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I'm learning about the band structure of semiconductors, and I learn that there're genetically two types of semiconductors .For n type semiconductor it has a shallow donor level whose electrons can be excited to the conduction band and thus the electrons are the dominant charge carriers . And here come my questions:

  1. why we just don't care about the holes on the donor level? Does it mean that the donor level is too flat so that the effective mass of the holes on the donor level is too big to make it delocalized?if it is the case ,then why donor levels are flat?
  2. How these levels(donor levels and acceptor levels) affect the Fermi level? For example ,if we make the donor level shallower ,does it mean that the Fermi level will become bigger?
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why we just don't care about the holes on the donor level? Does it mean that the donor level is too flat so that the effective mass of the holes on the donor level is too big to make it delocalized?if it is the case ,then why donor levels are flat?

My understanding is that the donor is localized because it's associated with a certain donor ion site, at a specific location in the lattice. The donors are just too far apart for merging of their associated orbitals. Typically the number of donors will be less than $10^{-3}$ times the number of host atoms, and non-degenerate doping levels (what we normally study at first) the impurity concentration is generally less than $10^{-6}$ as a fraction of the lattice sites.

The donor levels are spread out in k-space because they're localized in position space. That just comes from Heisenberg's uncertainty principle in the $\Delta{}x\Delta{}p$ form.

If there's a connection to effective mass (or if you even can define an effective mass for localized states), that's beyond my knowledge.

How these levels(donor levels and acceptor levels) affect the Fermi level?

Since the donor levels are above the intrinsic Fermi level, we know the donor sites are mostly unoccupied. Since there's nowhere else for those electrons to go, we know they go into the conduction band (producing a small but non-zero occupancy rate). Since some CB sites are now occupied, we know the Fermi level must be shifted up closer to the CB edge.

For example, if we make the donor level shallower, does it mean that the Fermi level will become bigger?

That does seem likely, but the effect would be small. With our typical donor impurities, the ionization of the donors is already nearly complete, so they have just about as much effect as they possibly can.

On the other hand, consider if the donors were much deeper below the CB edge. Then the fraction of them ionized would be smaller, and the number of CB states occupied would be smaller as well, indicating a smaller shift in the Fermi energy. But even if the donor level were at the intrinsic Fermi level, you'd achieve 50% ionization, and still produce some CB carriers, with some corresponding shift in the Fermi level (to the point that it would further reduce the donor ionization until things balanced out)

You could also consider the effect of temperature. A shallower donor level would require less thermal energy to ionize, so it would remain effective at providing CB carriers at lower temperatures than a deep donor level.

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  • $\begingroup$ Thank you so much for your detailed answer! I just have one last question :can donors' orbitals just to some extent merge with the orbitals of the host atoms? For example ,we just replace one Si atom with a P atom in a perfect crystal of Si .Since the orbitals of P atom and of Si atom are not drastically different ,so if I just calculate the band structure of such a doped Si ,will there be a flat impurity band introduced by the P atom? If so ,then how can I explain this from the point of view that the donors are just too far apart for merging of their associated orbitals? $\endgroup$ – meTchaikovsky Mar 31 '17 at 14:34
  • $\begingroup$ @Kangyu, Bands only form when a large number of individual levels interact. A single level (introduced by a single impurity site) can't turn into a level by itself. The orbitals of P and Si may be quite similar but it doesn't happen to put that last donor level into the conduction band. If you have the right reference (Landoldt-Boernstein?) you can look up exactly what the level is. $\endgroup$ – The Photon Mar 31 '17 at 16:55

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