diff --git a/src/14_doping_and_devices_solutions.md b/src/14_doping_and_devices_solutions.md
index 48851fc1e75547df501ea6edac07f3dbe3ff79a4..cbfa4d8752b245f8f21732652ae166aa2ed0d8da 100644
--- a/src/14_doping_and_devices_solutions.md
+++ b/src/14_doping_and_devices_solutions.md
@@ -19,16 +19,14 @@ $$ n_e - n_h + n_D - n_A = N_D - N_A $$
 Since $E_G \gg k_B T$, we can only use the law of mass action. 
 But the question offers us another piece of information - we are around $|N_D-N_A| \approx n_i$.
 That means that we are near the transition between extrinsic and intrinsic regimes.
-However, we also know that dopants energies are quite small such that $E_C-E_D \ll E_G$ and $E_A-E_V \ll E_G$. 
-That means that we expect $n_i \ll n_D$ and $n_i \ll n_A$ (since both $n_i$ and dopant ionazition depends exponentially of the corresponding energy differences).
-Therefore, we can confidently say that $N_D \gg n_D$ and $N_A \gg n_A$ so we esentially recover the dopant ionization condition.
-Thus, neglecting $n_D$ and $n_A$ such that they are both 0, the solutions to the charge balance are:
+Because the dopants stop being ionized at very low temperatures (see next exercise), we can neglect $n_D$ and $n_A$ in this exercise, just like in the lecture.
+Writing $n_e n_h = n_i^2$ and $n_e - n_h = N_D - N_A$ and solving these together, we obtain
  
 $$ n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
 $$ n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
-where $D = N_D - N_A$ and $n_i=n_{e,intrinsic}=n_{h,intrinsic}$.
+where $D = N_D - N_A$.
 
-Note that for N_D = N_A results for intrinsic semiconductors are recovered.
+For $n_i \gg |N_D - N_A|$ we recover the intrinsic regime, while the opposite limit gives the extrinsic expressions.
 
 ### Subquestion 3