# Why is the carrier distribution / density of states different in doped semiconductors?

How can it be explained that the carrier distribution is different in case of doped semiconductors? I understand it naturally, that the distribution of electrons is higher for n-type semiconductors. But is it also possible to explain it via the density of states or fermi level or so? I guess it should as the second column represents the density of states and there you can see that the number of electrons is rising with with the density of states which is rising in return with the energy.

I know the DoS for electrons is:

$$g_c(E) = \frac{2m^*}{h^2}^{\frac{3}{2}} \cdot \sqrt{E}$$

in case of electrons, $$\sqrt{E}$$ is $$\sqrt{E - E_C}$$

and then? What is E (exactly)? • The density of states in the conduction/valence bands is the same (for reasonable levels of doping). The occupation of those states differs greatly because of the doping. Why that is will be explained earlier in the chapter most likely... – Jon Custer Aug 16 '19 at 19:17
• I'm really unclear on what the question is. – Gilbert Aug 17 '19 at 12:25

In your formula, $$g(E)$$ is the density of states for an electron gas and $$E$$ is the energy of a state. An undoped semi-conductor cannot be approximated as an electron gas. To use this DOS, one needs to dope it and the extra holes/electrons in the valence/conduction bands could be considered as an electron gas. Then the formula can be applied.
• I'm not sure I understand your question, but the actual shape of the DoS is material dependant. For a doped semi-conductor, the DoS in the conduction band could be approximated as the one for an electron-gas where $g(E)\propto \sqrt{E-E_C}$. But this woule be only valid close to the band edge. Elsewhere it might be very different. – fgoudra Aug 17 '19 at 15:04