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Free carrier absorption in GaSb

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Given the carrier concentrations (1.4x10^16, 3.2x10^16 and 5.3x10^16 /cm^3), verify that the free carrier losses scale linearly.
    (use λ=10μm)

    2. Relevant equations

    Those concentrations are given by this paper: A. Chandola, R. Pino, and P. S. Dutta, "Below bandgap optical absorption in tellurium-doped GaSb," Semicond. Sci. Technol. 20, 886 (2005).
    (paper is attached)

    That paper gives the following formula for free carrier absorption (Drude-Zener):
    [itex]\alpha=(\frac{e^3}{4\pi^2 c^3 m_0^2 \epsilon_0})(\frac{1}{n(m^*/m_0)^2})(\frac{\lambda^2}{\mu})N\\
    where\ e=1.602 \times 10^{-19}, \epsilon_0=8.854 \times 10^{-12}, m_0=9.11 \times 10^{-31},
    n=3.8, E_g=0.726eV, m_h=0.4m_0, \mu_h=1000cm^2/V\cdot s[/itex]

    We are told to get parameters from the following website: http://web.archive.org/web/20120728125306/http://www.ioffe.rssi.ru/SVA/NSM/Semicond/GaSb/index.html


    3. The attempt at a solution
    I used the formula and values given above, but using matlab I got alpha = 7.5448e+019 which is far from the calculation given in the official solution*:
    [itex]\alpha=85(\frac{1.51 \times 10^{12}}{5.3 \times 10^{16}})cm^{-1}=2.5 \times 10^{-3} cm^{-1}[/itex]
    (I don't know what concentration that's supposed to be, though.) But I'd really like to know what formula that is. The factor in parentheses stays the same regardless of wavelength, and regardless of whether the calculation is for holes or electrons. The factor in front gets bigger as wavelength varies from 5 to 10 to 15 microns (50 to 85 to 90 for holes, 11 to 39 to 95 for electrons)

    * An unofficial solution has approximately 20, 40, and 85 cm^-3 as the alpha values corresponding to the three concentrations but this may be wrong because the units are wrong.

    Here's the matlab if anyone wants it, but it's probably wrong:
    Code (Text):
    e=1.602e-19;
    c=3e8;
    m0=9.11e-31;
    eps0=8.854e-12;
    pi=3.14159;

    n=3.8;
    me=0.041*m0;
    mh=0.4*m0;
    lambda=10e-6;
    mu_e=3000/10000;
    mu_h=1000/10000;
    conc=(1.4e16)*(100*100*100);


    alpha=((e^2)/(4*(pi^2)*(c^3)*(m0^2)*eps0))*(1/(n*(mh/m0)^2))*((lambda^2)/mu_h)*conc
     

    Attached Files:

    Last edited: Feb 4, 2013
  2. jcsd
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