer due to the fact that the electrons have left. Surely if somewhere in the n-type layer holes are generated somehow, they will be repelled by the positive space-charge layer and will not be able to cross the junction? I feel this is really a point where I don't understand something, so please could somebody explain (just qualitatively) what is going on here?

Thanks,
MajorMajor

 

[hiyok]

Probably you did not consider the difference between the chemical potentials on the p- and n-side semiconductors. Suppose you put two intrinsic (I consider intrinsic case for simplicity) semiconductors in contact. These two semiconductors have different chemical potentials, and thus carriers shall diffuse (as described in your post). Because of this diffusion, some electrons shall accumulate around the junction on the p-side, while some holes shall reside around the junction on the n-side. These electrons and holes shall form an electrostatic field that acts to impede further diffusion. So, basically there are two types of currents involved: the diffusion current due to chemical potential difference and the one due to this electric field. These two currents go in opposite directions. The system will reach equilibrium as these currents equal. Ultimately, an electric field is established inside the depletion layer. The reason why the electrons and holes must reside around the junction is because, semiconductors are conductive.