Update 14_doping_and_devices_solutions.md

src/14_doping_and_devices_solutions.md

@@ -8,23 +8,22 @@ from math import pi
 ## Exercise 1: Crossover between extrinsic and intrinsic regimes
 
 ### Subquestion 1
-Law of mass action (general):
-$$ (n_e + N_A) (n_h + N_D) = \frac{1}{2} \left(\frac{k_BT}{\pi\hbar^2}\right)^3 
+Law of mass action:
+$$ n_e n_h = \frac{1}{2} \left(\frac{k_BT}{\pi\hbar^2}\right)^3 
 (m_e^{\ast}m_h^{\ast})^{3/2}e^{-\beta E_{gap}}$$
 
 Charge balance condition:
-$$ D = N_D - N_A - n_e + n_h = (doping)$$
-
+$$ n_e - n_h + n_D - n_A = N_D - N_A $$
 ### Subquestion 2
  
-$$ n_{e} + N_A = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
-$$ n_{h} + N_D = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
+$$ n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
+$$ n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
 where $n_i=n_{e,intrinsic}=n_{h,intrinsic}$.
 
 ### Subquestion 3
 
 If $D<<n_i$, then the doping is not important and results of intrinsic are
-reproduced ($n_e \approx n_h$)
+reproduced.
 Contrarily, if $D>>n_i$, it's mostly the doping that determines $n_e$ and $n_h$.
 The thermal factor becomes unimportant.
 Check both cases with lecture notes approximated solutions by doing a Taylor expansion. 
@@ -37,8 +36,8 @@ Check both cases with lecture notes approximated solutions by doing a Taylor exp
 If all the dopants are ionized, the Fermi level gets shifted up towards the conduction band.
 This result can be obtained when using results in Exercise 1 - Subquestion 2 and the following:
 
-$$ n_D \approx N_D$$
-$$ n_A \approx N_A$$
+$$ n_D \approx 0$$
+$$ n_A \approx 0$$
 $$ n_e = n_h = n_i $$