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I am confused about Ec-mu. What about Nd and Na? I don't understand how to work with this?
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What are the units of Y?
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what happens to Mu when you take the derivative of ln[n(x)]?
ReplyDeleteAlso, wouldn't this have some time dependence, since as the electrons move around the electric field is changing, and also the rate of diffusion would change too?
Mu drops out because it is not a function of x--the beauty of taking the derivative is it reduces the function to only the x-dependent term.
ReplyDeleteUntil the equilibrium condition was reached, I think that n(x) would have time dependence. It seems like a flow of electrons would be necessary to stabilize things. I don't know what this would look like. Any thoughts?
You're right, I wasn't thinking about the nature of equilibrium, and an equilibrium CONDITION, which wouldn't be time dependent.
ReplyDeleteI also don't know what the time-evolution of the system would look like, besides a rough guess: the diffusion pushes e- left (in our classic picture), when the E-field then pushes them back when they start storming over to the left, then too many will be on the right again (but less than the original amount) and back and forth...
Here's my look at the matter; it seems like we're not taking into the account that there are 3 forces occurring (but the book does this exact same thing, so maybe I am missing something and someone could explain it to me).
ReplyDeleteIn your system, we have electrons drifting to the right and diffusing to the left. The diffusion is caused by the imbalance of densities of the electrons, and the drift is caused by the initial relocation of electrons from the n-droped conduction band to the p-doped valence band. If thats the case, how are we not considering the "affinity" for the electrons to normalize μ to a single value. Or this effect negligible after the initial combination?
I'm not sure what you mean by "affinity". In the picture I have in my head, there are two forces--the diffusion from the energy shift across the junction, and the current resulting from the electric field. The energy shift across the junction is a result of the shifting of the energy bands to ensure that mu is constant. Holes and pairs in the junction are able to combine and annihilate, creating a charge imbalance (from the dopant ions) that creates the electric field.
ReplyDeleteWell I suppose I should ask, since this was the way it was worded in class, is there a diffusion from the density gradient? Because if that is not accurate, then I agree with the model we have.
ReplyDeleteWhat I was saying tho is:
1)We have electrons from the N side moving to the P side to fill lower potential holes
2)We have electrons moving from P to N in the depletion region due to the electric field propagated by (1)
[This last one was suggested in class then, and would cause issues in my understand if its valid, i suppose]
3) There is electrons moving from the N side to the P side due to a concentration gradient (higher n(x) on the N side)