A short question relative to doping semiconductors: What defines the location of new energy states? For example, why does doping of n-type add energy states slightly below the conduction band (and not slightly above the valence band as it is in case of p-type?)?

edit: It's not quite directly connected but I found formulas to calculate the Fermi level in https://www.uwyo.edu/cpac/_files/docs/kasia_lectures/2-intrinsicanddopedsemiconductors.pdf (p. 10) But applying an external field also shifts the Fermi level, right?


I apologize as my previous answer to this question was wrong. Just to elaborate on the correct answer by @my2cts here is how to calculate the energy level in case of a donor impurity

First of all for an electron in a semiconductor near the bandedge, the electron behaves like a free electron except with a different mass, i.e., the effective mass $m^*$. Technically this effective mass has different values in different directions corresponding to the crystal structure but for simplicity assume it to be a constant depending on the material.

Now your donor atom donates an electron to the conduction band of the semiconductor and in turn becomes positively charged ion. Or viewed another way, this donor atom behaves like a hydrogen atom with a potential $$ U(r)= \frac{-e^2}{4\pi \epsilon r}$$ where we have replaced $\epsilon_0$ with the dielectric constant of the material instead. The ion becomes ionised when the electron reaches the conduction band (then it will be free to move around in the material) so instead of taking the ionisation to happen at $E=0$ for hydrogen atom, now we take it to happen at $E_c$ (the edge of the conduction band). Writing the Schrodinger equation analogously to the Hydrogen atom with our modifications we get $$\left[ -\frac{\hbar^2}{2m^*}\nabla^2 - \frac{e^2}{4 \pi \epsilon r} \right] \psi(r) = (E_d-E_c) \psi(r)$$ Where $E_d -E_c$ is the impurity level of the donor with respect to the conduction band edge level $E_c$ (which I believe is what you want to know). Now the solution can just be taken as for the hydrogen atom case with our modifications $$ E_d-E_c = -\frac{e^4 m^*}{2(4 \pi \epsilon)^2 \hbar^2} \, \frac{1}{n^2}, \quad n=1,2,\dots$$ For the ground state energy level, we get the following $$E_d=E_c- 13.6 \left( \frac{m^*}{m_e} \right) \left( \frac{\epsilon_0}{\epsilon} \right)^2 \, \text{eV}$$

A similar calculation can be carried out for acceptor case by taking the effective mass to be of a hole.

P.S. I have taken this explanation from Section 2.8 of Semiconductor Device Physics and Design by Umesh K. Mishra and Jasprit Singh

  • $\begingroup$ Indeed. Any good semiconductor physics book will have a chart of known impurity levels in silicon. Some are near band edges, some are mid-gap, and everywhere in between. Some impurities have multiple states in the gap, depending on their charge state. $\endgroup$ – Jon Custer Jun 10 '18 at 14:36
  • $\begingroup$ Thanks so far but what exactly characterizes the final energy state? Or the other way round: How can I know which impurity will add energy states closer to the valence or conduction band? $\endgroup$ – Ben Jun 13 '18 at 9:34
  • $\begingroup$ @Ben I have completely changed my answer to complement @my2cts’s answer as that is the correct one. The choice of a material whether it will be a dopant or acceptor depends on whether it can accept electron or donate one and then the relative position with respect to the band edge can be calculated by this method $\endgroup$ – ravjotsk Jun 17 '18 at 12:12
  • $\begingroup$ Thank you! Looks quite straight forward, however I'm not looking forward to calculate the Schroedinger equation myself :) $\endgroup$ – Ben Jun 18 '18 at 9:35

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