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Why are Fermi levels in metals lower than the Fermi level of $\text{p}$-substrates in metal-oxide semiconductors (MOS)? Specifically, why is there a difference of $qV$?

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The metal gate and the semiconductor, separated by an insulator in the MOS structure, can be each be considered to be a system in thermodynamic equilibrium und thus have a separate electrochemical potential, i.e. Fermi level. When you apply a voltage $V$ between the systems the Fermi levels will be separated energetically by $qV$. When the metal gate and the semiconductor of a MOS structure are short circuited, they constitute a single system with one Fermi level. This corresponds to the Fermi levels being aligned for an applied zero voltage $V=0$.

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  • $\begingroup$ Firstly thank you. i did't understand the part that when bi apply bias. Electric field can bend the band but it doesn't change fermi level (as far as i know.). But the metal side fermi level below dropped by qV. Bias voltage can change metal fermi level? $\endgroup$ – user196355 May 23 '18 at 9:40
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    $\begingroup$ @NailTosun - When there is a potential difference between two thermodynamic systems, of which each is in equilibrium (here the gate and the semiconductor bulk), the chemical potentials (Fermi levels) of these systems will differ by $qV$. $\endgroup$ – freecharly May 23 '18 at 22:35

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