Explain why a pure semiconductor crystal will always have equal numbers of electrons and holes present as electrical carriers. Explain why a crystal with additional donor impurities will norally have more electrons in the conduction band than holes in the valence band, still preserving the overall electrical neutrality of the crystal.
The Attempt at a Solution
Well, the first part is easy. A pure (intrinsic) semiconductor isn't doped with any acceptor or donor atoms so it can't have an unbalanced number of electrons and holes. It's electrically neutral.
But how can an extrinsic semiconductor be neutral? If it is doped with donor atoms, that means those atoms bring an extra electron when they replace a silicon atom in the silicon lattice. Electrons would then be the majority carriers and holes would be the minority carriers. Therefore, the material would have a net negative charge (n-type extrinsic semiconductor).
The only explanation I could think of was that the atoms were neutral (the silicon and the donor atoms) before there was doping, and so after doping they would still be neutral, because you can't create charge. For example, if phosphorus atoms are neutral before being doped into silicon, they would become positive upon integrating with the lattice (because they give up the fifth valence electron) and the electron would be free to move through the material (and thus its negative charge is still with the material). If this is the case, it contradicts the argument in the previous paragraph.
Am I crazy or is what I'm saying sound at all coherent?