Commit 7ad6fc28 authored by Isidora Araya's avatar Isidora Araya
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Update 13_semiconductors_solutions.md

parent 5259f41b
......@@ -106,18 +106,27 @@ This approximation indicates the chemical potential is "well bellow" the conduct
the valence band. This way, we only have a few electrons and a few holes per band, which allows us to
approximate Fermi statistics by Boltzmann's.
Assume $t_{cb}$ and $t_{vb}$ positive.
For electrons, do a taylor expansion around $k=0$ (min) of $E_{cb} = E_G - 2 t_{cb} [\cos(ka)-1],$.
For holes, do a taylor expansion around $k=0$ (max) of $E_{vb} = 2 t_{vb} [\cos(ka)-1]$.
Assume $t_{cb}$ and $t_{vb}$ positive.
Near $k=0$
$$ E_{cb} = E_G - t_{cb}(ka)^2$$
$$ E_{vb} = t_{vb}(ka)^2 $$
### Subquestion 3
Now, $g_e$ and $g_h$ can be calculated.
??? hint "How?"
The approximated bands are quadratic! We've seen these before during the lecture.
### Subquestion 4
$$n_e = \int_{E_{cb}}^{E_{cb}+4t_{cb}} f(\varepsilon)g_c(\varepsilon)d\varepsilon$$
$$n_h = \int_{-4t_{vb}}^{0} (1-f(\varepsilon))g_h(\varepsilon)d\varepsilon$$
### Subquestion 5
For an intrinsic semiconductor the number of electrons excited into the conduction band must be equal
to the number of holes left behind in the valence band, so $p = n$. Impose this condition with the results of
......
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