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E-K Diagrams, valence band, conduction band

  1. Sep 9, 2005 #1
    I'm having difficulty figuring out how to work this problem:

    Assume a material has a given E-K diagram:
    E(K)conduction = Ec + E1 * sin^2 (Ka)
    E(K)valence = Ev - E2 * sin^2 (Ka)

    E1 = 5eV
    E2 = 4eV

    I have to:
    * sketch the E-K diagram for the first brillouin zone (-pi/a < k < pi/a). label axes, label Ec, Ev, energy bandwidth.

    * find effective mass for the electron near bottom of conduction band

    * find effective mass for holes near the top of the valence band

    My questions:
    How do I determine Ec and Ev to draw the diagram? Do I need to draw two E-K sketches or is it 1 big diagram?? What is the energy bandwidth? Is it the same as Eg (band gap?) which is Ec - Ev ?

    How do I determine the effective masses?

    I don't want anyone to just do the problem, but I looked through the textbook (typical overpriced McGraw Hill crap that shows you one thing and has homework problems that are 20 steps ahead of the basic crap they show). :yuck:
  2. jcsd
  3. Sep 9, 2005 #2
    Why do you think you have to solve Ec and Ev? Just mark two points on the y-axis as those energies and plot the E(k) curves. And into the same diagram of course. I'm not sure what is meant by energy bandwidth but maybe google could help you. It could be the energy gap as a function of k but I'm not sure. The effective masses are hbar/the second derivative of energy wrt. to k.

    Do you have access to Kittel or Hook&Hall? Both are good intro level books for solid state physics.
  4. Sep 9, 2005 #3
    So I just treat Ec and Ev as variables and mark them arbitrarily on the y-axis for each of the curves?

    I graphed them on my calc and they appear to be just almost symmetrical of one another.

    So do I have to solve for a numerical value of the effective mass?

    Is there a difference between determining the effective mass of an electron and a hole?

    I'm using the book Fundamentals of Semiconductor Devices by the Andersons, so far it's a typical overpriced book that doesn't explain enough.
  5. Sep 9, 2005 #4
    That's how I'd plot them. They're asking for the m* at specific k values so yes plug&chug the numbers out. Same formula applies for both. I did a "practise paper" on MOSFET operation for a solid state class and used banerjee&streetman: solid state electronic devices for reference. It had a few chapters worth on basics of semiconductors so maybe that could be worth a look if you don't like the book you have now.
  6. Sep 9, 2005 #5
    I graphed them on the calculator and using Ec and Ev as variables (but assume they are zero with the calculator), the graphs are symmetrical about the x-axis (y=0). Does that make any sense?

    For the effective mass, how do you use the 2nd derivative? Isn't it just the slope?

    if so, how does one figure out where to pick points near the top/bottom of the band? The slope can be either neg/pos near the top/bottom of the curve depending on which side you use for the slope.

    I'm fairly lost - this is a lot of first/2nd year chemistry physics mixed in with semiconductor stuff that I never learned in undergrad school (although this is an undergrad 4th year course I have to take as a grad student).

    Is there a Semiconductor Physics for dummy book/webpage that is available?
  7. Sep 9, 2005 #6
    Just look for the maxima and minima inside the first Brillouin zone and take the second derivative at those points.

    Try this site, http://britneyspears.ac/lasers.htm, it comes with nice pictures and all.
  8. Sep 9, 2005 #7
    Solving for the 2nd derivative at the maxima/minima, I get 0. :confused: if you plot out the graphs, the maxima/minima are at the same K values for both curves?
    Last edited: Sep 9, 2005
  9. Sep 9, 2005 #8
    No you don't. Are you sure you did derivated it correctly? I'm getting something with sin^2(ka) and cos^2(ka) terms of which the sine term is nonzero at ka=pi/2.
  10. Sep 9, 2005 #9


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    Aren't we interested in the effective mass at ka = n\pi , where the cond band min and valence band max occur ?

    The second derivative looks like cos(2ka), doesn't it ?
  11. Sep 9, 2005 #10
    a = 0.5 nm

    I must have drawn the E-K diagrams wrong because the maxima/minima occur at K= -pi/a, 0, and pi/a
  12. Sep 9, 2005 #11


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    No, that is correct.
  13. Sep 9, 2005 #12
    Yes. And the extrema within the boundaries are at 2ka=-pi, 0 and 2ka=pi. And that's a good approximation of what I posted (<-better way of saying that I was inaccurate).
  14. Sep 9, 2005 #13
    I decided to pick K=0 for the minima of the conduction band and the maxima of the valence band (does this sound right?)

    solving for the 2nd derivative:

    E(K)c = Ec + 5ev * sin (K*a)^2
    E(K)c" = -10*sin(Ka)^2 + 10*cos(Ka)^2
    Evaluate m* = hbar^2 / E(K)c" = 1.11x10^-69 (units?)

    E(K)v = Ev - 4eV * sin (K*a)^2
    E(K)v" = 8*sin(Ka)^2 - 8 * cos(Ka)^2
    Evaluate M* = hbar^2 / (E(K)v") = -1.38x10^-69 (units?)

    How can I get negative mass? Does that imply the slope is just different?


    Thanks in advance :yuck:
    Last edited: Sep 9, 2005
  15. Sep 9, 2005 #14
    ok i know this isn't anything to do with the disscussion but u know this is the College level zone, by college does that mean university?! cause some people call university college. If it is itll explain y im finding things on here tough to understand.
  16. Sep 9, 2005 #15


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    alias : "college" is anything that happens after high school.

    metro : using k=0 is fine. And finding a negative effective mass is what I expected for the valence band - there's absolutely nothing wrong with that. It is not because the slope (dE/dk) is different but because the rate of change of slope (d^2E/dk^2) is different. While for the cond band, the slope increases from k<0 to k>0, in the valence band, it decreases.
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