It's very heavy on basically the last few lectures... I'm OK with that, though, considering the class is supposed to be focused on applications anyway.
Also, for 1B, I'd probably take a while trying to remember how to go about that...
I'm not sure what you mean by "different direction". Perhaps I'm missing something--can you clarify?
Zack:
The final looks pretty fair. The length looks reasonable as well as the content covered. Still, I agree with Dave that the material appears biased towards the latter half of the course. This does not seem too much of an issue, though, for these are the more interesting systems.
That looks like a really good overview. Although, the number of questions seems weighted to the last few weeks lectures, the work load seems more heavily weighted to the first problem. This seems like it would make a well balanced final. Plus Dave is right this class is focused on the physics of devices. I like how the first problem requires us to set the conditions and constants for the problem.
I'm a little confused by the graphing part of problem 1. Is it asking for a graph of the depletion region width as a function of voltage as well as a graph of the height of the potential step as a function of voltage? I have an expression that has all three variables in it. Should I be holding one constant to see how the other two effect each other? Do I have the wrong formula? Should I be able to get one in terms of one?
"Is it asking for a graph of the depletion region width as a function of voltage as well as a graph of the height of the potential step as a function of voltage?" yes. it is that simple.
The way i look at it, they should each depend only on V. Is that wrong?
I think Zack is right--I have two functions, one for the width of the depletion region, and one for the potential step; both depend on the applied voltage. Granted, the width depends on the potential step, but that in turn depends on the applied voltage. So, indirectly, the width depends on the applied voltage. I think this is akin to graphing f(x) and g(f(x)), where x in this case is the applied voltage, f the potential step, and g(f) the width.
zack- is it ok if we use a program like Grapher for mac to make the plots for part C, so long as we format them to be 2-3 inches, like the instruction sheet says? i really prefer using a computer to graph.
That's exactly what I'm saying. Our assumption ends up being that the integrated potential, which depends on x0, has to equal the potential jump. That's how we found x0 before. However, I'm saying that we can find the potential jump as a function of applied voltage. Then, we can use that function to solve for x0.
I like the first question. Its the system we've focused on for most of the class, but taken in a different direction.
ReplyDeleteIt's very heavy on basically the last few lectures... I'm OK with that, though, considering the class is supposed to be focused on applications anyway.
ReplyDeleteAlso, for 1B, I'd probably take a while trying to remember how to go about that...
Kelsey:
ReplyDeleteI'm not sure what you mean by "different direction". Perhaps I'm missing something--can you clarify?
Zack:
The final looks pretty fair. The length looks reasonable as well as the content covered. Still, I agree with Dave that the material appears biased towards the latter half of the course. This does not seem too much of an issue, though, for these are the more interesting systems.
That looks like a really good overview. Although, the number of questions seems weighted to the last few weeks lectures, the work load seems more heavily weighted to the first problem. This seems like it would make a well balanced final. Plus Dave is right this class is focused on the physics of devices. I like how the first problem requires us to set the conditions and constants for the problem.
ReplyDeleteI guess I just meant that I'd never examined the dependence on bias that closely. I think I'm going to find the graphs interesting.
ReplyDeletewhat does "one-sided" mean in number 2? I think the final seems pretty fair. I might have a little trouble with #3, but that's just me...
ReplyDeleteI believe that the consider it "one-sided" instruction refers to focusing on the electrons, rather than both electrons and holes.
ReplyDeleteI'm a little confused by the graphing part of problem 1. Is it asking for a graph of the depletion region width as a function of voltage as well as a graph of the height of the potential step as a function of voltage? I have an expression that has all three variables in it. Should I be holding one constant to see how the other two effect each other? Do I have the wrong formula? Should I be able to get one in terms of one?
ReplyDelete"Is it asking for a graph of the depletion region width as a function of voltage as well as a graph of the height of the potential step as a function of voltage?"
ReplyDeleteyes.
it is that simple.
The way i look at it, they should each depend only on V. Is that wrong?
Caleb:
ReplyDeleteI think Zack is right--I have two functions, one for the width of the depletion region, and one for the potential step; both depend on the applied voltage. Granted, the width depends on the potential step, but that in turn depends on the applied voltage. So, indirectly, the width depends on the applied voltage. I think this is akin to graphing f(x) and g(f(x)), where x in this case is the applied voltage, f the potential step, and g(f) the width.
Mike:
ReplyDeleteSo you're saying that we should be able to get an expression for the potential step that does not depend on the width of the depletion region?
yes. treat those both as "unknowns".
ReplyDeleteforget about the equations for a minute, and just think about a junction and what happens as you apply a voltage...
ReplyDeletezack- is it ok if we use a program like Grapher for mac to make the plots for part C, so long as we format them to be 2-3 inches, like the instruction sheet says? i really prefer using a computer to graph.
ReplyDeleteCaleb:
ReplyDeleteThat's exactly what I'm saying. Our assumption ends up being that the integrated potential, which depends on x0, has to equal the potential jump. That's how we found x0 before. However, I'm saying that we can find the potential jump as a function of applied voltage. Then, we can use that function to solve for x0.
Does that wording make any more sense?
Yeah I just figured that out last night. Thanks a lot for your help guys. I appreciate it.
ReplyDeleteIn our models of lasing, where is the confinement? It isn't exactly between the n-p and p-p+ junctions, is it?
ReplyDeleteHow about for diagram 8-18? Would the recombination occur in the space charge layer, as well as Exp[-x/D] within the p-type material?
Some rather intense work-ups (and diagrams) of pn junctions available through the ucsc library:
ReplyDeleteSemiconductor Radiation Detectors - Device Physics
Online ISBN: 978-3-540-71679-2
Best looking diagrams in Chapter 1 - "Detectors for Energy and Radiation-Level Measurement"
PS. You can turn it in to my mailbox.
ReplyDelete