Here is a plot which i think is proportional to J_drift (x) . What does the 16 reflect? Where iss the 1-x associated with? What is x_0.
Also, can you tell which drift current this could be? why?
When i see these graphs I think about what's next. I hope maybe you do too.
With that in mind, please consider the following semi-optional problem(s)/fun thing(s) to do and think about for Thursday:
What does this curve suggest about dn(x)/dt (=dJn(x)/dx) ?
(Like if this were the total Jn (x).....
Starting with this curve, plot all 4 J(x) curves.
How do they effect your thoughts about dn(x)/dt?
In general, what is missing from the equation:
dn(x)/dt (=dJ(x)/dx) ? (what else could effect n(x,t) and what form might it have?
PS. Any comment here (in the next 26 hours) is worth 5 bonus points on your midterm!
I believe this is showing electron drift, since we saw in class that the electron diffusion curve is just like this one, except mirrored over the x-axis, so that the total current cancels out to zero.
ReplyDelete1.) I think 16 reflects (ΔE/2)/kT, since the exponential term here comes from n(x) = Nc * exp(Ec(x)- μ), and Ec(x) is proportional to ΔE/2.
ReplyDeleteThe x-1 (I’m assuming that’s what you meant to type above) comes from the [(x-x0)/x0]^2 dependence of Ec(x). Therefore, x_0 = 1.
2.) This must be J_drift for electrons, since its magnitude increases and hits a maximum on the x>0 side, which corresponds to the n-side of the junction.
3.) Ok, this equation relates “the change in carrier concentration with respect to time” to “the change in current density w.r.t. x.” What does that mean? As the rate of charge flow increases as you move along the junction, the rate of movement of the carriers at that point increases too. Makes sense. Anyone else have possibly more in-depth responses for this one?...
4.) See .jpeg below. The light blue-green curve is the J_diffusion for electrons. The dark blue curve is J_drift for electrons. The red curve is J_diffusion for holes. The bright green curve is J_drift for holes.
file:///Users/brianvollbrecht/Desktop/Physics%20156/Homeworks/pre_MT/Graphs_of_J.jpg
5.) There needs to be an e (electron charge) in front of dn(x)/dt to make the units work out right. Also, I wonder if introducing some sort of (unitless) constat would help take into account the effects of mobility on n(x,t)?
i'm trying to upload that .jpeg more successfully...
ReplyDeletecan someone tell me how to post a small version of the actual jpeg on here like zack did?
It's interesting that dJ/dx = dn/dt, for J could have no time dependence but still have space dependence, so dJ/dx would be nonzero but not be a function of t. Thus, dn/dt would be simply a constant with respect to time, meaning either a regular increase or decrease of n at each point in space, regardless of the time. This seems like some sort of steady state situation, where the current density is not changing with time, and the number of carriers is going up (or down) steadily. However, the continuity equation assumes that no charges all of a sudden appear--that there is local charge conservation. But, we can create a hole-electron pair pretty easily. Then, those new free electrons could be swept away by the electric field, causing holes to be left-over, and so less negative charge than before.
ReplyDeleteQuantitatively, we have J(x)~(x-1)*Exp[-16(x-1)^2]. dJ/dx would be (-32(x-1)^2+1)*Exp[-16(x-1)^2] (I think), which is still a Gaussian, but modulated by the negative parabola -32(x-1)^2+1. So, it's max value is now at x=1, or at the boundary of the depletion region, meaning that the number density is changing most rapidly with time at the edge of the depletion region.--This seems intriguing, but I'm not sure what to make of it, or where to go from here. Any comments would be much appreciated.
I'm not sure if these comments are of any use, or very interesting, but I think this seems like a reasonable path to explore in lectures to come.
The 16 must come from the exponential for n(x). There isn't really another obvious reason for that number. This graph must be for the electron drift current.
ReplyDeleteIt would seem that Jn(x) would describe the flow of electrons as you move through the semiconductor, and it is rather interesting. Even though the electric field is the greatest in the center, there seems to be no movement in that part. (Or at least it is very little).
It is also interesting that there is the most dramatic change in movement at the edge of the depletion region. This doesn't make sense to me, but I am sure it will become clear soon...
Just to really play with things in the graph, the (x-1) term in the exponential term shifts its maximum to x=0. The multiplying (x-1) term counterbalances this, because it decreases to 0 as x increases to 1. With x0 = 1, I can then proceed to make a series of possibly crazy assumptions. The exponential term corresponds to density of charge carriers--largest at the boundary of the depletion region, and decreasing exponentially from there. The factor of 16 determines how fast the decay is. The multiplying term then corresponds to the strength of the electric field; maximum at the center of the depletion region, going to 0 at the boundaries. The current results from the interaction between the available carriers, the things which move, and the electric field, the source of the force which causes them to move.
ReplyDeletealso, why 26 hours? that's an unusually precise number.
ReplyDeletethe 16 pushes the graph to the right to resemble J_diff_hole. x_0 is equal to one. x-1 flips over the graph and makes (1,0).
ReplyDeleteThe d(jn(x))/dx is zero at approx .85. This would resemble to max drift and current. dn(x)/dx is zero because n(x) would be constant due to the drift and diffusion exchange of the current.
if this were all of J(x), then this would suggest that there is a negative drift from the n-side of the semi.
N_c will effect n(x)
Brian, you need to upload the picture to an image-hosting site, like imgur:
ReplyDeletehttp://imgur.com/
Then you can post a link to the picture. I'm not sure if you can just post a picture straight into a comment, but you can put the link.
is x_0 at 1? thats where the drift current goes to zero on the x-axis. outside of that there wouldn't be a depletion region.
ReplyDeleteis the 16 from the width of the valley in the graph--the exponential? I guess that would be because of the strenght of the electromagnetic forces from the charged particles. that would depend on the material.
would this be the drift current of the carriers/electrons? its negative current, and current is movement of positive charge, so negative current would be movement of negative charge of the carriers.
5 comments above, Kelsey said:
ReplyDelete"Just to really play with things in the graph, the (x-1) term in the exponential term shifts its maximum to x=0."
I think you meant x=1, no?
MikeK said:
ReplyDelete"Quantitatively, we have J(x)~(x-1)*Exp[-16(x-1)^2]. dJ/dx would be (-32(x-1)^2+1)*Exp[-16(x-1)^2] (I think), which is still a Gaussian, but modulated by the negative parabola -32(x-1)^2+1. So, it's max value is now at x=1, or at the boundary of the depletion region, meaning that the number density is changing most rapidly with time at the edge of the depletion region.--This seems intriguing, but I'm not sure what to make of it, or where to go from here. Any comments would be much appreciated.
"
That seems to make sense to me. Tomorrow perhaps we can consider an extra term such that:
dn/dt=dJ/dx - (n(x)-n0(x))/tau
where n0(x) is the equilibrium value that n(x) should have and tau is a characteristic time scale for creation or destruction of electrons (creation by "generation"; destruction via "recombination").
With this term we can let n(x) deviate from what it "should be" (according to mu), and indeed, that may have to happen where dJ/dx is non-zero.
(in steady state, what is dn/dt??)
26 hours? no reason really.
ReplyDeleteI definitely agree with what everyone is saying about the origins of the 16 in the exponential. I'm picturing this graph as the combination of the graphs of n(x) and E(x) ( in the donor doped side). The 16 is then from n(x)=10^16 component which must be where the combo of E and n had is maximum. As for what Kelsey is saying about the (x-1) terms I understand the math but not so much the implications or its physical grounding.
ReplyDeleteWhat I find interesting about the dn(x)/dt=dJn(x)/dx is that the only place I have seen a d/dt is in the books discussion about electron-hole recombination. I wonder if it would make sense to bring this into our discussion. Perhaps it could also account for why the drift and diffusion currents have maxima near the edges. The way I'm thinking about it is that at x=0, the free electrons and holes are brought close together and it would make sense that there would be a greater recombination rate there, this could then account for high drift and diffusion further form the center...
hmmm. I think the 16 comes from the unitless deltaE/kt ratio in the exponential.
ReplyDeleteThe (x-1) outside the exponential is from the electric field large and negative at x=0, goes linearly to zero at x=1==x_0. Does that make sense?
Yes. Oops. I meant to say "from x=0 to x=1".
ReplyDeleteqft: BrianV says"I think 16 reflects (ΔE/2)/kT, since the exponential term here comes from n(x) = Nc * exp(Ec(x)- μ), and Ec(x) is proportional to ΔE/2.
ReplyDeleteEmma: "is x_0 at 1? thats where the drift current goes to zero on the x-axis. outside of that there wouldn't be a depletion region"
ReplyDeleteThat's exactly right-we wouldn't have any current outside the depletion region. Like was shown in class on Tuesday, we saw that just inside the depletion region we saw the largest capability of current, since we had the largest number of carriers (orders of magnitude more than in the middle [x=0], where the E-field is the largest).
I like what Mike said:
"dJ/dx = dn/dt, for J could have no time dependence but still have space dependence, so dJ/dx would be nonzero but not be a function of t. Thus, dn/dt would be simply a constant with respect to time, meaning either a regular increase or decrease of n at each point in space, regardless of the time. This seems like some sort of steady state situation, where the current density is not changing with time, and the number of carriers is going up (or down) steadily. "
But that whole idea seems a little weird to me. Don't we have, in our case, a change in current as a function of x? Does this imply that an overall change in carrier density? I guess that would make sense, and I think Mike says why:
"However, the continuity equation assumes that no charges all of a sudden appear--that there is local charge conservation. But, we can create a hole-electron pair pretty easily. Then, those new free electrons could be swept away by the electric field, causing holes to be left-over, and so less negative charge than before."
If I get it, I REALLY get it now, but if I missed it, then I'm not really understanding much...
Sorry to post this so far after everyone else, but I'm literally going from place to place as fast as I can all day on Tues and Weds and am still getting places late and not getting home 'till late without stopping to check the blog. Anyway -
ReplyDeleteThe 16 has got to be from the exponential in the n(x) term, just like everyone said. I can't think of any better explanation. (1-x) is the (x_0-x) term that keeps popping up, in this case from the E(x) equation. Going off of that, x_0 = 1; this makes sense, considering the location of the minimum on the graph in relation to x=1.
What does this suggest about dn(x)/dt = dJn(x)/dx? I guess you could read this as "in order for electron density to change, electron current must change." Better still; "if more electrons move around, then electrons will be moving around more."
Though this seems like something Yogi Berra would say, it's significant in that it says something about how charge can't be created or destroyed here, and how it has to come from and go somewhere. As long as diffusion current is equal and opposite to drift current, like under zero bias, that isn't an issue, since nothing's moving around; since I'm assuming under bias they will be different, will this lead to some kind of capacitance as charges build up unequally? Farads already popped up once, and our "no free carriers" ansatz might result in sort of a dielectric between the two conducting sides, which is how you make a capacitor. Further going off of this idea (no idea if it's valid): since a diode is a p-n junction, might it's resistance to current flow in one direction have anything to do with this capacitance, in the same manner as a capacitor blocks DC current?
(After more thought: recombination keeps getting mentioned, and that makes more sense as a "charge-sink," or a place for the extra charges to go. Hm...)
I don't think I can be sure which current the graph represents, as I don't know which side is doped with what. Assuming it's the same model from class, though, with the right side being n-doped, I think that side would be electron drift.
As for the plots, my guess is that the original curve there will be flipped over the x axis for electron diffusion, flipped over the y axis for hole drift, and then flipped over both for hole diffusion. Electron drift will then be balanced by electron diffusion (and the same for holes), and it makes sense electron-related effects would happen on the n-side and hole-related effects on the p-side, since that's where most of each are. As for the change in concentration over time... I'm not sure what to think. I just keep thinking capacitance as charges start building up unequally; like water, if you have high current and then lower the current somewhere along the way - by adding a dam, perhaps - then you get a buildup behind that dam, like a reservoir. A capacitor is essentially an "electrical reservoir," so... yeah. (Now that I've considered recombination, though, I should add again that that sounds more likely. Also, Mike said it, so I'm just going to agree.)
For the last question, I have no idea. Going off of what other people say, I see how an extra e would be necessary, unit-wise... Though I agree that a mobility-related term would make sense, since charges have to move around. It probably wouldn't cancel, either, and might become squared if you combine the terms from either side; higher mobility would help increase current, but would fight concentration differences, so you might get blah on one side and 1/blah on the other.
Oh, and I think I switched hole drift and diffusion; forgot about the sign.
ReplyDeleteLatex test:
ReplyDelete$J_n$
$J_n (x) = \fraq {\partial x} {\partial d}$
ReplyDeleteSo, we can do $\LaTeX$ now?
ReplyDelete