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What does the Energy band diagram in a semiconductor look like if we introduced a doping gradient? What would it look like if after we added the doping gradient we introduced an electric field?

I know that an electric field in an intrinsic semiconductor tilts the band diagram. I also know that an electric field in a uniformly doped semiconductor tilts the band diagram but with the Fermi Level closer the conduction band. I don't however know what the band diagram would look like in the two cases I mentioned at the very beginning.

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    $\begingroup$ In equilibrium, the Fermi energy is constant. So, for the first part, how does the Fermi energy vary with doping (see for example Sze, chapter 1). Adding an external field does the same thing whether the material is doped or not. $\endgroup$ – Jon Custer Oct 22 '14 at 14:37
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The answer to your question can be a little unintuitive at first so I'll try and explain it as qualitatively as possible.

When you introduce a dopant concentration gradient into a piece of semiconductor, there will initially be a diffusion of carriers from the areas with higher carrier concentration to the areas of lower carrier concentration. Ignoring the diffusion of minority carriers, this results in a charge imbalance as ionised impurities are immobile and cannot diffuse together with the charge carriers.

As a result of this, the ionised impurities attract the charge carriers, establishing an electric field within the semiconductor at thermal equilibrium. Hence the conduction and valence bands are tilted.

However, due to the diffusion of charge carriers, the Fermi level throughout the semiconductor is constant. Due to the tilting of the conduction and valence bands however, its distance from the two bands varies throughout the semiconductor.

Hope this answers your question.

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