 ### 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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