About the energy bands, fermi levels in PN junctions?

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I'm having a bit of trouble understanding what's going on with the energy diagram of a PN junction. For your normal undoped semiconductor at zero bias, if you're plotting the energy as a function of distance along the material then you've just got the two horizontal lines where the upper one corresponds to the conduction band, and the lower is the valence band, and their separation is the energy gap.

Now you P-dope one side and N-dope the other to make a PN junction, and the diagram gets a kink in the middle. The conduction and valence bands on the P-doped side are shifted up, and on the N-doped side they are shifted down, like in this diagram:

http://www.matter.org.uk/glossary/images/p-n_junction.gif [Broken]

1) I don't quite get why it is drawn this way.
When you put in, say, some donor impurity, then they form a level just below the top of the energy gap because the non-bonding valence electron left over only has a small jump to cross into the conduction band. What I don't understand is why this somehow shifts the whole n-type region down a bit in the energy band diagram. The same goes for the upward shift of the P-doped region.

2) The Fermi level is the highest occupied energy state, right? At zero bias, this is drawn as one horizontal light straight across the diagram. I get why it is just below the conduction band in the N-type region, but it is also just above the valence band in the P-type region. This I am not too sure about, because shouldn't it lie actually inside the valence band?

The acceptor impurities will form covalent bonds in the lattice but still have the capacity to form one more right? They just lack the valence electron they need to be able to do it. Hence, the hole where that last bonding electron would be.

So why is the Fermi level not precisely the top of the valence band in the P-doped region?
Why do the Fermi levels of the N- and P- type regions perfectly align at zero bias?

3) Why do they get pulled out of alignment when there is a bias applied?
 
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Answers and Replies

  • #2
DrDu
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To point 1) Adding a small portion of dopant shifts the Fermi energy (the chemical potential) dramatically up and down in case of a n and p dopant, respectively. If two such doped regions are brought into contact, electrons will flow so long from n to p until the potential difference makes up for the original difference in chemical potential (so that the electrochemical potential becomes equal). The electrons that move from n to p stem from a small sheet at the surface of the n conductor and fill the holes in a small sheet of the p conductors (so called depletion or space charge regions). In the graphic you are showing these regions are not drawn (or replaced by delta functions) so that the potential really seems to jump.
 
  • #3
DrDu
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To 2) The Fermi level is the limit of chemical potential as T->0. It coincides only with the highest filled level, if it falls into a continuum (i.e. if it doesn't fall into a band gap).
Inside the band gap, the Fermi energy is very sensitive to doping while inside the band, it is only little influenced. Formally this is due to [tex]\partial \epsilon_F/\partial N=n(\epsilon_F)[/tex], i.e. the change of the Fermi energy with the total number of charge carriers is proportional to the density of states at the Fermi energy. At non-zero temperature, this relation gets somewhat washed out so that the derivative isn't actually infinite inside the gap but still very large.
 

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