I can easily understand the Schottky barrier height for Au/n-type Si or Au/p-type Si, but I have not found anything about intrinsic Silicon. Does intrinsic Si form a Schottky barrier? If so, what is the barrier height in this case?
Schottky contacts are formed between the metals and semiconductors. They could be rectifying or nonrectifying. In case of rectifier a Schottky potential barrier is formed at the interface between the two materials. This barrier controls the passage of the charge carrier from the semicondcutor to metal.
There are compote chapters in the books of electronic devices treating this topic in details.
There are also the p-n junctions which work as rectifiers. If the doping at both sides of the junction is high such that the two materials are degenerate, the p-n junction turns to tunneling one.
In case that one side is p-type and the other side is INTERINSIC a p-i junction is formed which behaves always rectifying.
There is also the n-i junction which behaves as the p-i junction.
These two junction when formed such that they build pin diode will be rectifying and has many applications in power devices, radiofrequency switches and photo and radiation detectors.
The pin diodes are popular devises with number of applications.
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The The valence band edge Ev of silicon has the level= 4+1.1=5.1 eV,
If we contact the two materials, they will get in thermal equilibrium meaning flat Fermi level. As a consequence, the barrier height from metal to the conduction band edge= phim - X,
The barrier height from metal to valence band edge= phim- x-Eg which is very small and equal to 0.2eV, this means that the electrons can flow fro the valence band to the metal leaving holes at the surfaced of the semicondcutor. These electrons and holes will reside at the surface from both sides building tunnel junction. So most probable under ideal surface conditions the junction may not be rectifying.
Consult [1] and note that the barrier height depends on the doping state of Silicon (it also depends on the thickness of the gold contact [2]). For the theoretical considerations, consult [3] (specifically Chapter 2). For an important overview, consult [4] and the references herein (such as that to the work by Heine [5]).
if you contact a semiconductor with a metal, then their chemical potentials adjust. The Fermienergie of Au is 5,53 eV (above the band of conductivity. The Fermilevel of undoped Si lies nearly in the middle of the gap (depends something on temperature), e.g. ~0,55eV. If you draw the band of conductivity near the contact, electrons have to overcome a threshold of 0,55 eV to come from Au to Si. This voltage you must apply to reduce the threshold for an ideal contact.
The second part of the question: Undoped Si is an intrinsic conductor. At room temperature Si behaves like an isolator. For higher T the resistance reduces and a low current could flow. If you need a switch for high temperatures, your diode could be useful. Otherwise you have no pleasure with such a construction.
Sara Stofela : Have you adjusted the question? I cannot recall "INTRINSIC (undoped)" in the original question. Questions should not be adjusted following the first response, or responses lose their meaning, which is a disservice to those who take the trouble to respond - their responses will look odd to later readers.