According to Fermi-Dirac statistics, in a metal, only certain number of valence electrons take part in conduction when they acquire an energy equivalent to KT for some temperature. Now my questions is: How is this different from conduction in a semiconductor where the concentration of conduction electrons also depends on the temperature due to breakage of covalent bonds? Why should resistivity due to electronic vibrations play a larger role in a metal than in a semiconductor? Note: I have just finished an undergrad course on quantum mechanics where we dealt with conduction in metals.
Electronic – Difference in conductivities of metals and semiconductors based on quantum theory
conductivitymetalsemiconductors
Related Solutions
That's not how I learned it. It has to do with energy gaps between valence band where the electrons at the highest energy are, and the conduction band, where an electron can come free of its atom. In metals these bands overlap and electrons move free within the metal's lattice, and that's what gives metals their typical shine.
Pure semiconductors are isolators at cryo-temperatures. But doping them with N-type material will cause doping atoms to bond with the semiconductor, and then there's one superfluous electron which, like in metals, can move freely through the lattice, and thus conduct electricity.
A useful way to visualize the difference between conductors, insulators and semiconductors is to plot the available energies for electrons in the materials. Instead of having discrete energies as in the case of free atoms, the available energy states form bands. Crucial to the conduction process is whether or not there are electrons in the conduction band. In insulators the electrons in the valence band are separated by a large gap from the conduction band, in conductors like metals the valence band overlaps the conduction band, and in semiconductors there is a small enough gap between the valence and conduction bands that thermal or other excitations can bridge the gap. With such a small gap, the presence of a small percentage of a doping material can increase conductivity dramatically.
An important parameter in the band theory is the Fermi level, the top of the available electron energy levels at low temperatures. The position of the Fermi level with the relation to the conduction band is a crucial factor in determining electrical properties.
(from this excellent site)
How can a charge neutral substance have a potential?
To say that a PN junction has built-in potential isn't to say that the PN junction has a potential relative to ground or infinity etc.
How can there be an inherent potential in a doped semi-conductor if it is charge neutral?
Charge has been separated within the PN junction and, thus, there is an electric field across the depletion region and an associated potential difference. A charged capacitor is neutral but there is a potential difference (voltage) across the dielectric. There are some similarities but...
However, I don't understand why we can't measure the voltage drop across the PN block with a voltmeter.
As explained, for example, here, the built-in potential is not readily measured with, e.g., a voltmeter. In other words, this question has been asked here several times (which means you are not the only one perplexed by this - most are at first) and there are good answers already available.
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Best Answer
Short answer: There is no band gap in metals so they stay as good conductors down to absolute zero. In a semiconductor there is a band gap, and even when doped the doping impurities lie in the gap, such that the carriers "freeze out" at some temperature. Here's something on doped semiconductors.