From the link below, they explain when an electron in the valence band is promoted to the conduction band it leaves behind an electron hole. But in the case of the image below, we can see when Si is doped with Boron its said to create an electron hole. But no electron leaves the valence band of Boron, but rather SI forms bonds with the 3 valence electron on Boron. So my question is, is the positive charge of the electron hole hear essentially just due to the preferential configuration of an octet on the valance band e.g having 4 covalent bonds on B is more stable? Because if it were to the case that a promoted valence electron to conduction band leaves a positive charge, then would Sb have a positive electron hole since an electron has left it?
2 Answers
I think you have the right idea. The four covalent bonds that the Si atoms make in the crystal is the preferential configuration. If you introduce a dopant like Sb, it has five valence electrons. Four of them basically act like the valence electrons of Si, and the fifth one is relatively free to conduct (out of the valence band). So, there is no "hole" left, because that essentially would mean a missing electron in one of the four covalent Si bonds. The Sb atom replaces the Si atom at the site in the crystal.
More details in Kittel Solid State (p. 209): https://archive.org/details/IntroductionToSolidStatePhysics/page/n228/mode/1up
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$\begingroup$ But what about the missing electron from Sb, is that not considered an electron hole anymore? $\endgroup$ Commented Jun 16 at 13:33
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$\begingroup$ That is correct. It would only be a hole if it was missing from one of the four covalent bonds (like the p-type situation). $\endgroup$– ad2004Commented Jun 16 at 17:31
Rather than a Lewis dot structure approach, the better approach to answer your question is through molecular orbital hybridization and band energy levels.
All bonds are covalent. The Si, Sb, and B are sp3 hybridized. Consider the n-type Sb versus the p-type B.
The Sb has one additional electron. This additional electron goes to an anti-bonding lowest unoccupied molecular orbital (LUMO) between Si-Sb. It is a bound electron and is not free to carry current. The occupied LUMO lies in energy just below the conduction (LUMO) band for Si. The electron is promoted out of the bound state into the conduction band and can carry current. The hole that is left behind is locked specifically to the Si-Sb bond and cannot hop to other locations; it does not carry current. Hence, in n-type, only the added electrons carry current after they are promoted out of their anti-bonding LUMO state.
The B has one less bonding electron. This absence creates a hole in the bonding highest occupied molecular orbital (HOMO) between Si-B. It is a bound hole and is not free to carry current. The unoccupied HOMO state lies in energy just above the valence (HOMO) band for Si. An electron is promoted out of the valence state into the hole. The electron is subsequently locked to the Si-B bond and cannot hop to other locations; it does not carry current. The hole that is left behind in the Si conduction band is free to carry current. Hence, in p-type, current is carried only by the holes that are created in the valence band after the hole in the dopant level is filled.
Additional notes with an energy level picture are posted at the link below.
