To be ready for our next class, please think about and discuss that here. For example:
- What is generation? What is recombination?
- which is dominant for x less than $\color{white} -x_o$?
- which is dominant for x greater than $\color{white} x_o$?
- (Hmm. Maybe this is a poorly considered question. Thinking about majority carriers may only cause trouble. Feel free to completely ignore this part (4) .) What do you think about the assumption of enhancement of both n(x) and p(x) in these regions? Would that make things a little simpler to think about g-r issues? In what way might it fail to capture the essence of what is happening in a junction, or not?
Regarding the in-class midterm, if you can stay ready for that, that would be great. No date is set yet: there will definitely be some notice in advance in a blog post here and probably also in class. It would cover everything we have covered involving:
1) homogeneous semiconductors
2) p-n junction physics:
a) in static equilibrium, where counter-flowing drift and diffusion currents make Jn(x) and Jp(x) each individually zero at all x (though the drift and diffusion parts separately have some interesting x-dependence, as we have been exploring),
b) in steady state with an applied voltage. (key things to understand might include: excess diffusion current, recombination, bias (forward / reverse), hmm... what else??? ...
Since no one has posted anything yet, I call the easy one:
ReplyDelete1. Generation (of an electron-hole pair) is the excitation of an electron to the conduction band, and recombination is the relaxation of an electron in the conduction band as it falls into a hole in the valence band.
2 and 3. I'm still going with the theory that recombination is the dominant process on both sides of the junction. I think I asked about this in class, and I don't remember the answer - but I don't see how you'd get any more generation in one area than in another if the whole thing as at the same temperature, so long as temperature is the only generation process we're considering right now. As for where all these electrons and holes come from, I'm going to go with our applied voltage source.
4. I have no idea. By enhancement of n and p of x, do you mean the idea that their concentrations begin to "deplete" before reaching the depletion region? That would seem to imply that they begin to recombine there. As for the essence of what's going on, I'd have to understand what's really going on to judge that.
And finally, recombinations that produce photons would imply that a lot of photons of energy E_g will be emitted from any recombination, which I believe is the process behind an LED. If that's true, then I guess you'd change the color by using materials with different E_g values.
And I just realized - this means that, if this were a research situation, we could answer our question about recombination regions by looking through some kind of microscope at an active p-n junction under bias and seeing which regions were brightest! Past that, we could shine a laser with photons of energy E_g at it and see where they get absorbed the most. Actually, if we could vary the wavelength of the laser, we could use the laser to find E_g in the first place - start at low energy, and increase until photons just start getting absorbed instead of passing through, and there's your E_g. Photons of energy < E_g will just pass through, right? Or do they get reflected? And is/was this technique actually used?
Nice comment! I hope this stimulates a lot of comments/discussion. (Kind of like stimulated emission of radiation.)
ReplyDeleteGeneration: formation of electron-hole pairs, via thermal excitation.
ReplyDeleteRecombination: the process of electron-hole pairs annhilating themselves, i.e., electrons filling the holes again.
Not totally sure, but here's my guess:
Since the bulk region on the n-side has Ec much closer to mu, it is easier to have electrons excited into the conduction band there, and likewise, in the p bulk region, mu is much closer to Ev, so one would expect electrons to re-enter the valence band there more easily. So, for x>x0, generation dominates, and for x<-x0, recombination dominates.
If we assume that a photon is produced for each recombination, then there is some potential for energy generation here... Thinking along the lines of solar cells, maybe (and this is just speculation) there is a some way to do the process backwards and have incident radiation (the sun, for example) generate EHPs in the material, and therefore a current across the junction?
also, i would agree with dave in imagining that the emission of these photons has something to do with how LEDs work...
An electron requires a "boost" of energy E_g to go from the valence band to the conduction band no matter what, correct? That's what leads me to think that generation and recombination are independent of mu. Unless I'm missing something, your theory assumes that crossing the energy band only requires a boost of (E_c - mu). Wouldn't that imply that the rest of the energy, (mu - E_v), is stored in some manner that isn't included in our energy band diagram, but can still somehow help electrons cross the energy gap?
ReplyDeleteMaybe my problem is that I'm still confused as to exactly what mu is. "Chemical potential" doesn't inherently mean anything to me; "electric potential" does, because it's potential due to an electric force. Is there something that works as an equivalent "chemical force?" Possibly the deep magic behind the Pauli exclusion principle or something?
Would it be accurate for me to think of our band diagram as a representation of "kinetic" or "active" energy, while mu represents the level that electrons have the "potential" to be at, but are not allowed to be at? Then your idea would make sense to me; electrons still require an energy E_g to cross the gap, but have a constant chemical (magic) pressure trying to get them up to a value of mu.
Is any of this valid??
I think that the photon emissions created from hole-electron pair recombination events also explains how lasers work. It doesn't necessarily have to be in the visible spectrum to be useful, either.
ReplyDelete