Why carriers are free to move from dopant ions?

Question

I apologise in advance for not finding the answer myself. I spent some time googling, but was only able to find a few related questions but no direct answer to my one. I also have to add that I'm not a student and study semiconductors in my free time.

I wonder why the majority carriers in a semiconductor have such a freedom of movement and not stay in the vicinity of the dopant atoms?

I'll provide a specific example: a Silicon crystal doped with Phosphorus. Phosphorous does indeed have an extra electron that can serve as a majority carrier, however, it also contains one proton more which attracts the electron.

Doesn't that mean that even if an electron manages to escape far enough for the electrostatic forces to be negligible, the extra proton not moving anywhere would constitute a local positive charge that would attract other passing by majority carriers and eventually trap one of them?

I would also imagine that the electrostatic force attracting electrons to the Phosphorus ion should be much greater in its direct proximity than any reasonable bias voltage applied to the far-away terminals.

Answer 1

• The electron in the $\mathrm P$ atom is strongly held on to the atom. It is very tough to remove it. This is evident from the band gap being large.
• As a dopant, we must remember the effect of electromagnetic shielding. The extra proton in the atom does indeed cause a larger attraction for the electron to stick around - but it also causes larger attraction of the neighbour atoms' electrons. They might skew around slightly, moving slightly closer. They then are in competition with the excess electron, who then feels a smaller resulting attraction.

All in all, from the excess electrons perspective, the attraction to that dopant atom is still larger than the attraction towards any other atom. But not much larger due to the above shielding effects. This is evident from the band gap now being smaller.

For a good, matching dopant atom, the band gap reduces to just the right size that matches which ever excitation you want to apply to your material. If no energy is added, then this slightly larger attraction will indeed cause electrons to "stick around" dopants in a slightly larger concentration. But as soon as a slight amount of, say, thermal energy or a matching low-energy photon or alike appears and adds energy to the material, then this excess electron easily "rips free" from this weak attraction and becomes excited.

Answer 2

Phosphorous does indeed have an extra electron that can serve as a majority carrier, however, it also contains one proton more which attracts the electron.

The energy level associated with the localized state around the phosphorus nucleus is very close to the lower edge of the conduction band (We basically choose phosphorus as a dopant because the states it provides line up nicely this way). This means

• It only takes a small thermal excitation to promote the bound electron into the conduction band.

• The energy is above the Fermi energy of the electrons in the lattice, therefore the bound state is likely to be unoccupied.

But these are really two ways of saying the same thing, since if the state isn't occupied, the reason for that is the electron got thermally excited into the conduction band.

If the state is unoccupied, then the only (if we did things right) states it can go to are the conduction band states. Even though the conduction band states are also above the Fermi level and thus unlikely to be occupied, there are so many more of them than there are bound states that the small fraction of c.b. states that are occupied can account for essentially all of the bound states being unoccupied.