For metal-oxide dielectric-semiconductor Schottky diode and metal-semiconductor Schottky dioode, I found that the turn-on voltage of the former was much larger than that of the latter. Is it due the band offset of dielectric and semiconductor only? In general, the breakdown voltage could be improved, and is there any other changes?
Hi Zeng,
For metal-oxide dielectric-semiconductor (MOS) Schottky diodes the quality of the interface between the metal-oxide and the semiconductor is very important, as well. As such, the number of traps at the interface will strongly influence the turn-on voltage of such diodes. Obviously, this is not the case for metal-semiconductor (MS) Schottky diodes.
I hope it helps!
Regards,
L.A. Majewski
The MOS structure is totally different from the M-S Schottky diode. The turn-on voltage of the former is composed of different components and different theory of operation. The MOS threshold voltage is given by:
VT = VFB + 2fiF - Qox /Cox = VFB + 2Vt ln(Nd/ni) +√(4e.epsilons.Na.fiF) /Cox
where VFB is the flatband voltage, Qox is the oxide charge and Na is the semiconductor doping.
Look at my free download Book (Fundamentals of Electronic Devices) in researchGate for details.
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