From bb3c0cd85bb3decfb6baef54d250e05a8fc39c83 Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Mon, 20 Jul 2020 05:02:26 +0000
Subject: [PATCH] Update 13_semiconductors.md - fix

---
 src/13_semiconductors.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/src/13_semiconductors.md b/src/13_semiconductors.md
index dfe0cbbf..d4911d21 100644
--- a/src/13_semiconductors.md
+++ b/src/13_semiconductors.md
@@ -133,10 +133,10 @@ Therefore we can approximate the dispersion relation of both bands as parabolic.
 
 Or in other words
 
-$$E_e = E_G + \frac{\hbar^2k^2}{2m_e},$$
-$$E_h = \frac{\hbar^2k^2}{2m_h}.$$
+$$E_e = E_c + \frac{\hbar^2k^2}{2m_e},$$
+$$E_h = E_{v,h} + \frac{\hbar^2k^2}{2m_h} = -E_{v} + \frac{\hbar^2k^2}{2m_h}.$$
 
-Here $E_G$ is the band gap, and the top of the valence band is at $E=0$.
+Here $E_c$ is the bottom of the conduction band and $E_v$ is the top of the valence band.
 
 Observe that because we are describing particles in the valence band as holes, $m_h > 0$ and $E_h > 0$.
 
@@ -189,7 +189,7 @@ $$ g(E_h) = (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3$$
 
 Applying the algorithm:
 
-$$n_h = \int_{-E_v}^\infty f(E+E_F)g_h(E+E_v)dE = \int_{-E_v}^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}\sqrt{E+E_v}\frac{1}{e^{(E+E_F)/kT}+1}dE$$
+$$n_h = \int_{-E_v}^\infty f(E+E_F)g_h(E)dE = \int_{-E_v}^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}\sqrt{E+E_v}\frac{1}{e^{(E+E_F)/kT}+1}dE$$
 
 $$n_e = \int_{E_c}^\infty f(E-E_F)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_c}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
 
-- 
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