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In energy band diagram of a semiconductor, do energy levels such as $E_c$, $E_v$ have negative values?

Also, why electrons in semiconductor have energy?

What is the formula for energy of an electron?

Thank you.

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    $\begingroup$ The conduction and valence bands are bound states in a semiconductor, so, yes, they are negative with respect to infinity. Why would electrons not have energy in a semiconductor? $\endgroup$ – Jon Custer Jun 20 '16 at 13:46
  • $\begingroup$ Thank you. I don't understand where the electron energy come from. Is it because electron vibrate around nuclear? $\endgroup$ – anhnha Jun 20 '16 at 15:14
  • $\begingroup$ Where does the electron's energy come from in an atom? Start there before going to a bunch of atoms together (aka a solid). $\endgroup$ – Jon Custer Jun 20 '16 at 15:26
  • $\begingroup$ Here is what I have in mind. Electrons carry negative charges and nuclear carries positive charges so nuclear will attract electrons. And this accelerates electrons and give them energy. Is it correct? $\endgroup$ – anhnha Jun 20 '16 at 15:34
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The values of $E_c$ and $E_v$ in the band diagram depend on the point of reference. So yes they can have negative values if you chose your reference that way. Keep in mind that their difference $E_g$ stays constant nonetheless.

Electrons are fermions and therefore governed by Fermi-Dirac statistics. That means that they have to comply with the Pauli principle. The distribution function reads:

$f(E)=\frac{1}{1+e^{\frac{E-E_F}{k_B T}}}$

This equation in this very form is only valid for semiconductors in equilibrium. $E_F$ is called the Fermi-level and determines where the turning point of the function is. In the following figure the Fermi level would be at 5 eV.

Fermi-Dirac distribution

(Source: http://www.doitpoms.ac.uk/tlplib/semiconductors/images/fermiDirac.jpg)

In an intrinsic semiconductor (=no doping) the Fermi-level typically lies in the middle of the bandgap.

The Fermi-Dirac distribution gives the probability that a state at a certain energy is occupied by an electron.

States are allowed combinations of energy, momentum and spin the electrons can occupy. The distribution (in terms of energy) of such states in the semiconductor is described by the Density Of States (DOS) $D(E)$, which depends on things like the lattice structure and binding energy. In the bandgap for example are no such states, which is why electrons can simply not be there.

These states are simply bound states in the semiconductor lattice, which is why electrons have energy. If you solve the Schrödinger equation for a periodic potential (see Kronig-Penney model and Bloch functions) you will get certain energy eigenstates.

Periodic potential

(Source: https://upload.wikimedia.org/wikipedia/commons/thumb/0/04/Periodic_square_potential_130707.png/600px-Periodic_square_potential_130707.png)

You can for example now calculate how many electrons are in the conduction band.

$n=\int_{E_c}^\infty f(E)D(E)\ dE$

(n is called carrier density)

Since the description of electrons in semiconductor heavily depends on statistics it makes no sense to ask for an equation for the energy of a single electron.

I hope this answers your questions!

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  • $\begingroup$ Thank you very much for the detailed answer. I have a question relating to periodic square potential above. Hope you could explain it more. Potential at a point is the work to bring an unit charge from infinity to that point. So for the positive ion as above, should potential all positive instead of negative as in the picture? $\endgroup$ – anhnha Jun 21 '16 at 3:12
  • $\begingroup$ That depends on the energy reference level, which you can set to be wherever you want. In the case of the figure the reference level was set to be at the upper edge of the potential wells. It would not make any difference for the calculations if the level was set somewhere else so that the whole potential structure has positive potential. $\endgroup$ – counter Jun 21 '16 at 7:57

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