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Seebeck effect and hot-probe test

  1. Apr 6, 2006 #1
    Hi,
    Can anybody tell me precisely the nature of the Seebeck effect which
    enters in the hot-probe test to determine the nature of doping in a
    semiconductor ?

    That is, I need the explanation of why we can measure a potential when
    the com electrode is heated when the electrodes are connected at two ends
    of a doped semiconductor.

    I've seen that if a Si wafer is doped with n_d=1e16 cm^{-3} of donor
    material, then at room temp. we have n=1e16/cc and p=1e4/cc so that

    np=n_i^2=1e20 at 300K

    at 400K n_i=1e13/cc, so n=1e16/cc and p=1e10/cc so that

    np=1e26 at 400K.

    here are some explanations that describe the exact opposite of the
    experiment, but I don't see why ! Can you help, please?

    a) The diffusivity of both holes and electrons decreases with T.

    So the flow of electrons in a n-doped Si wafer should be from the cold
    side to the hot one. So we should measure a negative potential.

    b) The chemical potential decreases as the temperature increases. So the
    electron should flow from the cold to the hot region, and again the
    optential should be negative.

    c) as the density of negative free charges is quite constant, what is
    important is the concentration of positive free carrier, that is p.
    the we should see a flow

    j = - q D (d_x p)

    with (d_x p) = (1e10-1e4)/cc /10cm = 1e9 /cc/cm
    for the gradient of density of the holes in a 10 cm long wafer, we should
    see a flow from the high density to the low density of (D=30 cm^2/s)

    j = - 3e8 q /cm^2 s

    so the hot point is depleted of holes and charged negatively. the voltage
    is again negative.

    The three explanations gives the same result, so what is wrong in my way
    of thinking ?
     
  2. jcsd
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