From dc2ddd228ef5fabe69e55b17376c222e3326a5c4 Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Mon, 20 Jul 2020 04:32:19 +0000
Subject: [PATCH] Update 13_semiconductors.md

---
 src/13_semiconductors.md | 16 ++++++++--------
 1 file changed, 8 insertions(+), 8 deletions(-)

diff --git a/src/13_semiconductors.md b/src/13_semiconductors.md
index 64e185b2..3d24335e 100644
--- a/src/13_semiconductors.md
+++ b/src/13_semiconductors.md
@@ -191,12 +191,12 @@ Applying the algorithm:
 
 $$n_h = \int_{-E_v}^\infty f(E+E_F)g_h(E+E_v)dE = \int_{-E_v}^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}\sqrt{E+E_v}\frac{1}{e^{(E+E_F)/kT}+1}dE$$
 
-$$n_e = \int_{E_G}^\infty f(E-E_F)g_e(E)dE = \int_{E_G}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_G}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
+$$n_e = \int_{E_c}^\infty f(E-E_F)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_c}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
 
 We need to solve $n_e = n_h$
 
 Simplification:
-Fermi level is far from both bands $E_F \gg kT$ and $E_G - E_F \gg kT$
+Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$
 
 Therefore Fermi-Dirac distribution is approximately similar to Boltzmann distribution.
 
@@ -204,8 +204,8 @@ $$f(E\pm E_F) \approx e^{-(E\pm E_F)/kT}$$
 
 Now we can calculate $n_e$ and $n_h$:
 
-$$n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_0^\infty\sqrt{E}e^{-E/kT}dE =
-N_V e^{-E_F/kT},$$
+$$n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_{-E_v}^\infty\sqrt{E+E_v}e^{-E/kT}dE =
+N_V e^{E_v-E_F/kT},$$
 
 with
 
@@ -218,18 +218,18 @@ the density of holes with energy $E<kT$ (compare with the rule above).
 
 Similarly for electrons:
 
-$$n_e = N_C e^{-(E_G - E_F)/kT},\quad N_C = 2\left(\frac{2\pi m_e kT}{h^2}\right)^{3/2}$$
+$$n_e = N_C e^{-(E_c - E_F)/kT},\quad N_C = 2\left(\frac{2\pi m_e kT}{h^2}\right)^{3/2}$$
 
 
 Combining everything together:
 
-$$n_h \approx N_V e^{-E_F/kT} = N_C e^{-(E_G-E_F)/kT} \approx n_e$$
+$$n_h \approx N_V e^{E_v-E_F/kT} = N_C e^{-(E_c-E_F)/kT} \approx n_e$$
 
 Solving for $E_F$:
 
-$$E_F = \frac{E_C + E_V}{2} - \frac{3}{4}kT\log(m_e/m_h)$$
+$$E_F = \frac{E_c + E_v}{2} - \frac{3}{4}kT\log(m_e/m_h)$$
 
-An extra observation: regardless of where $E_F$ is located, $n_e n_h = N_C N_V e^{-E_G/kT} \equiv n_i^2$.
+An extra observation: regardless of where $E_F$ is located, $n_e n_h = N_C N_V e^{-(E_c-E_v)/kT} \equiv n_i^2$.
 
 $n_i$ is the **intrinsic carrier concentration**, and for a pristine semiconductor $n_e = n_h = n_i$.
 
-- 
GitLab