From the textbook that is prescribed for my course as well as on google, I have been told due to the diffusion of electron from n->p and the holes from p->n, positive ions are left in the n-region and negative ions are left in the p-region, where there are no free charge carriers, thus causing the formation of an electric field across the pn junction which eventually acts to halt the diffusion of the charge carriers.

However, my question is that: as holes go from p->n and electrons go from n->p, the recombination that occurs on either side of the metallurgical junction causes the annihilation of the charge carriers, so shouldn't this mean that the ions (let's just look at the p-region for the sake of the argument) with net negative charge that were missing an electron in the valence shell, that has now been filled through recombination, shouldn't this just become an atom with no net charge? As all of the valence bonds are now stable? Of course my question makes no sense in terms of charge neutrality of the crystal structure as a whole as it will no longer be neutral, but I don't understand why I am wrong.

This exact question was posed in this thread in SE EE, however, I didn't understand the answers, hence why I am asking again.


  • \$\begingroup\$ Hmm this is really well explained in the question you've linked. The N region likes to give up an electron, but it's not actually ionized until it does so. The P region has holes that can hold up an electron, but until the acceptor molecule actually picks up an electron it's not an ion. Thats what I read anyway. \$\endgroup\$ – K H Jan 28 at 9:43

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