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If a metal and a semiconductor are connected, they are going to equilibrium. For this, the Fermi level has to be equal. So far I understood that this causes a charge exchange till said equilibrium has been reached. As a result, the band edges in the semiconductor are either bend up or downwards, depending on its doping. The Schottky barrier is the difference between the metal work function and the semiconductor electron affinity.

Why?

Why does the Schottky barrier need to remain constant?

JRE
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Tobias
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1 Answers1

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I have thought about it for a while and I think I have found the answer myself.

The Schottky barrier is defined by: \begin{equation} \Phi_B = \Phi_M - \chi \end{equation} When I apply a voltage to the metal, the Fermi-level gets moved either up or down. But applying voltage to a metal doesn't change its work function. The same with the semiconductor. Applying a voltage to a semiconductor doesn't change its electron affinity. If neither of the values change nor can the Schottky barrier value.

Tobias
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