Feb26-13, 11:42 AM
Hello, I'm studying digital integrated circuits and I'm new to solid state physics. I've studied PN junctions, drift and diffusion currents, now I'm trying to see these subjects in terms of energy bands and I'd really appreciate it you could explain to me a couple concepts.
When two materials contact each other, at thermal equilibrium their Fermi energies must equate. I'm ok with this if I see a band diagram and I understand that if this wasn't true a pn junction would not exhibit any built-in voltage. But what's the physical reason for this equation?
In a p-dope semiconductor, the Fermi energy is smaller than its intrinsic value; that is, it gets closer to the upper bound of the valence band.
I think of it like this: in a p-dope region it's easier, for an electron that leaves its atom (for example, a silicon atom), to get catched by a positive ion (for example, a boron atom) than to "jump" out of the valence band and become a mobile charge carrier. Is my interpretation correct?
My last question is about energy bands in a MOS system with a p-dope semiconductor.
The book I'm studying says:
"Because of the work-function difference between the metal and the semiconductor, a voltage drop occurs across the MOS system. Part of this built-in voltage drop occurs across the insulating oxide layer. The rest of the voltage drop (potential difference) occurs at the silicon surface next to the silicon-oxide interface, forcing the energy bands of silicon to bend in this region."
Here's the band diagram: Link
The metal and semiconductor Fermi levels match, but the oxide Fermi level doesn't show up; plus, the oxide work-function isn't even mentioned. Don't these two parameters have any influence on the bending of energy bands?
Thanks in advance.
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