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In a P-N junction depletion region is developed when electrons from N-region diffuse to P-region and hole from P-region diffuse to N-region. Here diffusion thereby development of depletion is caused by the movement of majority carriers.

Schottky diode has a metal and N-type semiconductors junction. In both metal and N-type semiconductor majority carriers are electrons. Then how a barrier is developed at the metal-semiconductor junction of a Schottky diode?

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This is the explanation I was given several decades ago when I was spotty PhD student - actually it might have been Neville Mott himself who gave the lectures as I think he was still active in the early 80s. I was actually working on silver-germanium selenide contacts, but the effect is common to all metal-semiconductor contacts including Schottky diodes. If you feel the urge to follow up this answer by asking me deep and searching questions note that (a) this was thirty years ago and (b) I'm not sure I ever grasped exactly why interfacial states would pin the band gap.

Anyhow, if you look at the band structure of metal and the semiconductor separately, i.e. not in contact, they look something like:

Schottky1

It doesn't matter exactly where the Fermi level lies wrt to semiconductor band gap, though in every picture I've see it's drawn near the top of the gap. The argument is that when you contact the metal and semiconductor this creates states at the interface that lie in the band gap, and these interfacial states pin the middle of the band gap to the Fermi energy of the metal. The bands in the semiconductor curve to accomdate this:

Schotthy2

and the result is that there is an energy step, $\Phi$, between the metal and semiconductor. So electrons from the semiconductor flow into the metal creating a depletion zone in the semi conductor. This behaves like a PN diode where the metal is analogous to the P side. Electrons flow freely from the semiconductor to the metal, but it's difficult for electrons to flow from the metal to the semiconductor because they need enough energy to jump the energy step, $\Phi$.

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  • $\begingroup$ The bulk region should be a good conductor. This is the reason why one dopes the semiconductor (positively or negatively). This shifts the Fermi-potential near the valence or conduction band. $\endgroup$ – Tobias Mar 19 '14 at 16:10
  • $\begingroup$ @Tobias: aha, thanks. Any explanations for how interfacial states move the energy bands so the centre of the band gap lines up with the Fermi energy? That's the bit I never grasped. $\endgroup$ – John Rennie Mar 19 '14 at 16:18
  • $\begingroup$ Only very roughly. The disturbation of the crystal structure at the interface zone introduces new admissible levels in the gap. These can be occupied by the conduction band electrons. Therefore, the energetic level where 50% occupation propability lies, i.e. the Fermi potential, shifts down relatively to the semiconductor bands (that is the first statement in my answer). $\endgroup$ – Tobias Mar 19 '14 at 16:29
  • $\begingroup$ @john If understood in the right way, Energy level in the semiconductor is shifted by the influence of metal. Then how it is possible for conduction band level to go above the fermi level in the metal. $\endgroup$ – tollin jose Mar 20 '14 at 3:55
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    $\begingroup$ @user129412: the $x$ axis is distance normal to the interface. However this is a schematic diagram so don't take it too literally. I don't know how the Fermi levels of the metal and semiconductor compare, but whenever I've seen these diagrams drawn the Fermi energy of the metal is always drawn somewhere in the semiconductor band gap. $\endgroup$ – John Rennie Jan 16 '16 at 6:36
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The band structure at the surface between metal and semiconductor is disturbed. This creates additional intermediate mid-gap states and the electrons from the conduction band near the surface "fall" into these levels. The Fermi-niveau drops down w.r.t. the conduction band. But, we know that the Fermi-niveau is constant at thermodynamical equilibrium. Thus, the higher distance between conduction band and Fermi-niveau can be displayed as a rise in the band potentials of the conduction band (and the corresponding valence band). Because of the potential barrier the electrons are driven back into the semiconductor and you get a positively charged depletion zone in the semiconductor bulk near the contact zone.

But, that is all described on the Wikipedia page on Metal induced gap states and the Wikipedia page on Shottky-barriers with nice pictures.

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  • $\begingroup$ Thank for your help. But your answer seems to be complicated for me. May be because my basic knowledge in this field is not good enough. $\endgroup$ – tollin jose Mar 20 '14 at 3:59

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