Thursday, April 22, 2010

HW #3, begin immediately, due Tuesday if possible

As we have discussed, one can start with a depletion approximation, in which we "pretend" n(x) and p(x) are both zero in the "depletion region", then, starting with that, on can calculate V(x) and hence Ec(x) and Ev(x) and then use them to recalculate n(x) from n(x) = Nc exp{-(Ec(x)-mu)/kt}, and p(x) from an analogous relation.

1) Suppose Ec-mu = mu-Ev = 0.1 eV, and Egap=1. eV (and kT=.025 eV)
a) Graph Ec(x). What is total variation in Ec(x) from left to right in your graph.
b) where are the boundaries of the depletion region?

2) Suppose Ec-mu =0.15 ; mu-Ev = 0.05 eV, and Egap=1. eV (and kT=.025 eV)
(How does that change things?)
a) Graph Ec(x). What is total variation in Ec(x) from left to right in your graph.
b) where are the boundaries of the depletion region?

3) (re)Calculate n(x) for each case.

4) For a homogeneous doped semiconductor, given n=4x10^17, m*=.2me, and hbar/tau=3 meV:
a) what is tau in sec?
b) what is sigma in (Ohm-cm)-1 ?
c) what is mu in cm^2/V-sec ?
d) What is rho? and what is R (resistivity) in Ohms for a "strip" 1 mm wide, 100 nm thick and 2 cm long?
e) What is I for that strip for a voltage of 10 microvolts end to end? How much heat will that generate?

5) how come nobody uses the concept of mean-free-path for semiconductors? (but they do for metals)

6) In a p-n junction, one can write J=n(x) e mu Ex(x) - e D dn(x)/dx (eqn 1) (c.f. eq 5.3 or so in Streetman), where n(x) is the density of electrons in the CB, mu is the mobility, Ex(x) is the electric field in the x direction and D is the diffusion coefficient. Turns out D = mu KT/e (Einstein relation).
a) Show that we can rewrite eqn. 1 (above) as:
J=n(x) e mu Ex(x) - n(x) mu KT d(ln{n(x)}/dx

b) take the derivative using the expression way above for n(x) = ...

c) evaluate at a value of x such as x=0 for a nice symmetric junction (as in problem #1). Show that this is, or is not, zero. [side points: What is Ec(x)? How is it different from and related to Ex(x) in terms of definition, units, physical relationship, etc. ]

7) Consider two different junctions, one in which Ec-mu =mu-Ev = 0.1 eV, and another in which
Ec-mu =mu-Ev = 0.15 eV. What is the diffusion current across the junction interface (x=0) in each case? How does the diffusion current depend on "delta mu"? It is a weak or strong dependence?

8) Suppose you apply an external voltage of 0,3 Volts. You may represent that by a linear dependence of mu through the depletion region (but mu is independent of x outside of that range). Do it in a way the reduces the offset of the band edges (so that Ec and Ev have less x dependence.)

a) How does this effect the extent (length scale, x_0) of the depletion region?

b) Sketch the band edges and mu

c) (major extra credit) Starting with the expression for J (in #6, above) consider, analyze and discuss how this effects the current through the p-n junction. Big change, small change...???

Questions and discussion are welcome and encouraged.

73 comments:

  1. This comment has been removed by the author.

    ReplyDelete
  2. in problem #1, we are given Ec-mu = mu-Ev... is this just referring to the regions outside of the depletion region? ie, we are still working with the same picture we had in class, correct?

    ReplyDelete
  3. brianv,
    I think you're right, we are working with the same picture as in class and that 0.1eV refers to the regions outside of the depletion region, in the bulk p- and n-type materials (where Ec-mu and mu-Ev are constant)

    ReplyDelete
  4. question from a student received by email:
    I'm having a problem with calculating the width in the depletion region.
    I tried to look at the equation Ec(x)=Eg+dE/2 (x-x0)^2/x0^2, and look at a region where we know Ec(x) -> at x=x0+ε. however, this doesn't really give me anything solvable as far as I am seeing.

    I then went back to an example in class where we had n(x) = Nd exp[-dE/2 (x-x0)^2/x0^2 /kt), but again I really don't see anywhere to go with this.

    Could someone please point me in a better direction?

    ReplyDelete
  5. to calculate the width I think you want to start with wanting to make mu constant.

    The equations you mention above use x_0, but can't be used to calculate it. Flattening mu, via charged depletion region -- E(x) ---- V(x) ---- determines x_0. Can someone talk about that more. I can't type much.

    ReplyDelete
  6. I think that we want to go back to the older equations to calculate x0. Since we want mu to be constant across the junction, we require that the energy shift = qV, where V can be written as a function of x0. You have to use Ec-mu (or mu-Ev) to get the dopant concentration, which is one of the terms in V.

    ReplyDelete
  7. I'm pretty sure we need more information to solve the for the depletion region width in problem 1. We need Na and Nd and/or Nc and Nv. Also, I am assuming we are working with doped silicon (for finding epsilon)? And Na/Nv=Nd/Nc since Ec-mu=mu-Ev, right?

    ReplyDelete
  8. Yeah, in order to get x_0, we need Nd. However, we can only find Nd/Nc (not Nd itself) if we only know kT and Ec-mu, for Nd = Nc*Exp[-(Ec-mu)/kT]. It's just a guess, but I'm thinking we can just use Nc=5*10^18/cm^3 like usual. Also, yes, I believe Ec-mu=mu-Ev --> Na/Nv = Nd/Nc.

    ReplyDelete
  9. Oh, and we also need epsilon_r for x_0, which is ~12 for Si...so, I think we have to assume that too.

    ReplyDelete
  10. hmmm. i guess you do need a lot of things to get the width. I didn't anticipate that. (I was just focused on qV.) Good work sorting that out. (We can use the additional input as specified by Mike.)

    Yeah, so sorry about that and good group effort analyzing that. I was trying to make things easier, sort of/in a way, cause at least this way you can see what energy shift is needed, right?

    ReplyDelete
  11. One more thing: we need to know if Nc=Nv in order to calculate the width of the depletion region.

    ~Lou

    ReplyDelete
  12. So, I used the following thinking to calculate the depletion region width on the donor side (x):

    qV=e*rho(x^2)/2epsilon=∆E

    -->x^2=(2*epsilon*∆E)/(e^2*n_d)
    where n_d=Nce^(-(Ec-mu)/kT)=9*10^22/cm^3
    ∆E=0.8eV

    And I got x=270

    Did anyone else get that? Does that sound right?

    Also, if Nc=Nv the total depletion region would just be twice the above.

    ~Lou

    ReplyDelete
  13. I agree with your formula, but I got x0 = 107 nm. I calculated the same values for the energy shift and dopant density, so I'm not sure why our answers would be different.

    ReplyDelete
  14. lou: i think you need to double-check your value of Nd. i used the same methods as you did, but i got Nd=9.2*10^16 cm^-3. i then calculated x0 (by the same method you suggested), and got 7.5*10^-8 m, which seems reasonable.

    also, a caution that could save you some time: remember to make sure that you get epsilon in units that include eV rather than joules- this held me back for a while before i realized my mistake!

    ReplyDelete
  15. Ya, sorry, I meant to say n_d=9*10^22/m^3, not /cm^3

    ~Lou

    ReplyDelete
  16. what does epsilon physically represent in our model? I missed class and never saw these equations last week!

    ReplyDelete
  17. Ahhhh! I see what went wrong! With x^2=(2*epsilon*∆E)/(e^2*n_d) you cancel one of the e's (fundamental charge) in the denominator when you convert ∆E from eV to V

    Just in case anyone else didn't notice this at first, I figured I'd mention it. :)

    I ended up with 108nm, so approx 110nm.... Seems close enough to your answer Kelsey. Brian, you got 75nm? Seems pretty different from Kelsey's and my answer... O.o Did you forget the 2 multiplying epsilon in the numerator? That is supposed to be there, right? The lecture where we solved for the width, I think Zach forgot to include it.

    Another note, we still need to know if Nc=Nv to figure out the entire width... ^_^

    ~Lou

    ReplyDelete
  18. brad: epsilon is called the permittivity of the material. you find the permittivity for, say Silicon, by multiplying epsilon_r, the dielectric constant of Si which can be found in appendix III of our book, times the "permittivity of free space," epsilon_0, which you might remember showing up in electrostatics equations.

    so, epsilon_r for Si is 11.8 according to Streetman, and epsilon_0 = 8.85*10^-12 F/m.

    as i mentioned in my above post, you will have to get epsilon_0 in terms of eV, rather than Joules, but that's a pretty simple conversion... hope that helps

    ReplyDelete
  19. lou- you're right, i did not include that extra factor of 2...but i don't quite understand where it comes from. could someone explain that to me?

    ReplyDelete
  20. This comment has been removed by the author.

    ReplyDelete
  21. I don't have the 2 in the numerator, and I don't think it needs to be there. When you find the function for Ec(x) for 0<x<x_0, we had 1/2*e^2*N_d/epsilon*(x-x0)^2. At x=0, this becomes 1/2*e^2*N_d/epsilon. By symmetry, this energy is just half the total height, so we need to double this value. Doing that cancels the one-half factor, and we are left with Delta E = e^2*N_d*x_0^2/epsilon. Using this I got 77nm, which is very close to 7.5x10^-8m.

    ReplyDelete
  22. For problem 2, we are seeing an asymmetry we haven't really worked with before in the P-N junction. Before I was able to, via symmetry, say what ∆E was. However now, I'm assuming we need to calculate ∆E directly.

    Would we use ∆E=eρx²/2ε? If so, what is ρ? Is it just the resistivity of the silicon? If thats the case, its just 6.40×10² for Si? But then what would we use for x? =/

    ReplyDelete
  23. brad:

    in that equation, rho is the charge density (C/m^3) in the depletion region, and it can be found by multiplying the charge of an electron by N_d.

    ReplyDelete
  24. In all of these problems, ρ is the density of charge, and will be Nd on the n-type side and Nv on the p-type side. Chapter 5 in the book has a good explanation of how to work with an asymmetric junction. Basically, you want to get the equation for V in terms of W, the total width, instead of x0 to either side of the junction.

    An equation similar to your ∆E one (more complicated, since ρ is different on different sides of the junction) is what gets used to get W. Once you have W, you can get x0 on each side of the junction by requiring that the charge density per area be constant
    (xn0*Nd = xp0*Nv).

    But, since the x0's are unknown until this equation is used, ∆E can't be calculated this way. Instead use the requirement that μ be constant across the junction. To meet this, it is necessary that
    ∆E = Egap - [(Ec -μ) + (μ - Ev)]

    ReplyDelete
  25. This comment has been removed by a blog administrator.

    ReplyDelete
  26. I'm not very sure about #5--is the mean free path not well-defined for semiconductors because the number of carriers can change with doping? Any thoughts?

    ReplyDelete
  27. I was thinking that it might have something to do with the sheer number of available states. Based on our discussions in lecture, it doesn't sound like any given electron in the conduction band is likely to encounter another electron. The same is true for the holes in the valence band. My understanding of mean free path is that it is related to the frequency of random collisions, and I'm not sure what the charge carriers in a semiconductor collide with. It seems like collision is a lot likelier with the electrons on the Fermi surface of a metal.

    The trap states from dopants might also affect the mean free path, I guess..

    ReplyDelete
  28. Zack:

    Did you make a sign error in #6? Should it be J=n(x) e mu Ex(x) + e D dn(x)/dx (+ instead of -)? This is what the text has in the front cover for e- (it's - for holes). If I use your version, I keep on getting that Jx(0) is non-zero, which didn't seem to make sense since we're at equilibrium. If, however, we had a plus sign, the sign issues would disappear.

    Kelsey:

    So, are you saying that the mean free path doesn't make sense because there is nothing for the e-/holes to collide off of, and so no real mean free path? Aren't there phonons in a semiconductor, too, which would cause some scattering?

    ReplyDelete
  29. Given the phrasing of the question, I'm trying to figure out what the significant difference between semiconductors and metals is. In the kind of model we're using, the main difference seems to be the large number of states available with thermal excitation. I'm just trying to find a way to relate that to mean free path. I'm not sure how to think about phonon modes in a semiconductor, or how to integrate them with our model.

    ReplyDelete
  30. i am confused on number 2b:

    Does Nv=Nc for this problem? and, does Na=Nd for this problem? i am having a little trouble understanding the implications of the fact that mu is not exactly centered between Ec and Ev...

    ReplyDelete
  31. Regarding number 5, I remember Zach specifically saying that the MFP on a semiconductor due to the crystal structure was on the order of mm. In that case, would the MFP not be on the same order of the device itself, and therefore, negligible to think about?

    ReplyDelete
  32. For 2b, I am assuming that Nv = Nc. The fact that μ is not centered between Ec and Ev means that Na =/ Nd. Different energy gaps lead to the same probability of occupation at μ outside of the junction.

    ReplyDelete
  33. In number 4, is the me in m*=.2me referring to the mass of an electron?

    ReplyDelete
  34. pretty sure. at least, that's what i used.

    ReplyDelete
  35. 4e. Is the heat generated equal to the power? P=R*I^2

    ReplyDelete
  36. Yeah, P=I^2*R (or you can use P=IV, or V^2/R--they're all equivalent)

    ReplyDelete
  37. What kind of values are people getting for I? I got something really small (3*10^(-10)), which makes me suspect an error in my math.

    ReplyDelete
  38. For 5, i was thinking that for semiconductors scattering time and mobility are meaningful "intrinsic" properties, but that there is an issue with defining a mean free path that is not an issue for metals. (i could be wrong.)

    For 6, sounds like there is a sign error. Can you be more clear and definitive in correcting that. I think that would help other people. (Thank you)

    ReplyDelete
  39. like in #4, try to calculate the mfp? what is the problem??

    ReplyDelete
  40. In #6 the equation should read J = n(x)*e*mu*Ex(x) + e*D*dn(x)/dx

    ReplyDelete
  41. PS. For extra credit, explain the origin of p-n junction I-V behavior, especially the exponential increase of I as a function of V.

    ReplyDelete
  42. 6a.) How the heck do you go from dn(x)/dx to d{ln(n(x))}/dx?

    ReplyDelete
  43. lou: set equation 1 equal to the new equation (the rewritten one). the first term is the same for each, so all we really need to do is get the two second terms equal:

    e*D*dn(x)/dx = mu*kT*n(x)*dln(n(x))/dx.

    we know D=mu*kT/e, so we are left with dn(x)/dx = n(x)*dln(n(x))/dx.

    divide both sides by n(x), and you end up with a simple derivative result; the derivative of ln(any function) = (1/function)*derivative of that function... in this case:

    (1/n(x))dn(x)/dx = dln(n(x))/dx

    ps- i wish there was a better way to write equations on this blog :/

    ReplyDelete
  44. *bump

    I wish the blog read LaTeX code

    ReplyDelete
  45. 6b) Are we supposed to take the derivative ∂J/∂x? Where n(x)=Nd*d^[(∆E/2kT)*((x-x_0)^2/(x_0^2))]?

    ReplyDelete
  46. No, I think he wants us to evaluate d(Log[n[x])/dx. This turns out pretty nicely because you can split the log into separate parts, some that are constant and one term that is not (with respect to x, that is).

    Re explaining the sign error in 6:

    First, for my notation, e is positive (the magnitude of the electron charge). So, we should have:

    J_electric_x = e*n(x)*mu*E_x(x), since the current is in the same direction as E (J = sigma*E is just Ohm's law).

    J_diffusion_x = e*D*dn(x)/dx, since the current goes opposite the way the electrons flow (the electrons go in the -dn/dx direction, or in direction of maximal decrease of density).

    So, J_total_x = J_electric_x + J_diffusion_x, or:

    J = e*n(x)*mu*E_x(x) + e*D*dn(x)/dx.

    Note for holes the plus would become a minus, since the current is in the same direction as hole flow and so J_diffusion_x would be -e*D*dp(x)/dx. However, J_electric_x would still be positive, since the E-field isn't changing, and J is parallel to E (through ohm's law).

    ReplyDelete
  47. Can someone tell me there method of finding 2b because now that its not doped evenly on both sides the depletion region should not be from -x_0 to +x_0, which means that we can't find the region like we did in 1b, right?

    ReplyDelete
  48. This comment has been removed by the author.

    ReplyDelete
  49. This comment has been removed by the author.

    ReplyDelete
  50. This comment has been removed by the author.

    ReplyDelete
  51. Here are some interesting relationships:
    current density = n e velocity,
    = sigma E
    = n e mu E
    and so:
    velocity = mu E

    E has units of Volts/cm .

    (Please post correction if I have made any mistakes here.)

    ReplyDelete
  52. Nevermind, I think I figure it out.

    ReplyDelete
  53. @ Zack
    For 4, you've said "1 mm wide, 100 nm thick and 2 cm long"

    Does that mean A=L*W and V=L*W*T?

    ReplyDelete
  54. judging by the title of this topic, i guess the deadline for homework 3 might be flexible? does anyone else feel like they could benefit from a little more in-class coverage of topics like (but not limited to): non-symmetric doping, diffusion, external voltages applied to p-n junctions...

    i wonder(zack) if maybe we could turn in this homework on thursday?

    ReplyDelete
  55. I'm close to finishing, but I would appreciate more in class discussion before we turn this assignment in. We really haven't talked much about 7 & 8, and while I think I could answer them semi-accurate (the book does a decent job explaining it), I think I would benefit from in class discussion.

    ReplyDelete
  56. Please do all you can by tomorrow at noon. 8 hours of sleep is also a good idea imo. I think you can handle the non-symmetric doping; ask questions about that if you are stuck.

    On the other hand, the quest to understand external voltages applied to p-n junctions never ends. It is the holy grail.

    ReplyDelete
  57. For #3, how are you supposed to solve for n(x) when you don't know x? O.o'

    ReplyDelete
  58. @ lg.
    I just plugged in everything but x

    ReplyDelete
  59. what do you mean you don't know x? x is a variable. Do you mean x_0 ? no. i guess not?

    Oh. Hmmmmm You are supposed to write the expression for n(x). Mike already did it above, didn't he?

    ReplyDelete
  60. assuming scattering time is a well-defined quantity (and that is a big asssumption) what would you need to get a mean free path from it? Is that intrinsic to the material?

    ReplyDelete
  61. research indicates that sleep strongly and significantly enhances learning of new concepts. I encourage everyone to get a good nights sleep before tomorrow's class!

    ReplyDelete
  62. Looks like it is due Thursday. Please keep the comments and questions flowing.

    ReplyDelete
  63. For #3, maybe recalculate is not the best word there. What that really is trying to ask is for you to write the formal expression for n(x), in terms of Nc and an exponential derived from the Fermi function (which involves Ec(x), right). In fact, probably it is the same expression for 1 and 2, just the x_0's and the Ec(x0 are different. Is that true?

    ReplyDelete
  64. i know we tried to do this today in class, but i got even more confused- how do we find n(x) in problem 3, for the case in problem 2 (where Nd does not equal Na)?

    ReplyDelete
  65. also, a clarifying question on #6:
    for part b, is it referring to the derivative of n(x), and for part c, is it asking us to evaluate J at x=0, or dn/dx at x=0?

    ReplyDelete
  66. #6, part b) It is referring to the derivative of d(ln(n(x)/dx ? (why that and not dn(x)/dx ??)

    c) it is asking about J at x=0.

    PS. Your discussion related to that (above) is excellent and I will reproduce it here:

    "set equation 1 equal to the new equation (the rewritten one). the first term is the same for each, so all we really need to do is get the two second terms equal:

    e*D*dn(x)/dx = mu*kT*n(x)*dln(n(x))/dx.

    we know D=mu*kT/e, so we are left with dn(x)/dx = n(x)*dln(n(x))/dx.

    divide both sides by n(x), and you end up with a simple derivative result; the derivative of ln(any function) = (1/function)*derivative of that function... in this case:

    (1/n(x))dn(x)/dx = dln(n(x))/dx "

    ReplyDelete
  67. "i know we tried to do this today in class, but i got even more confused- how do we find n(x) in problem 3, for the case in problem 2 (where Nd does not equal Na)? "

    With regard to n(x), there are two distinct parts. One is to calculate Ec(x). The other is to express n(x) in terms of Ec(x). Does that division seem clear? Which part are you confused about?
    (Also, please see my comment 4 comments up and just above yours.)

    ReplyDelete
  68. PS. To brianv and everybody:

    i wouldn't overthink or overemphasize problem 2. True it is computationally a bit more difficult than #1 because it has less symmetry. That means that deltaE is not divided equally between the two sides. But aside from that it is straightforward. You assume a form for rho(x), get an E(x), then a V(x) from that, and you have two unknown widths, related by a charge neutrality condition, etc.

    Conceptually it has very little in it that is different from #1. Just one extra equation and one extra unknown (since x_0p and x_0n are different and related by Na x_0p = Nd x_0n . In your preparation try to continue to think in convergent and divergent styles and don't just try to wrestle down the grungiest problem around cause that it not what will be on the take-home.

    ReplyDelete
  69. ok thanks zack. i guess the part i was really stuck on was calculating Ec(x) for the non-symmetric (Na not equal to Nd) case. the rest of it makes sense to me.

    ReplyDelete
  70. also, a little late in the game, but if anybody wakes up tomorrow, or is also a night owl, "Cool stuff from Kelsey" helps out quite a bit for #6 (and subsequently 7) for anybody who is still having some problems with it. For 6c, think about what SHOULD be happening at equilibrium, and hope that you can reproduce it with numbers.

    ReplyDelete
  71. I have a term clarification question that I hope Zack won't consider cheating on the midterm. He lately started referring to a p-n junction as "unbiased" and I haven't really understood what this meant. I felt like I was following the material well enough that I would pick it up as we continued but now it's proving to be a problem. Does unbiased mean that a p-n junction is equally doped on both sides or is it more technical than that? We had previously been calling that a symmetrically doped semiconductor so I think this is where my confusion comes from. Is it referring to the lack of voltage across the semiconductor? I hope that someone will still be checking this blog regardless of the fact that we don't have homework this weekend.

    ReplyDelete
  72. good question. unbiased means zero externally applied voltage.

    (symmetric, on the other hand, means equally doped on both sides, as you say.)

    "Is it referring to the lack of voltage across the semiconductor?"
    Yes.

    ReplyDelete