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There are no spare positive/negative charge carriers on the metal side of the diode, so the spare positive/negative charge carriers on the semiconductor side can not diffuse into the metal and recombine with spare negative/positive charge carriers. Therefore no potential barrier will be formed by ions on each side.

So how does a Schottky diode work?

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  • \$\begingroup\$ "Therefore no potential barrier will be formed by ions on each side." Thanks for playing, here's your consolation prize. \$\endgroup\$ Jul 29, 2014 at 21:46
  • \$\begingroup\$ @JYelton I'm not sure if this is a duplicate. The linked question is asking what it is and where it is used. This is asking how it works \$\endgroup\$
    – Funkyguy
    Jul 29, 2014 at 22:40
  • \$\begingroup\$ @Funkyguy A valid point, it may not be a duplicate after all. \$\endgroup\$
    – JYelton
    Jul 29, 2014 at 22:41
  • \$\begingroup\$ @JYelton we'll see what others think \$\endgroup\$
    – Funkyguy
    Jul 29, 2014 at 22:42
  • \$\begingroup\$ The potential barrier is formed by the work function difference in the two materials, not from "spare" carriers. The diffusion of carriers towards and away from the junction formed from dissimilar materials is what causes the depletion regions, externally applied fields is what bends the bands by the potential barrier is a material effect. \$\endgroup\$ Jul 29, 2014 at 22:52

2 Answers 2

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First it depends on the type of semiconductor(SC) n and p types are different. I'll just do n-type. (p-type is the mirror image of n-type). Then, in theory, it only depends on the relative work function of the metal and the SC. If the metal is below the SC then electrons can flow and no diode. But if the metal is above the SC then you get a depletion region in the SC, and a diode.

Oh, in practice other things, like surface states play a role. (At least that's my limited understanding.)

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You can explain it in more better way. In the metal which acts as the anode an in the n-type semiconductor which act as the cathode, electrons are the majority carriers. In the metal, the level of minority carriers (holes) is not significant. when the metal and semiconductor are joined to form the junction, the electrons in n type material will immediately flow in to the adjoining metal. This will establish a heavy flow of majority carriers.In the conventional p-n junction, the minority carriers are injected into a the adjoining region.But here electrons are injected into a region where electrons only are the majority carriers. In schottky diodes, the conduction entirely is due to the majority carriers. That's how it differs from the conventional P-N junction diode. I hope this helps.

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