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I read in an article that the addition of acceptor impurities contributes hole levels low in the semiconductor bandgaps so that electrons can be easily excited from the valence band into these levels leaving mobile holes in the valence band. My question is- what will be the energy level corresponding to the holes created after electrons from the Silicon atoms occupy the holes due to acceptor atoms. Is the energy level the same as the valence band? If so, then those holes can be easily occupied by other electrons of the valence band. Am I correct in this concept?

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Okay, these are two energy bands which electrons can occupy (each band is actually a set of unique states). In a pure (undoped) semiconductor crystal, the "valence" band is chock full, and the "conduction" band is empty. As such, the bonding between atoms in the crystal is by sharing electrons in the valence band. These electrons are tightly bound to the crystal matrix and can't go far.

If you add a few atoms from the next element on the chart, for example, you'll have an electron on each one that doesn't fit into the valence band and are forced into the higher (conduction) band by the exclusion principle. They're not part of the crystal matrix and, although they have a simple ionic attraction to the larger nuclei, they're pretty free to roam about if an external field is applied. This is an N-type semiconductor.

If you add a few atoms from the previous element on the chart, you'll find that the valence band isn't completely full. The "holes" are spots in the valence band that electrons could fill. As such, the "holes" can move by the process of empty bonds "borrowing" an electron from a neighboring bond. The process continues in a chain under an external field, with the effect that the "holes" appear to move in the opposite direction. You can think of it kind of like bubbles in water--it looks like the bubble is moving through still water, but in fact the water is flowing around the bubble. The analogy isn't perfect, since water bubbles are in fact filled with air, but it's more of a metaphor.

So N-type conduction happens straight up in the conduction band, and P-type conduction happens one step at a time in the valence band.

Does that help? If not, I'll try another approach.

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  • \$\begingroup\$ So, the conclusion sums it all. I was also thinking of the same. I have one more question. It is said that in a p-n junction diode, the free electrons from the N side due to diffusion move to P side and combine with holes. My question is on stating -ve charge on the p side in the depletion region. A hole can be either on Silicon atoms or acceptor atoms (eg. Boron), right? If the electrons from n side diffuse into holes of silicon, then silicon atoms become neutral but if they combine with holes of Boron atoms, then Boron atoms will have -ve charge on them making them ions. Am I right? \$\endgroup\$
    – Sandeep
    Jan 16, 2020 at 16:24
  • \$\begingroup\$ You seem to be talking about the depletion region. Despite the ionic attraction of the conduction band electron to its nucleus, it is also attracted to a lower energy state, and a nearby hole represents that opportunity. The area where the lower energy state is close enough to overcome the ionic attraction is the depletion region, where N-type free electrons find homes in P-type holes. This recombination results in an area around the junction where the crystal bonds are all occupied, but there is an electric field caused by the charge displacement. \$\endgroup\$ Jan 16, 2020 at 16:42
  • \$\begingroup\$ @ThePhoton Sure (assuming you meant 0K), but for most practical cases it's a good enough approximation under about 150C (for silicon). \$\endgroup\$ Jan 16, 2020 at 16:44
  • \$\begingroup\$ Wow, I need coffee. \$\endgroup\$
    – The Photon
    Jan 16, 2020 at 16:45

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