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Normally, no electron energy states exist in the band gap, the gap between the valence band and conduction band in a semiconductor.

However, if we dope the semiconductor, i.e. add donor (n type) or acceptor (p type) atoms to it, we introduce new electron energy states in the band gap!

Take for example silicon, in which we introduce phosphorus, which is a group V element and thus a donor atom. This will introduce extra filled electron states just below the conduction band.

Now, this all happens at 0K, so no current can flow (this is logical as electrons don't move at this temperature, even with an electric field applied). But if we raise the temperature e.g. until room temperature at 300K, the electrons gain energy and can jump into the free energy states in the conduction band. These electrons in the conduction band can now conduct electricity.

Now, my question is: why can't the electrons in the band gap conduct current? Why do they have to be in the conduction band in order to conduct electricity?

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The electron states in the band gap are localised, whereas the states which contribute to the bands are not.

The electron states in a solid are not simple, there's a lot of non-trivial quantum mechanics going on. The electron states in a free atom are localised around the atom - the electrons in those states can't leave the atom without a lot of energy, so they can't conduct anything. When you pack lots of atoms together, the surrounding electron states overlap and mix. Which states mix with each other is dictated by the energy: similar energies means more mixing. You end up with a new set of states which extend over the whole block of material. If the material has a periodic lattice, these electron states group together into bands.

Every state in a band has some velocity (called a Fermi velocity) associated with it, and an electron in that state can be thought of as moving through the material with that velocity. The Fermi velocity of electrons in the conduction band is very large, but because the electrons are all going in different directions, there is no net current. An applied electric field moves some electrons from states which were going in one direction, to states going in the other. In a metal, one of the bands is part full, so there are plenty of nearby states to move electrons into. In a semiconductor, there is a gap between a full band and an empty one so it's much harder to push electrons into the higher band.

When dopants are added, they don't form a nice periodic lattice and they are much more spread out than the silicon atoms that host them. This means that the electron states around the dopant can't mix with states from other dopants to form a band. Since the energy levels of the dopant states are different from the silicon states, they don't mix (much) with them either. Instead, the electron states are localised around the dopant, much like the states around the free atom. An electron in that state can't conduct in the way one in a band can. It either has to jump up into the band, or jump to another nearby dopant. The former happens in semiconductors, the later is known as incoherent transport, and appears in some other materials.

I'm not sure how well I've explained this, but if you don't get a clear answer here, you could try the physics stack exchange. This definitely feels more like condensed matter physics than electrical engineering!

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  • \$\begingroup\$ So, basically, these newly introduced states are discrete steps and are all occupied (so no electron can jump to another state), while the conduction band is continuous so there are a lot of states where electrons can jump around in? Thanks a lot! I didn't know about physics stackexchange, will post there in the future :D \$\endgroup\$ – Robbert May 17 '16 at 21:07
  • \$\begingroup\$ Yep, that's the crux of it. For electron donor dopants anyway. For p doping, they are all empty and just above the valance band, but it's the same idea. \$\endgroup\$ – Jack B May 18 '16 at 9:54

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