Update 14_doping_and_devices_solutions.md

src/14_doping_and_devices_solutions.md

@@ -13,12 +13,12 @@ $$ n_e n_h = \frac{1}{2} \left(\frac{k_BT}{\pi\hbar^2}\right)^3
 (m_e^{\ast}m_h^{\ast})^{3/2}e^{-\beta E_{gap}}$$
 
 Charge balance condition:
-$$ D = N_D - N_A$ - n_e + n_h $$
+$$ D = N_D - N_A$ - n_e + n_h = (doping)$$
 
 ### Subquestion 2
  
-$$ n_e = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
-$$ n_h = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
+$$ n_e + n_A = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
+$$ n_h + n_D = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
 where $n_i=n_{e,intrinsic}=n_{h,intrinsic}$.
 
 ### Subquestion 3
@@ -35,14 +35,22 @@ Check both cases with lecture notes approximated solutions by doing a Taylor exp
 ### Subquestion 1
 
 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:
 
-### Subquestion 2
-
+$$ n_D = \approx N_D$$
+$$ n_A = \approx N_A$$
+$$ n_e = n_h = n_i $$
 
 
+### Subquestion 2
+Now,
+$$ n_D^{\ast} = N_D (1-\frac{1}{e^({E_D-E_F})/k_BT+1})$$
+$$ n_A_{\ast} = N_A (1-\frac{1}{e^({E_F-E_A})/k_BT+1})$$
+$\ast$ indicated non-ionized concentrations.
 ### Subquestion 3
 
-
+Use Germianium Fermi Energy at room temperature and perform the
+[$\tebtit{key algorithm of describing the state of a semiconductor}$](13_semiconductors/#part-1-pristine-semiconductor)
 
 ## Exercise 3: Performance of a diode