Talk:Indirect bandgap

Is this entirely correct: "Indirect bandgap semiconductors can absorb light, however this only occurs for photons with significantly more energy than the bandgap."

wull, there's a slope there. I'd put it: "Indirect bandgap semiconductors do absorb light, however they do so weakly for photons with only slightly more energy than the bandgap. The absorption rate (i.e., in %/micrometer) increases a few orders of magnitude as the photon energy becomes significantly more than the bandgap." Remember distance matters: black glass sliced very thin looks mostly clear, and a 10-mile ball of your clearest quartz would probably look black. jimswen 10:16, 8 March 2007 (UTC)

Couldn't an indirect semiconductor absorb a photon with the bandgap energy, if the momentum vector of the photon was just right? --User:Dgrant

No, because the momentum of a photon is very small compared to the momentum offset between CB and VB. Radiative transitions appear as very nearly vertical lines on the E-k diagram. Radiative recombination and emission is possible with the help of a well-placed phonon, but such events are very unlikely. In practice, conduction band electrons in silicon are de-excited via traps. -- Tim

E-k space

Can someone explain what the E-k space / k-vector is? I don't get it. Thanks, --Abdull 12:14, 30 May 2005 (UTC)

k-vector is about the same as velocity vector, except it doesn't follow the same square-law E vs v curve as a free electron in a vaccum. Inside the crystal the atoms make a bumpy potential field, somewhat like driving over a dirt road with wash-board ripples. So there are energy components dependent on velocity, adding to classical kinetic energy. Parts of the E-k curve near k=0 (standing still) are parabolic, about like kinetic energy (E=1/2mv^2). jimswen 09:56, 8 March 2007 (UTC)

Words do not constitute an explanation. That also seems to be the problem with this article. Graft | talk 21:35, 29 May 2007 (UTC)

Absorption Spectrum (Silicon comment)

Recommend revision to include: "visible wavelengths correspond to photons with energy greater than that of silicon's bandgap." Perhaps with a link to Planck's constant, Silicon's bandgap energy, etc. It currently is not immediately obvious that the statement applied to Silicon (Eg = 1.14) is true.

application

Do such semiconductors make good sound-emitting diodes? :) -lysdexia 02:43, 3 October 2006 (UTC)

Not usually. They don't emit well straight away from the face, and they don't expand or contract much with temperature, and any phonons generated are unlikely to be coherent with each other. I did read somewhere that lateral current in a surface layer could cause travelling-wave amplification of surface-waves, which could couple to grazing-emission sound. jimswen 10:04, 8 March 2007 (UTC)

Image

I've made an image of an indirect band gap which shows it as the combination of the optical transition with phonon assistance to create the necessary change in momentum. Anyway it matches the one of the Direct bandgap page. Does anyone think this would be a better image?

Profjohn 20:09, 2 May 2007 (UTC)

This is much clearer than the trio of images included, which don't even have a legend. What are they for? Who knows. Your figure is much clearer... I am going to put it in. Graft | talk 21:36, 29 May 2007 (UTC)