# Why and how do holes in the valence band of the p-type material of a p-n junction diffuse through to the valence band of the n-type material

tl;dr -- See the question title

I quote from this Wikipedia article on the depletion region:

By definition, the N-type semiconductor has an excess of free electrons (in the conduction band) compared to the P-type semiconductor, and the P-type has an excess of holes (in the valence band) compared to the N-type. Therefore, when N-doped and P-doped semiconductors are placed together to form a junction, free electrons in the N-side conduction band migrate (diffuse) into the P-side conduction band, and holes in the P-side valence band migrate into the N-side valence band.

Pardon my naivete (if the phrasing or question suggest any), Physics is not really my of area expertise; but my understanding prior to this had been that the diffusion of holes in the p-type region to the n-type region was really a consequence of the n-type conduction band electrons "filling" the former positions of the p-type holes, and conversely, n-type valence band electrons equivalent in quantity to the now diffused electrons occupying (somehow) the evacuated positions of the diffused electrons in the conduction band--thus "creating" the appropriate number of "diffused holes" in the n-type region.

However this would also suggest that the conduction band of the n-type region is continually replenished--which is apparently wrong by this answer:

Higher energies are occupied on the left than the right, and what happens? Answer: some of the higher-energy electrons on the left "fall into" the lower-energy states on the right. Yes, this leaves a + charge on the n-type side and a − charge on the p-type side with an associated voltage. [...] The result: on the right hand side all of your valence band is filled in; on your left hand side none of your conduction band remains, and you've got a "depletion region".

What I'm failing to understand is how the p-type holes would then diffuse to the n-type region if (1) the n-type electrons diffuse to the p-type conduction band and (2) the conduction band in the n-type material is left devoid of electrons. Would appreciate your responses.

• Steven Mould, 'How diodes, LEDs and solar panels work' This video explains the whole thing pretty well. It has animations!! – R.Crane Mar 8 at 4:57
• First, note that there is a population of holes in the n-type region, and similarly a population of electrons in the p-type region. And that, in equilibrium, detailed balance requires that $np = n_{i}^{2}$. In your top quote, if you place p-type next to n-type, as the carriers start to diffuse (as they always do), then $np >> n_{i}^{2}$ and recombination will occur. As the carriers annihilate each other, the underlying charged dopants contribute an electric field. The build up of this field ultimately forms the depletion region. – Jon Custer Mar 8 at 15:28
• @R.Crane that video summed it up pretty succinctly. Good catch – Garikai Mar 10 at 17:47
• @JonCuster I really get the gist of your answer as it relates particularly to the relations/equations which have to be satisfied in the upshot, but I want a bit of clarification on two things: (1) So recombination will stop occurring when np = ni2?; and (2) My original inquiry was particularly on the mechanism behind the whole "diffusion" of holes phenomenon which I'm still trying to get a grip on. The linked video seems to have left that part out or if I observed correctly, doesn't happen at all – Garikai Mar 10 at 18:01
• Recombination, like carrier generation, is occurring all the time. $np=n_{I}^{2}$ is the balance point where the forward reaction equals the reverse reaction. – Jon Custer Mar 11 at 14:02