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I have been chewing up some time ago the Schottky-Mott theory of Schottky Barrier height (which ignores the surface states). All the deduction seems to ground on fundamental thermodynamical principles (as the equality of Fermi levels- i.e. equality of chemical potential in equilibrium) but there is something which I can't clearly see and is one of the key points to calculate the height barrier: Why the energy bands on the semiconductor side just at the interface are assumed to be the same that the ones of the isolated semiconductor? It is clear that the band have to bend (because of the electric field) but I see no reason to why the bands values should "start" to bend from the original values (the values of the isolated semiconductor).

Thanks.

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  • $\begingroup$ To put it in other words, the question could be restated: why the electronic affinity is assumed to be the same at the interface than at the bulk of the semiconductor? $\endgroup$
    – Fernando
    Commented Jul 21, 2014 at 23:02

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Because that is the obvious place to start - the mental experiment of bringing two "bulk" materials together and seeing what happens. That is the straightforward, simple calculation to explore. I'm sure Schottky and Mott knew that it did not capture all possible issues, but you do the simple first and see how bad it is. As it turned out, the theory was pretty good at explaining a number of Schottky barriers generated in the lab. It wasn't good at explaining them all, hence the more complex theories including pinning at surface states, etc. that then had to be developed. But, those are built upon the underpinnings of Schottky-Mott theory.

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  • $\begingroup$ I somehow figured out that this heuristic approach was the case, it seemed too naive to me tough and I struggled to refuse it. Your answer, which I'm inclined to embrace, gives me an awful taste of not having a solid theoretical ground to determine the height of the barrier; additionally, I'm not comfortable with the claiming that the electronic affinity is the same even when another conductor is influencing the surface effect; it's just a non physical assumption. Thanks for the answer. $\endgroup$
    – Fernando
    Commented Aug 1, 2014 at 11:12

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