Friday, February 22, 2013

Homework 8:

suggested activities:  Watching the videos in the Feb 21 class notes post. Also, it may be helpful to read about moderately doped semiconductors and carrier generation and recombination processes in semiconductors. Also, p-n junctions, but just the most superficial things. I think we can understand them in a simple yet accurate manner. We will focus on p-n junction LEDs, lasers and solar cells.

8. $v= (Bbc/2\hbar c) sin(kb) = (Bbc/2\hbar c) (\pi/20)$
     = (4 eV 1 A/(2 1972 eV-A))  (pi/20) c
     = 4.8x10^6 cm/sec


9. One more problem! (the focus of this problem is to help you build your skills of actually calculating numbers.)
Consider an electron in the conduction band of a crystal for which the energy is given by:
$E_c (k) =  E_o - (B/2) cos (kb)$.
Then confirm that,
$v(k) = (Bb/2 \hbar) sin(bk)$,
and, for the case b = 1 angstrom and B=4 eV,
a) calculate v(k) at $k = \pi /(20 b)$ in Angstoms per second and in cm/sec. Get numbers!   (Soon we may try to show (numerically) that this is the electron speed associated with this state.)


Due Friday, 3 PM

1. a) If a semiconductor has a band gap of 1 eV what frequency of photon would you require in order to create an electron-hole-pair excitation? What color is that?

b) What color of photon would a semiconductor with a 1.9 eV band gap emit in a transition from the bottom of the conduction band to the top of the valence band?



2. Suppose you have a 1-dimensional semiconductor that is 2 cm long. Suppose the left half is doped with 10^4 acceptors per cm and the right half is doped with 10^4 donors per cm.

a) graph the density of electrons in the conduction band and the density of holes in the valence band (empty states) as a function of x. You can call the density of electrons in the conduction band, n, and the density of holes in the valence band, p.
(hint: don't make this too complicated. It is really simple!)



3. Consider a 1D semiconductor, our typical one, with a band gap of 0.5 eV between the top of the valence band and the bottom of the conduction band. Suppose you substitution dope the right-hand half of it with 10^4 donor atoms per cm and then apply a voltage (electron field) that pushes the mobile electrons in the conduction band (on the right) toward the left, into an region where they all fall into the valence band with each one emitting a photon. Let's assume that, in response to the applied voltage, the electrons move with a speed of 3 $10^6$ cm/sec.
a1) sketch a picture of that...

a2) Would the region on the left where the downward transition takes place need to be doped at all? why or why not? In what way would you dope it to allow downward "recombination" transitions. 

b) How many photons would be emitted per second?

c) what would be the energy of each photon (more or less)?

d) how much total power would be emitted? (in eV/sec, and Joules/sec = Watts)



4. Now consider a 3D semiconductor with a band gap of 0.5 eV and doped on the right-hand half of it with 10^17 donors/cm^3. Apply a voltage (electron field) that pushes those electrons toward the left (at a speed of 3 $10^6$ cm/sec) into a region on the left where they all fall into the valence band with each one emitting a photon.

a) would the region on the left need to be doped? In what way? Why?

b) How many photons would be emitted per second? (are the units the same or different than in #3? Why?)

c) what would be the energy of each photon (more or less)?

d) how much total power would be emitted? (what are the units of that?)

5. a) What semiconductor band gap would be best suited to creating a red LED?
b) a) What semiconductor band gap would be best suited to creating a green LED?

6. Discuss how substituting N in GaAs might possibly help you obtain a green LED (i.e., help you change a red LED material into a green one?)

7.  Consider a 1D semiconductor with an energy gap of 0.8 eV. Suppose the density of states near the bottom of the conduction band is g(E) = $2 \times 10^9$ states/(eV-cm). Further, suppose we treat g(E) as constant within the band and zero in the band gap).
a) sketch a plot of this approximate g(E) (which is either zero or $2 \times 10^9$ states/(eV-cm)).
b) Integrate the product g(E) times f(E) over the range of the conduction band,  where f(E) is the Fermi function. Put the Fermi energy $E_F$ right in the middle of the gap and set KT=0.025 eV = 1/40 eV.
(important hints:
 hint 1) What is the largest value of f(E) for E in the range of the conduction band? If it is small, then you can ignore the +1 in the denominator and the integrand becomes much easier. Perhaps if you are lucky the integrad becomes a constant, g(E), times a simple decaying exponential? (Do you feel lucky?)
  hint 2) The upper limit of integration (from the top of the conduction band) will give you a very, very, very small number. I would suggest that you totally ignore it.)
c) What are the units of your result from b)?
What is the numerical value of your result? (Do you need to have anything else given?)
d) Explain why this calculation might give you the density of electrons in the conduction band at room temperature for this semiconductor when it is un-doped (intrinsic).

8. a)  What is the relationship between the structure of diamond and the structure of Silicon?
b) Comparing the C-C bond in Diamond, and the C-C bond in Graphene: which is strongest? Why do you think that might be?

Appendix:
The Fermi function is:
$f(E) = (e^{(E-E_F)/KT} + 1)^{-1}$.
It tells you the probability, in a thermal physics/statistical physics sense, that a particular state at energy E is occupied. Note that it can't be bigger than 1 (Fermions) or less than zero.
You are given that KT= .025 eV (so you don't need to know K) and also where $E_F$ is (in the center of the gap).  Note that whenever $E-E_F$ is greater than about 3KT you can approximate f(E) by a simple exponential.

48 comments:

  1. I was wondering if there was any chance that we could get a video or notes pertaining to this homework posted. I find that most of the time we don't cover material for the homework assignments until the following tuesday/thursdays after the assignments are posted, and I'd find it so much more helpful if I didn't to wait all that time and put the weekend to good use.

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    1. These are good points you make. Did you see the Feb 21 class notes and videos? I think those might be helpful. Let me know what else you would like to be covered in addition to that and I'll work on it. (Maybe make another video.)

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  2. For 3b I think we need the velocity of the electrons to figure out how many photons are being emitted per second. It would be $v\cdot 10^4 \text{cm}^{-1}$. What should we use as the velocity?

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  3. Will we use the same speed for problem 4?

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  4. For 7b. Are we integrating with respect to K? i.e when it says "over the range" of the conduction band, does this mean from K = -pi/b to K = pi/b or something else? The info at the bottom seems to indicate something else.

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    1. I was thinking that the graphs and integral would be as a function of energy, E. That is much easier.

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  5. I suppose it also bears asking whether the K in number 7 is the same as the k we have heretofore been dealing with (crystal momentum)?

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    1. That K is fixed, not a variable. Part of KT=.025 eV. It is a constant that converts temperature to energy.

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  6. For problem 6, we need to increase the band gap between the conduction and valence bands from ~1.8eV to ~2.4eV, but wouldn't substituting N in for As in GaAs decrease the potential and thus the energy level differentials? Wouldn't this lower the band gap?

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    1. well, let me ask this: what has a bigger gap, diamond (C), Si or Ge? What are their gaps? What is the trend for those materials?

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  7. Could you give us a bit more information about the range for the integral in 7b? I know we are supposed to integrate over the conduction band, and we know it's centered at $E_F$ but I don't know the width.

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    1. 4 eV. (but ignore that upper term after you get it.) comment back later if you see what i mean or not about the upper limit term...

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    2. PS. don't make it too hard. With the right approximations this is an integral you can in your sleep.

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    3. I just don't know how to find the actual limits of integration. Are we supposed to figure out what they are or are they given and I'm just not seeing it?

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    4. "I know we are supposed to integrate over the conduction band, and we know it's centered at EF but I don't know the width."

      If I understand this, then i think it is a completely incorrect starting belief. (Though i may be misunderstanding.)

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  8. The key point if this problem is to set up the integration. Once you have that, it is straightforward. So you are definitely focused on important issues.

    You are asked to integrate over the conduction band, meaning starting at the lower edge, Ec, and going up to the upper edge (Ec+4 eV). Here would be my suggestion. Draw an energy axis. Indicate on that where Ec is, where the top of the conduction band is, and where E_F is.

    Where E_F is is very important. It is not in the middle of the band. Where is it?

    Please let me know what you are picturing regarding where E_F is relative to Ec. Thanks.

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    1. Right, I misread the problem. E_F is below E_c, which is not in the conduction band. Ec=E_F + .4eV because we know the width of the gap.

      If we are integrating from Ec to Ec+4eV wouldn't we need to know the value of Ec?

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    2. That just depends where you define E=0 to be. I think if you pick a spot for that, everything will start to make sense. Let me know.

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    3. Thanks, I got it now. I didn't think to set my energy values to whatever makes the calculation easiest.

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    4. Wherever you put the zero of energy, I think the calculation is the same (has the same result). It is sort of like a coordinate system. You have the freedom to put E=0 wherever you want, and then you have to work consistently with that choice.

      Do you see how the upper limit of integration is negligible compared to the lower one?

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    5. Yes I understand the two hints you gave us. I assumed those would explain themselves once I figured out the limits.

      Thanks

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  9. In regards to number five, I have a green laser that emits photons of 523nm wavelength, and it has a peak power of <5mW. I was wondering what the normal contact surface area of a p-n junction would be? That way I could do a rough estimate of how many donors/cm^3 are actually in there (I know it's a little more complicated for a laser, but it still would be cool to get an order of magnitude).

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    1. I am not sure about the area, but maybe you could work the other way - assume an electron density of about 10^18 cm^-3. then what would the effective area have to be?

      Lasers can enhance efficiency and are coherent, but there is no magic in the power conversion relation. It is still one photon per electron that drops down to the Valance band. No different from an LED. A laser is just and LED with phase that produces a coherent beam. (Something called stimulated emission is involved. maybe when can cover that as part of spectroscopy.)

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    2. Sweet, so maybe I was over thinking the stimulated emission part, since it must be a one-to-one correspondence between electrons and photons. I also looked around about threshold voltage to see the consensus, and it seems to be the same. It is because of the exponential function related to the current through the LED as you had said. A small difference in voltage means a jump from .045mA to about 700mA and continues to rise.

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  10. I get the impression that the lone pair electrons from the nitrogen could somehow be sequestered by the GaAs. I have not actually checked out the band gaps suggested for investigation by a look up on wikipedia, but based on that data it looks like the band gap decreases down the group. That makes sense to me knowing how the total energies of the hydrogen atomic states behave as n increases.

    I also tried playing around with the Rydberg formula, and I got a ratio (1/(n^2)-1/(m^2)) of about .2 (though that was for Z=1). I worked that out to approximately a transition from n=4 to n=2 which would suggest to me that the electron is going from the third to the first excited state. That seems to me then almost like the electron is falling down into an unoccupied state somewhere in 2p land. At the same time I could be absolutely mistaken in every step I've gone through, and to be totally honest I am pretty skeptical about my reasoning.

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  11. When I solve number 7 I get a very interesting value and set of units. I'm not sure if this correct, but I think it might help if I know what this integral is supposed to represent.

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    1. Well you multiply g(E) in states/(cm-eV) by f(E) which is in electrons per state (and is a very small number (much less than 1. exponentially small i would say). So the units of the integrand are electrons/(cm-eV). Then you integrate over energy (eV) and then get electrons/cm. That is: electron density in our 1D semiconductor.

      And that is what it represents. the density of electrons in the conduction band for an undoped semiconductor at room temperature. I hope it is more than 1 electron/cm...?

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    2. okay that makes sense. And I did get above 1 e/cm but it was still quite low, in the single digits. That's why I was concerned.

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    3. That seems reasonable to me. Because it is undoped. So that is just the "background" number of electrons due to finite temperature.

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  12. Just a comment about the format of this homework. Having everything in bold makes the assignment difficult for me to read. Could we return to the format (not everything is bold) of the previous homeworks in the future?

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    1. Sorry about that. That is not deliberate, normal bolding. Something weird happened when I was editing and it all went bold. Then I was afraid to mess with it; afraid of losing the whole post. I should have explained.

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    2. When you are embedding latex weird things can happen like that. sometimes things won't post.

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  13. For problem 3c, is the photon's energy simply that of the band gap that it falls through? Or is its energy related to the kinetic energy of the electron that emits the photon?

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    1. I think I've answered my question actually. Do the electrons emit photons through Bremstrahlung radiation? i.e., the kinetic energy is translated to the energy of the emitted photon?

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    2. i think just the band gap energy, since it starts near the bottom of the C band and ends up near the top (of the V band).

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    3. for part D do we multiply the energy by the amount of photons per second? i get a pretty big number, even in eV :/

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  15. 8b is interesting! Graphene is the new and cool thing in physics. Most of the info that I get says that graphene is the strongest material known. According to Chemistry & Physics Of Carbon, Volume 29 edited by Ljubisa R. Radovic: Graphene has on the average a bond strength of 452kJ/mol and diamond has C-C bond of 347kJ/mol. This suggests a stronger bond in graphene. The author Morinkovic goes on to say that it is because of a smaller interatomic distance in the graphene compared to diamond (.142nm compared to .154nm). What would this mean in terms of quantum states? Why would a smaller interatomic distance matter? Does it relate directly to a stronger 1/r potential, since the protons are closer together?

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    1. Both graphen and diamond have really stron covalent bonds. pretty close to the strongest found.

      My belief is that the reason graphene bonds are even stronger arises from a fundamental difference between sp3 and sp2. Basically that in sp2 the off-center value of $\bar{x}$ is bigger. I think that is the basic origin of the extreme strength of these covalent bonds.

      In sp3 the radial state has to be split 4 ways. In sp2 it is only split 3 ways. I think one can show that the larger coefficient on the Psi2r state, leads to the larger off center value of x, and that that is the origin of the stronger bond.

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  16. For part D on problems 3 and 4 i am getting really big numbers, for example i got 3*10^23 for 4D, i got this by multiplying 10^17 (photons/cm^3) by (3*10^6)cm/sec. Was that the wrong method to obtain the photons per sec?

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    1. sounds fine. i think you mean: photons/(sec-cm^2)

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    2. doesn't the final answer have the units you mentioned above?

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