# More physical explanation of impurity energy levels in a doped semiconductor?

I'm reading about doped semiconductors in Ashcroft and Mermin. They tell you that when donor impurities are added to a semiconductor, their energy level $E_d$ is just slightly below the conduction band energy $E_c$, like in this picture:

Their explanation for this is as follows:

1. The donor electron is weakly bound to its donor ion, so its binding energy is very small.
2. The donor electron's state is formed from conduction band states.
3. Thus, the donor level's energy has to be within the binding energy of the conduction band energy, because that's what energy the donor electron has.

Part 1 makes total sense to me. It's mostly parts 2 and 3 that don't -- I don't really know what they mean by "...the conduction band levels from which the bound impurity level is formed..." means. Earlier, as an explanation for why the binding energy of the donor electron is very small, they also say:

I don't really get what they mean by the "Superposition" statement. I can kind of guess: the conduction band states form a complete set, so it's formed from them. But I don't really know why, even if that is true. Does anyone have a more intuitive, physical explanation of this?

Thanks!

• Can you clarity statements 2 and 3, do you have reference for that so I can understand better? – boyfarrell Dec 2 '13 at 10:30
• @aboyfarrell, it's on pages 578-579 in Ashcroft and Mermin. – YungHummmma Dec 2 '13 at 14:24
• I don't have a copy. Could you summarise the point in a bit more detail? – boyfarrell Dec 2 '13 at 14:44
• @aboyfarrell, page 578: i.imgur.com/yysfbs7.png and page 579: i.imgur.com/usBcSw7.png – YungHummmma Dec 2 '13 at 15:26