Minority carriers in a Schottky diode

[AMsolar]

Say I have a schottky diode with n type semiconductor. Why can't electrons from the metal flow to the valance band of the semiconductor?

If I reverse bias the schottky diode, the valance band may be higher than the fermi level at the contact, and electrons should be able to flow from metal to the semiconductor. Consequently the reverse saturation current should increase. Could anyone explain?

Thanks

 

[spf]

Say I have a schottky diode with n type semiconductor. Why can't electrons from the metal flow to the valance band of the semiconductor?

The electrons cannot flow from the metal to the valence band of the semiconductor because the valence band of the semiconductor is full. It's a n-type semiconductor, so it has an excess of electrons and thus electrons in the conduction band but no holes in the valence band.

In case you meant the conduction band and not the valence band of the semiconductor : the electrons cannot flow from the metal to the conduction band of the semiconductor because a potential barrier forms at the contact between the metal and the semiconductor. In theory, it's height can be calculated as the difference between the work function of the metal and the electron affinity of the semiconductor. (In practice, the height depends more on the microstructure of the contact layer.) In order to cross this barrier, the electrons either have to tunnel through it or be thermally activated to jump over it. Only extremely few electrons manage to do it.

If I reverse bias the schottky diode, the valance band may be higher than the fermi level at the contact, and electrons should be able to flow from metal to the semiconductor. Consequently the reverse saturation current should increase. Could anyone explain?

It doesn't matter how much you reverse bias the diode, the potential barrier height (as seen from the side of the metal) always stays the same as it is caused by the surface charge double layer at the contact. Thus, the Fermi level of the metal which raises with increasing reverse bias pulls up the edge of the conductance band correspondingly. This is for example shown in Figure 2.9 in

http://gorgia.no-ip.com/phd/html/thesis/phd_html/node3.html#SECTION00321000000000000000

However, even if the height of the potential barrier stays the same, I guess (but I'm not sure) that its width becomes smaller making the tunneling more probable and thus increasing the current. (Maybe somebody else can comment on this?) Finally, if you increase the reverse voltage high enough. a breakdown occurs which usually destroys the diode, see

http://en.wikipedia.org/wiki/Avalanche_breakdown

Very interesting is also

http://www.dei.unipd.it/~gauss/Download/SCHOTTKY.pdf

Hope that helps.