I'm currently doing some independent research on silicon based devices and I have come up with one question I can't seem to find the answer to elsewhere.
I understand that when an n-type and p-type silicon are put back-to-back the free electrons from the phosphorus diffuse over the barrier and combine with the boron atoms free valence shell holes. I understand that this comes to equilibrium and an electric field opposed to further diffusion is created, the remaining free phosphorus electrons are opposed by the positive boron ions and that force is greater than the pull from the boron holes further away.
Now what I don't understand is why then the electrons on the p-type side that have combined with the boron atoms cannot diffuse further over the p-type silicon.
The only thing I can think of is that the boron atoms are happy being negative ions as this satisfies the octet rule and the boron holes in the p-type haven't got enough energy to break them out of the valence shell.
One last smaller question is do the boron negative ions oppose the n-type free electrons or are the boron holes too far away to further attract the electrons? I presume the former due to the fact the n-type free electrons don't diffuse into the n-type side depletion zone. This also applies to the boron holes, I presume they are repelled by the positive phosphorus ions.
Thanks in advance for anyone's help and sorry if this relates more to physics than EE.