will it fall back immediately by releasing same amount of energy or will it stay there.
electron is said to be free in the conduction band. If only the external stress (optical) electron will stay for a time dependent on material characteristics and come back to lower energy state. you can find more answer about your interest in the book of optical properties of semiconductor "Mark fox"
In case of a photon hitting a semiconductor, the energy of the photon will create an electron-hole pair. So, if the energy of the photon is greater than the band gap of the semiconductor, it will create an electron-hole pair. Not just the electron is excited from the valence band to the conduction band, but the 'hole' that it leaves behind in itself is capable of carrying current.
From there, it depends on your semiconductor. I mean, general semiconductor theory applies. Note that in semiconductors, in general, recombination (that is the process of fallback) is generally only present at interfaces and at defects in the lattice. So, suppose its a good quality semiconductor, then recombination will not be dominant. (if its a lousy semiconductor, expect it to be dominant.)
In a solar cell, there is a junction. A solar cell in essence is a diode biased in the forward direction, but not enough that there is significant current flow. In the solar cell, the 'hole' will drift to the P material, and the electron will drift to the N material, and they both become majority carriers there. The net result is the generates a light-induced solar current. (it is a solar cell.)
Note that there are other light-sensitive semiconductor devices, but in general recombination is not the dominant process. (unless you have a very poor performing device, of course)
Cheers,
Henri.
The photo-electron will not fall back immediately to its previous energy level; it will also not stay indefinitely at the excited level. First it will loose its kinetic energy to phonons by several collisions and fall to the conduction band edge (Ec). If there is an electric field present, then the photo-electron will move opposite to the direction of the field; otherwise, it will recombine with a hole by direct (band-to-band) or indirect (trap-assisted) recombination.
Thanks, to all of you for your kind answers which helped me to clear my troubling dought. Special thanks to Tillmann Krauss for sharing the useful paper links.
Dr. Samares Kar sir, I have one more doubt.
As you mentioned the concept of recombination taking place, so in this phenomena is there chance of photon emission?
Dear Alok,
If recombination takes place, there is chance of photon emission in case of a semiconductor with a direct band gap. The wavelength of the photon corresponds with the bandgap energy. Silicon has an indirect band gap, so for silicon there is usually no photon produced. For other materials, like gallium arsenide, gallium indium arsenide, gallium nitride, they have direct bandgap, and there photon emission takes place.
Note that recombination creating photon emission is the operating principle of the semiconductor laser and the light emitting diode. In these devices, the material is flooded by minority carriers (electrons in P region, holes in N regions) because of the diode conducting current. Due to the diode conducting current, the majority carriers cross the junction, and become minority carriers at the other side of the junction. There, they fall back, creating photons. The energy of the photons corresponds to the bandgap of these semiconductors. There has been a search for large bandgap semiconductors to create these devices, because then the energy is short wavelength (blue up to ultraviolet), very useful for lighting purposes. 5For white LED's, they essentially emit ultraviolet light, which is converted to visible light using phosphors, just like in a conventional TL lamp.)
Cheers,
Henri.
Dear Mr. Henri
Your prompt and detailed answer is very much helpful.
Thank you very much
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