1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: N-carriers, number density, energy gap E_g, temperature

  1. May 3, 2009 #1
    hi folks, I've gotten most of this problem but for one part. I've learned quite a bit reading about the physics involved here on the web, but for all I've learned when I apply it, it doesn't work (ie give the right answer). So, I'm missing something.

    1. The problem statement, all variables and given/known data
    In silicon, E_g = 1.12 eV and the effective mass of the n-carriers is m* = 0.31*m_e, where m_e is the electron mass. Find the number densities of n-carriers at 100 K and at 300 K.

    ans = _____ (n-carriers/cm^3)

    2. Relevant equations
    n = N_c * e^-(E_c - E_f)/(kT)
    n = N_c * e^-E_g/(2kT) (is this ok to use?)

    N_c = 2 * ((m* * k * T)/(2*Pi*h-bar^2)) ^ (3/2)
    not a typo here, it's m-star times k times T, etc.

    3. The attempt at a solution

    I can get the answer at 300 K, because the energy gap E_g given works for room temperature, ie 300 K. what I can't get is the answer for when T = 100 K, where in my understanding the value of E_g is slightly larger. I've researched and found a Varshni empirical formula, three constants which are material properties specific to silicon. but this didn't end up helping, that is, my slightly larger value for E_g did not yield the correct answer.

    my question is, am I right in thinking the central part of my mistake in the T = 100 K case is how E_g changes with temperature? or does it not change in this problem? Or is there another thing going on here that i'm ignoring/misinterpreting?

    Tips or suggestions are very much appreciated on this one.

  2. jcsd
  3. May 3, 2009 #2
    You can use 100K in that formula. Just because E_g is larger will only mean that there will be very few conduction electrons found in the system (in fact exponentially smaller than at 300K). Once the temperature becomes greater than the gap, then the carrier density shoots up.
  4. May 3, 2009 #3
    nickjer, I guess I'm not sure I understand.

    i've tried just using the above formula (for n, in terms of N_c and E_g) with T=100, and it doesn't work.

    I don't doubt the possible truth of what you wrote, but I'm not sure how it explains why just using T=100 does not yield the proper result.
  5. May 3, 2009 #4
    What result are you getting, and how do you know it is not correct?
  6. May 4, 2009 #5
    the result I'm getting is, n=5.10299*10^(-11), but it should be 5.5*10^(-11). it's off by a little bit, and I can't explain how. I have the answer for this textbook problem, but I have to figure out How to get it.
  7. May 4, 2009 #6
    Are you using:

    [tex]n_i = \sqrt{N_v N_c} e^{-E_g/2kT}[/tex]

    You need to take into account the hole's effective mass as well which is slightly larger.
  8. May 4, 2009 #7
    ok, but how do i calculate N_v without information about m*_p? the effective mass of hole. you say slightly larger, but, how do I calculate it with what's given in the problem?
  9. May 4, 2009 #8
    You can't calculate [tex]m_p^*[/tex] based on what is given in the problem. Unless you look it up in a table somewhere. Besides, both the electron and hole effective masses are roughly the same for silicon, so you should get the roughly the same answer. Just double check your work and units.
  10. May 4, 2009 #9
    well, you were right about them being roughly the same answer. the prof changed the answers on us, so that my previous submission (where I just used T=100 and the effective mass of the n-carrier) that had been marked incorrect is now correct.

  11. May 4, 2009 #10
    Good to hear that.
  12. May 7, 2009 #11
    does that mean that ((m_n*m_p)/me^2) = (0.31)^2 since we know that m_n & m_p are almost equal to each other?
  13. May 7, 2009 #12
    I'm not sure but possibly.

    I am more certain that Sqrt[N_v * N_c] is approximately N_c.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook