Notes: a) V== Vext refers to an externally applied voltage which can cause current to flow across a p-n junction.
b) This is a very important assignment!
c) Please check here frequently for comments. Also, you will get a 5 point HW bonus for your comments. Also, you can use LAtex in your comments! WooWoo! (thanks to Brad)
1. For equilibrium (V=0) show that the electron drift current in a p-n junction can be expressed as a function of x using the: electric field at x=0 (negative), the potential step and depletion length scale $\color{white} e \Delta \Phi & x_0$, also kT, etc (see "Image" above. Also, correction note: Jn(x) should refers to just the electron drift current, not the total Jn(x). The label is misleading. Sorry.)
a) what is the range (x) of validity for each line of the 3 lines of the eqn for Jn(x) ? Why?
b) For $\color{white}\mu =5000 (cm^2/V sec) , N_d = N_a = 10^{17} (cm^{-3})$ and $\Delta \Phi =0.8 eV$ and $E_g = 1.0 eV$ (no tricks, as you see), find the values of E(0) and $\color{white} x_0$ and use them to obtain largest value (max or min) of the drift current. (Is it positive or negative?)
c) graph this drift current vs x (for x in the range of validity of lines 2 and 3 of the eqn for Jn(x) which is...?)
2. a) Graph all 4 currents for this system. Why do you not really need to do any more calculations to do this and, for example, why do you not really Need to calculate Jn(x) drift for x less than zero in order to do an okay graph. (Answer here. That will help people.)
b) Why do we not worry much about generation and recombination (or dn/dt) for the unbiased p-n junction?
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For a biased p-n junction:
3.
a) explain why and when we can neglect the time derivative term in the continuity equation?
b) for x greater than x_0, explain how we can simplify the expression for hole current (and why).
Then combine that with the continuity equation to obtain a solve-able diff eq for p(x). (You may post your result here, if you like.) (latex is enabled.)
c) Assume $\color{white} p^0 (x) = (n_i)^2/N_d$ . Justify and discuss that here. (PS. What is $\color{white} p^0 (x)$. Elucidate that here for an extra bonus.)
d) Assume $\color{white} p (x_0) = ((n_i)^2/N_d) e^{eV/kT}$.
What is the significance and meaning of this assumption or ansatz? This is worth a lot of thought! (and a major bonus: what is $\color{white} p (x_0)$?
4. With the above assumptions, solve for the hole current for x greater than x_0.
5. a) For V=.05 Volts, how big is this hole diffusion current at x= x_0? How about for V= 0.1 or 0.2 Volts?
b) Just out of curiosity, how does it compare in magnitude to the (individual) currents you calculated in problem 1 (and 2)?
6. Discuss the x-dependence of this hole diffusion current for x gt x_0? Describe and explain what is going on?
7. Think about the dependence of J on x in the depletion region (and outside). Discuss any thoughts or speculations you have about the dependence of J on x in the depletion region (or anywhere else) (comment here).
8. Think about the dependence of n(x) and p(x) on x, both in and outside the depletion region. Discuss and comment here.
Sweet script!
ReplyDeleteFor number 1b) we're supposed to calculate E(0). Going back to the solutions from the previous homework, $\color{white} E(x) = \tfrac{-e N_d}{\varepsilon } (x+x_0)$. Looking at the units tho, we get $\color{white} \frac{[coul][cm^3][cm]}{\frac{[coul]}{[V][cm]}} = [V][cm^5]$. I see no way to work around this to make the units work. Am I missing something or is the equation incorrect?
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ReplyDeleteAlso, if anyone needs help generating code for LateX, this website is useful. There is a preview at the bottom so you can see what you are doing. When you are done just copy the code in like this
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ReplyDeleteWell I agree the units are wrong, thats the issue im having. I'm using the E(x) equation that Zach posted on "HW #3 problems 1 and 2 solutions: Please comment and critique"
ReplyDeleteJust realized N_d was cm^-3 ... don't know why I was thinking cm^3. Silly mistake
ReplyDeletebrad-
ReplyDeletethe units of that equation for E(x) do work out. remember that the units of Nd are cm to the MINUS 3rd power. also, i found it more intuitive to put epsilon in units of (C^2)/eV*cm, but i guess thats just a matter of personal preference.
o woops i guess i didnt refresh my screen soon enough- glad you figured it out though
ReplyDeleteso i am confused on the very first part of number 1 :/
ReplyDeletewe are trying to find ELECTRON drift current, so wouldn't we want to use the expression for the electric field in the region 0<x<x0 ?
in my notes, that means that the field would be given by: E(x)=E(0)*(x-x0)/x0, but in zack's equation above, we have the negative of that...
can anyone explain this to me?
well i am not sure. There could be an error in the equation above, or could it be because the electron carries a negative charge (so a negative E-field produces a positive current)?
ReplyDeleteok yea i guess equation one is the generic equation for drift current, so for electrons, it would have a minus sign in front of it. makes sense!
ReplyDeletewell i think you are being very nice about that.
ReplyDeleteActually, I think Zack is fine since E[x=0] is negative, and so if we factored that negative out, our slope would be (-)(1-x/x0), which is a positive slope, exactly what is right for our E[x]. I had actually worried about this as well, until I realized that E[x=0] is inherently negative.
ReplyDeleteSo, with #1b, $\Delta\Psi$ should be in volts, so that $\Delta E$ is in eV, right? Should we change it to say that $\Delta E=1.0eV$? Otherwise the units don't work out and it doesn't really make sense.
ReplyDelete~Lou
I think it should be $\Delta \Phi = 0.8 Volts$. Does that seem better? (and then $\Delta E = 0.8$ eV.
ReplyDeletePS. Phi, not Psi)
Ok, I'm confused... $x_0=\sqrt{\frac{\epsilon\Delta\Phi}{eN_a}}$ \\
ReplyDeleteand $E(0)=-\frac{\rho}{\epsilon}x_0$ \\\\
The units work only if $\epsilon=8.85\times10^{-12}\frac{C^2}{J\cdot m}$ in the equation for $x_0$ and $\epsilon=8.85\times10^{-12}\frac{C^2}{J\cdot m^2}$ in the equation for $E(0)$. (Look at the m vs. m^2 in the denominator of the units)
What is up with this!?
~Lou
I got something abysmally high for E(0) on the order of $6.6\times10^{25}eV$
ReplyDeleteO.o' That just doesn't sound reasonable to me.
~Lou
P.S.: Note on the post just above, $\epsilon=11.8(8.85\times10^{-12})$
what are your units for things?
ReplyDeleteSometimes E refers to electric field (Volts/cm) and sometimes to energy (eV).
I am not sure which one you mean.
Usually E(0) refers to the electric field at x=0, whereas $\Delta E = e \Delta \Phi$ is an energy.
I would suggest:
ReplyDelete$\epsilon_0 = 8.85 x 10^{-14}$ coul/(volt-cm)
Can anybody else help out here?
PS. I think your equations (for x_0 and E(0), will both be fine if you use that.
ReplyDeleteE(0) (electric field) has units of Volts/cm.
Ahhhh, of course! We're looking at the electric field... makes much more sense. Silly me thought E(0) was energy for some reason... thanks.
ReplyDeleteI got E(0)=10^5 V/cm
Seems like a lot... did anyone else get that?
~Lou
Whether that is reasonable, or not, can be estimated by considering: a)you know that you want a voltage change of about 0.8 Volts, b)you get by integrating over a region of width 2x_0.
ReplyDeleteSo how wide x_0 is tells you essentially how large the electric field needs to be.
Comparing your value to other common values makes sense. I remember E_0 typically being on the order of 10,000 V/cm.
ReplyDeletecan someone tell me what the continuity equation is referring to for 3a?
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ReplyDeleteThe continuity equation is the 4th line down in the picture at the beginning.
ReplyDeleteHere's a good way of describing it:
"The continuity equation describes a basic concept, namely that a change in carrier density over time is due to the difference between the incoming and outgoing flux of carriers plus the generation and minus the recombination."
(http://ecee.colorado.edu/~bart/book/book/chapter2/ch2_9.htm)
So basically, it says that electrons have to come from and go somewhere, or current = change in concentration of carriers.
I think question 3a is related to 2b, in that they are related to the contunity equation:
ReplyDeletedJ(x)/dx=dn(x)/dt
so 3a is just asking, when will thing no longer be chaning over time, which to me indicates a steady state for the junction.
For 1c do we have to do a derivative, or do we just put in values for x in order to graph drift.
ReplyDeletethanks. I am wondering for a biased p-n junction where do the charge carriers actually pile up before they reach steady state?
ReplyDeleteDoes anyone know how to go about doing 5. I don't know what to do with t_p.
ReplyDeleteFor 1b, you can take a derivative and set it to zero and then realize the exponentials can be factored out and ignored, or, an equally good approach is just to calculate a few values and home in on the region of the max (or min) value.
ReplyDeleteRodney said: "... I am wondering for a biased p-n junction where do the charge carriers actually pile up before they reach steady state?"
I think that is a very important question!
Rodney: "Does anyone know how to go about doing 5. I don't know what to do with t_p."
ReplyDeleteZack: How did it go with 3 and 4? Those come before 5. Can you rephrase your question with reference to your results from 3 and 4??
Assuming I did everything right (which might be a big assumption ...), my final equation has a $ \tau_p $ & D dependence. What should we do for these values?
ReplyDeleteAnyone want to get together tomorrow at the science library before class (~10am)?
ReplyDelete~Lou
Mine has a tau dependence as well, but I simplifies the D into (mu)KT/e which we have values for. (I'm assuming this is for number 4?)
ReplyDeleteLou, I'll be there. I need to work on this stuff.
ReplyDeletewhy don't you try $\tau_p$ of about 10^-6 seconds or 10^-5 seconds. How does your result depend on tau_p. Is it proportional or what?
ReplyDeleteI just cleaned up the eqn in 3d. were you able to make sense of that before? Is this what you thought it said?
ReplyDeletealso i fixed a latex problem in 1b.
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ReplyDeletefor 3 doesn't dp/dt go to zero outside x_0 b/c there is no e-field. This simplifies the eqn.
ReplyDeletefor 4. Are we using the simplification and eq 4 from way above to solve current(plugging in p^0 and p(x)) or does eq 5 need to be used here?
for 5a, can we use the mobility value that was given?, because the mobility for electrons is not the same as holes. I was assuming the mobility value given was for electrons.
ReplyDeleteLou:
ReplyDeleteI got E[0]=-1.1x10^7 N/C for my E-field, which corresponds to -1.1x10^5 V/cm, in excellent agreement with your value.
My value for Jmax seemed high---9x10^5 C/cm^2-s. Does this seem comparable to others' values?
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ReplyDelete@Mike
ReplyDeleteI got something very similar for E[0], but I got $ 2*10^(15) C/(s*cm^2) $ for $J_(max)$, which I fear is too high as well
I got a magnitude of 1.1*10^4 V/cm for my E-field and my Jmax was 95 A/cm^2.
ReplyDeleteI've been having trouble with basic algebra and orders of magnitude, though. At least we have the same numbers, even if they are orders of magnitude off!
Brad, I was getting a similar value for a (very long) while, and I realized that I had an extra power of e in my exponential.
ReplyDeletefor number 6 what does x gt x_0 mean?
ReplyDeleteIt means "x greater than x_0"
ReplyDeletethanks. i thought greater than would be >, so i thought it was something else
ReplyDelete@ Brad-I got the same value for my Jmax-seems ridiculously high. I still can't figure out how to fix it.
ReplyDeleteWhat exactly is the simplification we're supposed to get for 3b? Just that our dp(x)/dt term gets killed?
and by that I of course mean the E(x) term drops out since we're outside the depletion region, IN ADDITION to what I said above which was related to 3a.
ReplyDeletetthompso:
ReplyDeleteI don't think dp/dt goes to zero outside x_0 because there is no E-field (I think dp/dt is zero even in -x0 to x0, where the E-field is nonzero). It is zero because we are assuming a steady state, when all transients have died out and the carrier number isn't changing.
Justen:
ReplyDeleteYeah, I think the simplification is that E=0, so we have only the diffusion current term. Then, we can take x-derivatives of both sides and get dJ/dx = -eD*d^2p/dx^2; from the continuity equation we have dJ/dx=-dp/dt+(p-p0)/tau. We assume steady-state, so dp/dt=0, and we have (putting continuity and diff. eqn together): -eD*d^2p/dx^2 = (p-p0)/tau.
Kelsey:
ReplyDeleteFor the E-field, I'm using: $E(0)=\frac{N_dex_0}{\epsilon}$. First of all, I put everything into SI, and then converted back to V/cm at the end. So, $N_d=10^{17}/cm^3=10^{23}/m^3$. I found $x_0=73nm=7.3\times10^{-8}m$. We have $e=1.6\times10^{-19}C$ and $\epsilon=\epsilon_r\epsilon_0$, where $\epsilon_r=12$ and $\epsilon_0=8.85\times10^{-12}C^2/N-m^2$. Putting these all together, we have:
$E(0)=\frac{(1.6)(7.3)}{(1.2)(8.85)}\times10^{-19-8+23-1+12}=1.1\times10^{7}V/m$.
This is in Volts/m. To get Volts/cm, simply divide by 100, yielding $E(0)=1.1\times10^5V/cm$, only 1 order different from your value.
Note that my value for E(0) is a magnitude (it's actually negative in sign), and that the units (though cut-off) are V/m; ie, $E(0)=-1.1\times10^7V/m$.
ReplyDeleteRegarding 7, I believe we are assuming that electron-hole pairs are not being created or destroyed in the depletion region. Moreover, we're assuming a steady state. The first assumption means p(x)-p^0(x)=0, and the second means dp/dt=0. But, these are the only two terms in the continuity equation. That is:
ReplyDeletedJ/dx = -edp/dt-e(p(x)-p^0(x))/tau = 0.
So, inside the depletion region, dJ/dx=0, meaning J (and so I) is a constant. So, the current density is a constant in the depletion region.
Well, I should be more specific--the current density due to excess carriers is a constant in the depletion region (the part due to drift and diffusion for unbiased are non-constant, but cancel identically and so are ignored).
Regarding #8:
ReplyDeleteIn the deep n-type (x>>x0), n(x)~Nd and p(x)~p^0(x0)=ni^2/Nd. Near x0, but still past it (x>x0), n(x) is still basically Nd but p(x) is more like p(x0)=(ni^2/Nd)*Exp[eV/kT]. In between x0 and x>>x0, p(x) decays exponentially to the constant value ni^2/Nd. For x<<-x0, and x<-x0, and in between x<<-x0 and x=-x0, just switch n and p, and Nd to Na (but they are equal for our problem), because of the symmetry of the problem.
Inside the depletion region, n(x) and p(x) have to be continuous, and so have to connect their endpoints at the boundaries of the depletion region. It seems plausible that they will follow the same form as before (Exp[-k(x-x0)^2]) inside the depletion region, but now the jump is smaller because the potential difference phi has been reduced by the bias voltage V. So, it seems like p(x) and n(x) plots will look similar to the unbiased case for inside the depletion region, but just a bit shallower because of the lessened effective voltage, phi-V.
Do these comments make any sense? Any comments are welcome!
Thank you, Mike. I managed to forget how many cm are in a m.
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ReplyDeletehow exactly did you guys go about solving number 4? i am really stuck on that one.
ReplyDeleteIn addition to WOW!, i have a few thoughts/suggestions:
ReplyDeleteFirst, regarding $\mu$, the mobility. I think we should use the same value for electrons and holes even though that is not really correct. (Though i would be interested in how different values of mu and m* would effect our results after we do examine this simpler model.)
Second, I wonder if we might think of the region of the junction, including both the depletion region and the diffusion (or recombination) regions surrounding it, as "hot", or non-thermal, with reference to their electron distributions??
Third, how would tau_p's (recombination times of 10^-3, 10^-6, 10^-9 (all in seconds) effect our results? (all are longer than the collision tau (is that about 10^-13 seconds i think (based on the mobility). How long does it take an electron to get across the depletion region? Why might that be relevant?
$ 7 $ For the current dependence on J inside the depletion region, we have seen that the flow is not constant. This result to a building up of particles in some region inside the depletion region. This is as expected, looking at the equation $ J(x) = e \mu \epsilon (x) n(x) $. We have n(x) which is exponentially increasing during the depletion region, and E(x) which is linearly decaying to 0. This combination results in a peak of particle flow which eventually tapers as a result of the E-field going to 0.
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ReplyDelete$ 8 $ We know the equations for both p(x) and n(x). $ n(x) = N_d e^{-(E_c(x)-\mu)/(kT)} $. this results in an exponential growth within the region of the depletion region, from ~10^3 at $x=- \infty $ to $ -x_0 $, and ~10^17 from $ x_0 $ to $ x=\infty $. Outside of the depletion region both n and p are constant, which simplifies things outside the depletion region. Outside we have no dependence on n(x).
ReplyDeleteThe reason for the ease of graphing 2a) is because there is no bias the drift and diffusion are mirrored images.
ReplyDeleteFor the bonus: $p^0(x)$ is the hole concentration without bias, or what the hole concentration is "supposed" to be.
ReplyDeleteI know it may be rather obvious to some people, but I am confused about how to get the hole current in # 4. Can someone point it out to me?
ReplyDeleteIn addition to Dave's comment, the http://www.openmaths.org/cgi-bin/mathtex.cgi?\usepackage[usenames]{color}\color{cyan}%20\gammacorrection{1.3}\png%20\normal%20p^0(x) is also the hole concentration outside of the depletion region.
ReplyDeleteoops, I meant p^0(x) not all that gibberish.
ReplyDeletedoesn't $mu$ become $mu(x)$ when we apply a voltage? Won't this affect our $n(x)$ quite considerably? Like on the left side $x<0$, we'll have even less (less than $10^{3}$), and on the right side, even more (more than $10^{17}$)?
ReplyDeleteIn lecture we discussed an ansatz that the chemical potential is not a single value for both carriers within the depletion zone, that instead each carrier is subject to a chemical potential equal to that at the boundary of the side on which it is majority carrier for the side of the depletion zone on which it is majority carrier. Therefore, in the region -x0<x<0, the hole concentration would seem to remain at the dopant level. Similarly, in the region 0<x<x0, the electron concentration would seem to be constant at the dopant level. However, we also have a current across the junction, so carriers are moving. Then there are the minority carriers, the concentration of which we know is nonzero even in the region just outside of the depletion zone.
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