When we dope a silicon lattice with phosphorus the extra electron in phosphorus finds itself a state in the conduction band while no hole is created thus increasing electron density while not changing hole density. However, the nucleus of the phosphorus also has an extra proton, does this have no effect? I suppose it would perturb the band structure by a small amount (probably not noticeable) but does it not register as a hole because it is local?
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1$\begingroup$ Well, the electron finds itself in a state just below the conduction band, and can the hop easily into the conduction band. But, clearly, the state it is originally in is perturbed from the rest of the crystal. $\endgroup$– Jon CusterCommented Dec 12 at 15:29
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3 Answers
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The binding energy of the electron to the phosphorus is 44 meV. Phosphorus behaves like a scaled hydrogen atom with the complete set of excited states. At low temperatures the electron is bound but at room temperature the point defect is ionised.
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Regarding the extra proton in the phosphorus nucleus, you're correct that it does have an effect, but a small one. The additional positive charge of the proton slightly perturbs the potential energy landscape of the lattice, which can affect the band structure.
However, this perturbation is typically localized around the phosphorus atom and doesn't significantly impact the overall band structure. The reason it doesn't register as a hole is that the extra proton doesn't create a localized state within the valence band; instead, it slightly modifies the existing band structure.
This effect is often neglected due to its relatively small magnitude compared to the dominant effects of doping.
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the extra electron in phosphorus finds itself a state in the conduction band while no hole is created thus increasing electron density while not changing hole density
Not true. Through the law of mass action the hole density decreases then you dope with phosphorus. (When non-degenerate)
However, the nucleus of the phosphorus also has an extra proton, does this have no effect?
It has an effect: a fixed charge is added to the semiconductor lattice. This is particularly relevant, for example, in the depletion region of a pn junction.
