Why do we assume complete ionization of donor energy level (any doping level) even if we know that when Ef will be closer to it, the probability of electron finding will increase?

Context- DISCUSSION: You may have noticed that as Na increases, E_f rises toward E_d and the probability of nonionization can become quite Targe. In reality, the impurity level broadens into an impurity band that merges with the conduction band in heavily doped semiconductor (i.e., when donors or acceptors are close to one another). This happens for the same reason energy levels broaden into bands when atoms are brought close to one another to form a crystal . The electrons in the impurity band are also in the conduction band Therefore, the assumption of n Na (or complete ionization is reasonable even at very high doping densities. The same holds true in P-type materials.


1 Answer 1


Donor and acceptor levels are close in energy to the conductor and valence band, respectively. Thermal energy can completely ionise the levels and promote electrons to the conduction and holes to the valence band. This is a good assumption at room and moderately cold temperatures.

Indeed, when temperature of a extrinsic semiconductor is lowered, thermal energy can no longer ionise the dopants and the carrier density reduces. This is called dopant freeze out.

See the section here on https://ecee.colorado.edu/~bart/book/book/chapter2/ch2_6.htm

So think of $E_f$ as being determined by the dopants in the first place, not as a free parameter which can cause occupancy to occur near a given state.

Bands in semiconductor are a result of the periodic potential of the crystals lattice. At very high doping/impurity concentration wavefunctions of neighbouring dopant/impurity atoms begin to overlap and become quasi periodic. Pauli exclusion principle applies and the energy levels split and band formation occurs.

  • $\begingroup$ Would be interesting to post a reference so I can understand better. $\endgroup$
    – boyfarrell
    Commented Dec 24, 2018 at 20:15
  • $\begingroup$ That additional context makes things a lot clearer in terms of the physics. So what are you asking as the question? What’s the things that does not make sense? $\endgroup$
    – boyfarrell
    Commented Dec 25, 2018 at 10:36
  • $\begingroup$ Added some commentary on that $\endgroup$
    – boyfarrell
    Commented Dec 25, 2018 at 12:14
  • $\begingroup$ Just mean to say exactly that. $\endgroup$
    – boyfarrell
    Commented Dec 26, 2018 at 19:57

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