Friday, April 11, 2014

Homework 3 & solutions link.

Here is a link to solutions:
https://drive.google.com/file/d/0B_GIlXrjJVn4YnByRnF3bDJaQVE/edit?usp=sharing

Please comment on any errors you find or any questions or thoughts you have. I recommend starting soon. I doubt this set this set can be done in one sitting. 2, for example, is a bit difficult and time consuming I think, but you may learn a lot from it. 5 and 6, which involve diffusion current, are way harder, and also interesting imo. Note the break in this post, and that there are problems below the break!! As always, a lively ongoing discussion is the best way to tackle these problems.


1. Consider a semiconductor with a 1 eV gap and with: \( D_2 = 0.6 \times 10^{22} \, states/(eV-cm^3)\) and \( D_3 = 0.3 \times 10^{22} \, states/(eV-cm^3)\).
a) Suppose it is doped to a level of \(n = 10^{17} \,electrons/cm^3\). What is \(\mu\)?
b) Suppose it is doped to a level of \(p = 10^{17}\, holes/cm^3\). What is \(\mu\)?
c) Which \(\mu\) is larger? By how much? Are they both in the gap? What are the units of \(\mu\)?
d) Which  \(\mu\) is closer to "its band edge"? why?


2. When two things are in intimate contact,  \(\mu\) has to be the same in both (in equilibrium). In this sense \(\mu\) is like temperature. ( \(\mu\) is to carrier density as T is to entropy.)      Suppose you put the material from 1a) into intimate contact with the material from 1b). (Say there is an interface at x=0 and for \(x \le 0\) you have the material from 1a) and for \(x \ge 0\) you have the material from 1b).
a) What could happen near the interface to enable \(\mu\) to be the same on both sides?
b) Okay, that is pretty difficult. Maybe we can break that difficult problem into parts? Maybe a bunch of electrons from the left side sneak over to the right side? If that happened, would the crystal still be neutral (uncharged) everywhere? If not, qualitatively describe and sketch the net charge density as a function of x.
c) Hmm, Let's make a radical approximation. Let's assume that, within a distance d of the interface, all the electrons that would normally be in the conduction band on the left go to the other side. Let's assume that they fill any empty states in the valence band. With that assumption, plot the net charge density as a function of x. What is its maximum value?
d) Calculate and plot the electric field (associated with that charge) as a function of x. [hint: Gauss's law (GL) can be used to get the value of the E field at x=0 and also that E is zero outside the charged region.] Where is the maximum of E and what is the form of the dependence of E on x? (Feel free to discuss this here.)
e) What is the maximum value of E? Does that depend on d?
f) What is the difference in the potential energy of an electron from one side to the other due to the charge layers? (in eV) Does that depend on d?

3. When you have found the difference in potential energy (2f) that equals the difference in mu (1c), then you have found the equilibrium state of the junction.
a) do that and thereby determine d.
b) Would d be different if the doping levels were different? Discuss.
c) (x cred) email me your cogent discussion of the influences of doping on d.

4. Did we leave something out that you need to solve problem 2? something unit-less? something related to E&M? (It is no big deal if you solved problem 2 without that. It is a pretty easy correction to put in after.)

5. a) Why does a "barrier" form around the interface anyway? What role does it play?
[hint: I think that the reason it forms is to inhibit, stop and/or cancel out what is called "diffusion current". What is diffusion current? Diffusion current refers to flow from regions of higher density to lower density.  In this case it is an electrical current associated with an electron density that depends on x. Focusing on the CB, \( J_{diff}(x) = q D_n \frac{dn(x)}{dx}\), where q is the charge of an electron, \(D_n\) is the diffusion coefficient, n(x) is the density of electrons in the CB and J(x) is current density (coul/(sec-cm^2)) in the x direction.]
a') What are the units of \(D_n\)?
b) Using the electric field, E(x), that you calculated in problem 4, calculate the associated potential V(x) for the left hand side of the junction. To make this mathematically less messy and easier, let's do this only for the left side, and let's (for this problem) move our zero of x to the left hand side of the junction region. That way V(x) is proportional to \(x^2\), I believe.
c) Use this to calculate n(x). [mu is constant now and the band moves away from mu as x increases.] Your range of x for this is 0 to d, i.e., from the left edge of the "junction region" to the center. n(x) is proportional to an exponential involving the energy difference between the band edge and mu.)
d) graph n(x) vs x.
e) calculate the diffusion current and, f) graph it as a function of x.
g) which way does this current flow? which way do the electrons flow?

6. a) For the same range of x, calculate the "regular current" (proportional to the electric field, E(x)), i.e.,  \( J_{E}(x) = q^2\, \tau \, n(x) E(x)/m^*\).
b) graph this current vs x.
c) What is the direction of electron flow for this current? How does it compare with the electron flow from diffusion?
d) Is there a relationship between \(q \, \tau/m^*\) and \(D_n\) that would make these currents equal and opposite at every value of x so that they perfectly cancel?

optional:
7. How would you describe the exponential growth of current in an appropriately biased n-p junction? That is, what happens to our perfect equilibrium? Does one current becomes dominant or do both change or ...?

8. Looking at the actual density of states for Si, posted below, one can see why we would just use a constant density of states approximation. However, there is another possible approach that might be equally reasonable. That is, one could assume that near a band edge there is a single quadratic minimum in E vs k that dominates the DOS (near the edge).
a) Calculate the density of states in 3D in terms of m* using that approach. [I think it has a simple dependence on E, perhaps proportional to \((E-E_c)^{1/2}\) for the CB, for example?]
b) Calculate the density of electrons in the conduction band as a function of mu, for mu in the gap and not too close to the CB, using that DOS.
c) which approach (constant DOS or this one) do you prefer? (email me if you like)

36 comments:

  1. you don't, t3 and t2 get canceled out in the integral. you have an D3*(-KT)*exp-[E3-2t3-A(Eg)/KT] from E3-2t3 to infinity which becomes D3*(-KT)*( 0 - exp[-A(Eg)/KT]) which is D3*KT*exp[-A(Eg)/KT]). A similar thing happens for the other integral using D2.

    ReplyDelete
  2. I was a little bit vague with A on purpose. I just integrated from E3-2t3 to E3+2t3 (where E3-2t3 can be set to infinity with no major error in your answer), I set mu to be some unknown distance A(Eg) behind the conduction band (A is just a coefficient that will tell me how far behind the conduction band mu is relative to the gap width, which is nicely 1eV) so mu = E3-2t3-A(Eg). Then the integral's solution literally what he got in class D3*KT*exp[-A(Eg)/KT]) except A would have been 1/2 since mu was exactly in-between. For the valence band I did use (1−f(E)). But this is the same as 1/1+e^(-x). you can make the same approximation you did for the original fermi function, but for really small values of X (negative) and just integrate from -infinity to E2-2t2. I got some really nice numbers out of this!

    ReplyDelete
    Replies
    1. http://www.wolframalpha.com/input/?i=1-%281%2F%281%2Be%5Ex%29%29

      Delete
    2. that was 1-f(E)
      and this is the other one (listed in alternate forms)

      http://www.wolframalpha.com/input/?i=%281%2F%281%2Be%5E-x%29%29

      Delete
  3. If you wish to put zero somewhere, often the top of the valance band is best. That way everything is positive. But really, your way is fine also. No results will be effected at all by either choice.

    ReplyDelete
  4. I think we are okay without BW. As Tyler says, the outer limits of integration never matter anyway...

    ReplyDelete
  5. Well it is true the D(E) is uniformly displaced by a energy that depends on x, but I think what you really want is to plot the energy of the lower edge of the CB as a function of x. That is a simpler plot and easier to visualize. E on the vertical axis, x on the horizontal axis.

    ReplyDelete
  6. Anyone want to work on this at S&E?

    ReplyDelete
  7. Professor Schlesinger has decided to give everyone a day extension for the homework. In other words the homework is now due Friday.

    ReplyDelete
  8. 1. Be a little careful? What is \(n_c\) and what is \(n_v\)? on each side, indeed at each value of x, there is an n and a p and they are not equal. I think I understand what you mean, but caution is recommended.

    I would say d is related to doping, not that doping depends on d. Just an point of perspective, not that one is right or wrong.

    Your intuition re: higher doping implies smaller d: " More charge density, stronger electric field, stronger electrical potential, more potential energy, perhaps faster rate at which Evn transitions to Evp, shortening the depletion zone?" makes sense to me, but I I am not sure it is a simple inverse proportionality.

    ReplyDelete
  9. I'm on number 5c. How would you calculate n(x) using the voltage from 5b?

    ReplyDelete
    Replies
    1. \(n(x)= (KT) N_3 exp{-(E_c(x) - \mu)/(KT)}\)

      Delete
    2. and E_c(x) depends on V(x), right? It has a term -qV(x) added to it.

      Delete
  10. Also for number 6, I'm assuming that equation is for the drift current density. What do tau and m* represent?

    ReplyDelete
    Replies
    1. electron scattering time.
      electron effective mass.

      Delete
  11. For 1d I got mu1 closer to its band edge, but I can't figure out why that is the case. Does anyone have any thoughts about it?

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

    ReplyDelete
  13. If anybody wants to join I have a study room reserved from 4pm to 6pm. The room is S&E139.

    ReplyDelete
    Replies
    1. I saw this post to late. I am going to finish this homework tonight and we can meet up tomorrow to check our answers.

      Delete
  14. What is meant by "a single quadratic minimum" for number 8? Do we take this to mean that we can approximate the density of states by a quadratic function near the edge of the conduction band?

    ReplyDelete
    Replies
    1. i.e we can probably approximate the top of the valence band or the bottom of the conduction band to be parabolic. Streetman provides a derivation for the DOS via the 3D schrodinger equation but I'm assuming that isn't the route you are suggesting we take here...

      Delete
    2. Also, are you sure that the dependence on E isn't (E + E_c)^\frac{1}{2} ?

      Delete
    3. Excellent points. I edited 8 a bit to clarify that. quadratic minimum refers to E vs k.

      "Also, are you sure that the dependence on E isn't (E + E_c)^\frac{1}{2} ?" I think it is what you say but with a minus sign. Wouldn't it go to zero at Ec due to limited "phase space"?

      Delete
    4. Yes, I goofed on the sign.

      Delete
  15. In 1.c I wound up with a a mu that didn't depend on d, which conflicts a bit with 3. Can anybody offer any suggestions? Thanks.

    ReplyDelete
    Replies
    1. Nevermind, I realized the energies in there were d dependent.

      Delete
    2. I'm not very competent with latex so I'll type it out normally, Zack gave a few bounds for energy in class today.

      Ec(x) =Ecn x<0
      =Ecn + qv (x^2)/(2d^2) 0<x<d
      =Ecn + qv (1-((x-2d)^2)/(2d^2)) d<x<2d
      =Ecn + qv 2d<x

      and I'm thinking we can grab either the 0<x<d case or the d<x<2d case and just sub in d

      Delete
    3. though now that I'm looking at it it kind of looks like doing that would just cancel d

      Delete
  16. Is the barrier referred to in problem 5 related to Schottky barriers?

    ReplyDelete
    Replies
    1. Also on number 5 I believe part a is listed twice, once at the start and again just prior to asking for units of D.

      Delete
    2. It is closely related; a Schottky barrier is the limiting case where one side is metal and then "d" on that side is essentially zero.

      Our case, on the other hand, is a junction where both sides have the same d, because both sides are doped to the same level (10^17 cm^-3).

      Our case is quasi-symmetrical; d is the same on the p side and on the n side. The Schottky case is, I think, related to ours by letting the doping on one side go to infinity (or 10^22 cm^-3, whichever comes first).

      Delete
    3. I should have said: "a Schottky barrier forms in the limiting case where one side is metal ..."

      Delete
  17. I'm having some trouble with 4/5.b, what factor do we multiply our E field by? Are we just using E/epsilon where epsilon is some arbitrary constant that we don't plug in?

    ReplyDelete