From d28cdd8d566d05b04c6f20824210239d7e8d3709 Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Wed, 10 Apr 2019 20:34:46 +0000
Subject: [PATCH] Update 14_doping_and_devices.md - typo

---
 src/14_doping_and_devices.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/src/14_doping_and_devices.md b/src/14_doping_and_devices.md
index 50251319..54871e9c 100644
--- a/src/14_doping_and_devices.md
+++ b/src/14_doping_and_devices.md
@@ -74,7 +74,7 @@ draw_classic_axes(ax, xlabeloffset=.2)
 ```
 
 All donor/acceptor states at the same energy:
-$$g_A(E) = N_A \delta(E-E_A),\quad g_D(E) = N_D \delta(E- (E_G - E_D))$$
+$$g_A(E) = N_A \delta(E-E_A),\quad g_D(E) = N_D \delta(E- E_D)$$
 
 How large can $N_D/N_A$ be? The distance between donors should be such that the states don't overlap, so the distance must be much larger than 4 nm. Therefore **maximal** concentration of donors before donor band merges with conduction band is $N_D \lesssim (1Ã…/4\textrm{nm})^3 \sim 10^{-5}\ll N_C$.
 
@@ -86,7 +86,7 @@ $$n_e - n_h + n_D - n_A = N_D - N_A$$
 
 We already know $n_e$ and $n_h$.
 
-$$n_D = N_D \frac{1}{e^{-(E_D-E_F)/kT} + 1}, n_A = N_A \frac{1}{e^{-(E_A+E_F)/kT} + 1}$$
+$$n_D = N_D \frac{1}{e^{(E_D-E_F)/kT} + 1}, n_A = N_A \frac{1}{e^{-(E_A+E_F)/kT} + 1}$$
 
 Simplification:
 
-- 
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