The problems below, for Tuesday, 4-20, I think will warrant some discussion. I don't think you will just look at them and say, "oh, no problem, I can get drunk and do those Monday night". So feel free to say, "wow, I am stuck. what ...?" It is okay to ask any questions. I think other students will appreciate your attempt to articulate your stuckness; I believe I will too.Let's try posting all questions and thoughts as comments to this post since that other one has dropped down quite a bit.
The accompanying notes summarize some of what we need for our "symmetric" p-n junction problem. If you complete the 2nd half of V(x) (the x greater than zero part), then you have a change in potential across the depletion region, which is associated with the space charge regions on either side of the interface. You can use that to create a model picture in which the chemical potential is constant (i.e., does not depend on x), and the band edges are continuous, as a function of x, have the "correct" proximity to the chemical potential in the large x regions (large and negative, large and positive).
Please try the following:
(Assume Nv=Nc=5x10^18 cm-3 and Eg = 1.0 eV)
1. Guess (use your intuition to consider) which has a wider depletion region: the above model with a doping of 10^16 cm-3 on each side or with a doping of 10^17 cm-3 on each side. Explain your reasoning.
2. Calculate and compare the widths of the "depletion region" (2x_0) for dopings of 10^16 and 10^17 cm-3, respectively.
3. For the 10^17 cm-3 case calculate n(x), that is, the density of carriers in the conduction band. On what does that depend? (I imagine this is probably best done numerically since it requires exponentiating a more-or-less quadratic function. You can choose about 9 equally-spaced points covering the region from -x_0 to x_0, if you like, (an odd number allows you to "hit" x=0), and make a table of x, Ec(x)-mu, and n(x).
4. Graph n(x) on a linear scale. How does it look?
5. Do the same for p(x).
6. Here is an interesting conundrum. This is supposed to be equilibrium, right? But at x=0 there is a big, fat electric field, and the carrier densities are not exactly zero right? So would there be current flow? In what direction? If so, what kind of equilibrium is this? If not, why not???
Do not hesitate to ask clarifying questions. Are these problems not vague?
Temporary office hour modifications: (in the interest of reducing the clutter of "too many different posts", i'll put this here.)
Today and next Friday I will have office hours at 3:10 PM. Also, next week my Thursday office hours are canceled we will only have office hours on Friday at 3:10 PM.
I'm trying to figure out n(x) in the depletion region, and I was thinking of using n=N_c*Exp[-(E_gap-mu)/kT], where E_gap-mu is a function of x (and is equal to DeltaE + mu, using our notation from class yesterday). Does this make sense to people?
ReplyDeleteZack--what do you mean by calculating numerically? Symbolically Exp[f(x)], where f(x) is some quadratic, isn't that odd (it's a Gaussian shifted depending on the linear term in f(x)). However, haven't we always been calculating things numerically--why else would we care what the values of Nc, etc are?--or am I misinterpreting something?
Regarding the 2nd paragraph, yes, i pretty much agree. I was just saying maybe you will have to pick a grid of x points and make a table before you can do a plot, but maybe that is obvious.
ReplyDeleteRegarding your 1st paragraph, I'll let other people comment on that. If you approach the depletion region from the right, you might want to "connect" to N_donor, mathematically speaking, and then see how things diminish from that.
Assuming that you set the chemical potential as 0 for your energy axis, I agree.
ReplyDeleteSo, by connect, I guess you mean you want n(x) to be continuous, and it better equal N_donor at the center.
ReplyDeleteCan we plot using a computer program instead of by hand?
at the center of what?
ReplyDeleteStarting with the current assignment, would you each write at the top of your first page (next to your name) the number of questions, explanations or comments that you posted here?
ReplyDeletePS. Your questions/comments do not have to be amazing. Anytime you spend a little time looking at, thinking about, or trying to do the problem(s), you can post whatever is in your mind.
One of the (many) confusing things in this class is: where is zero??
ReplyDeleteLike without the problems we were doing before ("homogeneous" semiconductors) we would have never considered setting mu = 0 because that was the thing we were calculating. Mu depended on doping; sometimes it was up near the CB edge, sometimes it was down near the VB edge, sometimes it was in, or near, the middle of the gap.
But now putting our zero at mu doesn't seem that completely crazy because it is like the only thing that doesn't change as a function of x. Iam not saying it is the right thing to do or the best thing (maybe it is), but just noting how ironic it is that we are even considering that.
One fortunate thing: it seems like we never got that attached to where zero was. Our models were specified by a gap, and Nc and Nv: all co-ordinate free entities, and we could put zero wherever and then calculate things and express them in terms of proximities (i.e., mu is 0.1 eV below the CB edge) rather than in terms of absolutes, which we see now are a little ethereal...
For the n(x) problem, just based on units and functional forms, i am guessing one factor (the x-dependent one) might be of the form:
ReplyDeleteexp{-E*(x-x_0)^2/kT(x_0)^2} where E is is some unknown energy scale that you would have to calculate. So in this way you can maybe sort of simplify or separate the graphing, which will depend on E/kT but otherwise be kind of "generic", from the other parts of the calculations.
By center I meant where x=0, since mu is exactly in between the valence and conduction bands, and Nc=Nv.
ReplyDeleteBy the way, Kelsey and I were talking at the physics barbecue/softball game today, and we definitely think we should hold some of our classes outside this quarter. Does this seem feasible (comments from anyone are great)?
I wouldn't mind that at all.
ReplyDelete"I'm trying to figure out n(x) in the depletion region, and I was thinking of using n=N_c*Exp[-(E_gap-mu)/kT], where E_gap-mu is a function of x (and is equal to DeltaE + mu, using our notation from class yesterday). Does this make sense to people?"
ReplyDeletefor calculating n, this seems perfectly fine, but it seems there is a lot of detail missing in the f(x) that we would need to come up with. μ should be just some calculated constant at the end, as it is a line of slope 0. For ΔE, I'm not quite sure what you mean by that. Are you talking about Ec or Ev with its respect to μ?
So thinking about it more, I went through and drew the P-N Junction and wanted to confirm with others that this should be the nature of our p(x) and n(x)
ReplyDeletep(x) && n(x)
i have a repetitive motion injury so i can't type much. hopefully, other people will respond.
ReplyDeleteregarding coordinste system, one could try setting zero where Ev is on the far right. then Ec on th far right would be at Egap, and Ec(x) would start increasing quadratically at x_0. Just a thought. There is no one right way...
ReplyDeleteI don't know if I'm doing it quite correctly, but I tried to make (E_gap-mu) a function of x by relating deltaE to Potential function of x we derived in class(deltaE=e*(ρ/ε)*(x_0)^2).
ReplyDeleteI would love to have classes outside, but we would need to get a chalkboard (or just bring chalk outside and use the side of the building)
Egap and Ec are not exactly the same thing here. One depends on x. I would suggest not trying to answer anything at first, but instead start by defining a coordinate system (for energy) and a picture (graph) of E's vs x.
ReplyDeleteI found it useful to just look at differences. When we calculated n before, using n=Nc*Exp[-(Ec-mu)/kT], we were only interested in the separation between Ec and mu. Since we know how far Ec and mu are in the n-type material far from the depletion region, and we know the function that connects the Ec band for the p-type with the Ec band for the n-type, we can figure out how far the Ec band and mu are in any part of the material, and then just apply the original formula, n=Nc*Exp[-(Ec-mu)/kT]. Does this make sense to people? Are there contrary opinions--am I going in a wrong direction? Any input will be appreciated.
ReplyDeleteBrad:
ReplyDeleteRegarding your graphic, did you switch the labels on p(x) and n(x)? I think n(x) should be greater in the n-type and p(x) greater in the p-type. Also, I think there is a much steeper drop in carrier population as you pass through the depletion region. Your plot makes it appear linear, but since we're dealing with Exp[-f(x)], where f is a quadratic, we're going to have some drastic decreases--it'll be a constant in the n-type until you hit x0, and then decay like the left half of a Gaussian, at least until you get to 0. Then I think it still decays very fast, ending up with a very small constant in the p-type region far from the depletion region.
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ReplyDelete@ MikeK:
ReplyDeleteI definitely reversed p(x) and n(x). And also, yes they should not be linear. I just put that as a general transition.
I've gone and adjusted the file, do you agree more with this?
Also just looking around for more information than just the from the book and in-class notes, Hyper Physics & Wikipedia have some good diagrams that may help others.
ReplyDeleteI missed class on Thursday, so I'm not sure if someone suggested this and I'm simply re-iterating an idea that was already proposed and was deemed incorrect.
ReplyDeleteYou'll need to look at this while reading what I am suggesting.
Quick explanation, the blue Gaussian distributions are the mean free path (MFP) of the electrons on the n-doped region, and the red Gaussian is the MFP distribution for the holes in the p-doped region.
Essentially what I am suggesting is that the mobility of the holes/electrons in the conduction band are directly related to our depletion region, and that the linear superposition of the MFP Gaussians are a result in the width of the depletion region (ie, an electron in the conduction band to the very far right of my diagram does not have a MFP gaussian that allows it to travel to the p-doped region. The reverse goes for the holes).
By combining the superposition of the guassians (for the appropriate region, shaded green), we obtain a single distribution which we can then integrate.
Good idea or completely off base, and if I'm right, who wants to do it? ^_^;;
Ps: i think we'd be off by a factor of 2 by not including the left/red/p-doped guassians.
MikeK, I think I get what you're saying, and I like it. Setting it up like this:
ReplyDeleteHave a coordinate system such that x=0 on the far left of the depletion region and x=w (where w is the width of the depletion region) on the far right of the depletion region so you have a function of x that starts on the left and continues to the right until you've gone over the depletion region width. For x<0 and x>w, the function is fixed and acts like a p-type or n-type, respectively, w/o the junction being considered (depletion approximation). In this case, for n(x)=Nc*e^(-(E_g-mu)/kT) the value (E_g-mu) would be your function of x.
Another benefit of this, it seems like you could solve for p(x) by noticing n(x)=p(-x). Does that make sense?
Other things of note that i'm not sure about. For n(x), it seems like at x=0, n=0 and at x=w, n=10^17... but I'm a little confused about this. When you have a donor doped material with 10^17 donors, are all of the donated electrons in the conduction band at room temperature? That kind of makes sense... but I'm not sure.
So... how do you find your F(x) for E_g-mu??? hrmmmm......
More thoughts later. :)
Correction: At x=0, n≠0. It would equal whatever it would normally equal for an acceptor doped with N_a=10^17 material at room temperature. Right? i.e.: something REALLY small
ReplyDelete~Lou
Lou - I think your correction is right. If I'm not mistaken, that "something really small" is the n_i value of 10^10/cm^3, which is also equal to the hole concentration at that point.
ReplyDeleteIf I AM mistaken, then at least I have a post on here, thereby increasing the probability of finding a cookie in my near future.
What's a good value for epsilon, the permittivity of out material? I know it's the relative permittivity * absolute permittivity, but I don't know what to use for the relative part. All the info I can find indicates that it varies with doping; the only value I can find, though, is 1.04*10^-12 for intrinsic Si. I don't even know what the units are on this; Farads/cm is what I'm assuming.
ReplyDeleteAnyway, I'm approaching the n(x) problem by trying to relate voltage to charge concentration. If you can do that, then you can just use the given V(x) in the image up there. Haven't worked it out yet, though.
...And I just realized that ro is what I need to be looking at in that V(x) equation. It represents overall charge concentration, right?
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteWhen solving for n(x)(or p(x)), when separating an equation by zack was mentioned. Does this possibly imply separate equations for each of the three "regions" of our n(x) values, x<-x_0, -x_0 < x < x_0 and x > x_0? Or if that is not meant could this work. I am having trouble coming up with one single equation that connects all three regions.
ReplyDeleteLike lou said, " For x<0 and x>w, the function is fixed and acts like a p-type or n-type" (in his coord. system.) So the outside equations would be relatively simple, and the middle a little more complex if broken up into different equations. (Like the equations for V(X) on the above handout)
If not could someone help with connecting these regions in above equations.
Tyler,
ReplyDeleteYes you have separate equations for each 'region.'
Since the equation for V(x) is piecewise, it is easiest to maintain this for the equation for Ec-mu. (Ec-mu referring to what Mike posted earlier)
i have been working with this assumption so far, but for (abs. value of x) > x_0, n(x) is effectively zero, right?
ReplyDeleteOk... N_v=N_c and thus the depletion width on the p-type and n-type sides are equal, plus it seems we've been using the depletion approximation (basically, assuming the region between -x_o<x<x_o is totally depleted of mobile carriers), here is what I am thinking... probably wrong, but whatever.
ReplyDeleteThe number of holes in the p-type half of the depletion width is equal to the number of donors in the n-type half of the depletion width, so all the electrons from the n-type side should basically fill in all the holes in the p-type side and thus, why don't we just estimate it like undoped silicon in the depletion region such that n(x) will be constant in the depletion region and equal to the intrinsic carrier density... o.O'
Granted, there would be a transition region near the edges of the depletion region (that we haven't been considering in class) where the magnitude of the field is such that the tendency of electrons to diffuse from the n-type to the p-type is counterbalanced by the tendency of the electrons to drift in the opposite direction under the influence of the electric field (this transition region is where the equilibrium is gradually formed and the hole/electron density gradually returns to the density outside the depletion region. Here there would be a varying n(x)) It's like, the velocity probability of the electrons would lead some of them to have enough energy to get back to the n-type side, but this would taper off as x decreased through the edge of the depletion region, gradually increasing the number of holes across that region to 10^17cm^-3
Of course, the net charges on the n-type and p-type side would also affect the location of mu, but how the heck are we supposed to account for that...
So yeah.... Moral of the story: is anyone else as lost as I am?
~Lou
P.S.: If anyone actually gets this and could meet with me tomorrow before class to explain it, that would be awesome. Maybe we could even set up a group study.
Brian,
ReplyDeleteat room temperature, aren't there conduction electrons due to the silicon? 10^10 e/cm^3 right?
We usually ignore this number because we've been doping 10^17 which is much larger than 10^10, but we can't ignore this value when n(x) is small.
I realized that I don't understand the potential function that we got. V(-x_0)+(ro/2epsilon)(x+x_0)^2 I know that this function works for -x_0<x<0 but I don't understand how this could be the function to give us the correct potential for this situation. What is V(-x_0) and how does it contribute to the curve. I feel like I need to understand this before I can proceed. Can anybody help me out?
ReplyDeleteActually that last post helped a lot.
ReplyDeletegeno: thanks, that kind of clears some things up for me.
ReplyDeleteccaldon: in class, i am pretty sure that we just set V(-x_0) as zero on our plot of V(x) vs. x. then the rest of the V(x) function just looks like a backwards version of the curve for E_C or E_V from our Energy band diagrams, but with V(-x_0)=0 and V(x_0)=V_0
If anyone is meeting before class about this, let me know: 916-607-3845.
ReplyDeleteDoes ro change with x? Is ro equal to the doping level?
ReplyDeletei'm pretty sure rho (the charge distribution) is constant across the depletion region, because we are assuming an abrupt cut-off point at x_0. i.e., the donor electrons from the n-side can make the transition to the valence band on the p-side up to a certain point(x_0), but past that point, none of them can make the transition.
ReplyDeleteTo lg_magius:
ReplyDeleteyou said- "there would be a transition region near the edges of the depletion region"
I talked to Zack in office hours last week and he said that our approximation is accurate; the curvature near the edged of the depletion region is very small.
Also, when solving for n(x), it might help to redraw (or just look at) the graph for Energy vs. position that we made in class.
Remember that this was derived by integrating charge density (which gave us that upside-down triangle graph) to get the electric field, then integrating the electric field to get the potential function. Then multiplying the potential by the charge of an electron gave us our equation for energy.
(The reason that our energy equation is piecewise is because of the form of the charge density equation.)
When solving for n(x), (Mike said this at 1:08pm today) it is easiest to first solve for the energy difference in the exponential's numerator, Eg-mu. Once you have that then you just plug numbers in for x and make a table (as Zack suggested we do).
To ccaldon: rho is a constant and is equal to the doping level. This is because by assuming that all charges from 0 to X_o magically move to the region -X_o to 0, this means that 10^17 e/cm^3 moved from the right to the left region. Thus there is a net negative charge in the left region due to the extra electrons and a net positive charge in the right region due to the loss of electrons. Remember that even when a material is doped, the net charge is still zero.
Also, to further understand how we derived the equations for V(x) and E(x), you can quickly start from a graph of charge density
p.s. chapter 5 in the book has some useful explanations and graphs.
This may be a very basic question, but it seems to me that the depletion region region should be wider with more doping but the number crunching tells me differently.
ReplyDeleteTo Kelsey:
ReplyDeleteI thought so too! But here is the logic I'm thinking: with less doping, the potential energy across the depletion region is going to be less b/c the charge density will be lower, so it is more likely that electrons farther away from the n-type side will have enough energy to get back to the n-type side and the region will be wider. Does that make sense?
~Lou
kelsey: i also thought that initially. i was thinking by the logic that the depletion region extends farther into the lesser-doped side of a p-n junction that ISN'T in equilibrium. however, since we have N_a = N_d here, increasing their CONCENTRATION (cm^-3), we are actually increasing the DIMENSIONS of the depletion region; electrons farther out from x=0 will be able to make the crossover from the n to p side.
ReplyDeleteSo is it like there needs to be more electrons to fill the e field so the depletion region needs to extend out farther when there is less doping?
ReplyDeleteI agree I was initially thinking backwards. As the doping increases so does the the V_0, however, the depletion region shrinks. This is due to the 1/Na +1/Nd.
ReplyDeleteBTW Kelsey and I are in Baskin Lounge working on this right now if you want to stop by.
ReplyDeleteI'll be on campus before 11, so I'll probably stop by :) hopefully you're still there. if you move, sent me a text 831-332-1544
ReplyDelete