diff --git a/src/14_doping_and_devices.md b/src/14_doping_and_devices.md
index 55d241cb1d7e70f960c5b8384a727536d3869de4..e5eb35dd9ef503c453cc2712e795bfc2b5811a8c 100644
--- a/src/14_doping_and_devices.md
+++ b/src/14_doping_and_devices.md
@@ -35,32 +35,45 @@ m_h, m_e = 1, .5
     <iframe width="100%" height="315" src="https://www.youtube-nocookie.com/embed/1u305H4UiVs" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
 
 
+In the previous lecture, we learned how to deal with partially filled bands.
+The concept of electrons/holes established the foundations needed to understand semiconductors.
+We saw that the filling in semiconductors can be controlled by tuning the temperature.
+However, Fermi level control through temperature is still far too constrained and leads to equal electron and hole densities $n_e = n_h$.
+The full utility of semiconductors is achieved through another Fermi level control method - **doping**.
+In today's lecture, we will take a look at how doping allows the fine control of Fermi level and the practical applications that come with it.
 
 
 ## Adding an impurity to semiconductor
 
-* Typical semiconductors are group IV (Si, Ge, GaAs).
-* Unfilled shell of group V atom (donor) has 1 extra electron and its nucleus 1 extra proton
-* Group III atom (acceptor) lacks 1 electron and 1 nucleus charge
-
-Extra electron (or extra hole) is attracted to the extra charge of the nucleus.
-
-In H the energy levels are:
-$$ E_n = - \frac{me^4}{8\pi^2\hbar^3\varepsilon^2_0n^2} = -R_E /n^2= -\frac{13.6\text{eV}}{n^2}$$
-
-Bohr radius (size of the ground state wave function): $4 \pi \varepsilon_0 \hbar^2/m_{\mathrm{e}} e^2$
-
-In a semiconductor $m\to m_{\text{eff}}$, $\epsilon_0 \to \epsilon\epsilon_0$.
-
-An impurity creates a very weakly bound state:
-$$E = -\frac{m_e}{m\varepsilon^2} R_E = -0.01 \text{eV (in Ge)}$$
-$r = 4$ nm (vs $r = 0.5$ Ã… in H).
-
-Binding energy smaller than room temperature (0.026 eV).
-
-So a donor adds an extra state at $E_D$ (close to the bottom of the conduction band) and an extra electron.
-
-Likewise an acceptor adds an extra state at $E_A$ (close to the top of the valence band) and an extra hole.
+In order to understand doping, we need to remember some basic chemistry.
+Most semiconductors are made up of group IV elements (Si, Ge) or binary compounds between group III-V elements (GaAs).
+In both cases, there are 4 valance electrons per atom.
+If we want to increase the average number of electrons per atom, we can add a group V element that has an extra valance electron. 
+We therefore refer to group V elements as **donor** impurities.
+However, the extra donor electron is bound to the impurity because group V elements also have an extra proton. 
+In order to estimate the binding strength, we treat the lattice as a background and only consider the system of an electron bound to a proton.
+We immediately recognize this system as a Hydrogen model with energy levels
+$$ 
+E_n = - \frac{m_e e^4}{8\pi^2\hbar^3\varepsilon^2_0n^2} = -R_E /n^2= -\frac{13.6\text{eV}}{n^2}.
+$$
+The extent of the ground wavefunctions is given by the Bohr radius:
+$$
+r_B = 4 \pi \varepsilon_0 \hbar^2/m_{\mathrm{e}} e^2.
+$$
+However, the extra valance electron moves in the semiconductor's conduction band and not free space.
+Therefore, the electron's mass is the conduction band's effective mass. 
+Furthermore, the interactions between the electron and proton are screened by the lattice.
+As a result, we need to introduce the following substitutions: $m_e \to m_e^*$, $\epsilon_0 \to \epsilon\epsilon_0$.
+We thus estimate the energy of the bound state created by the impurity:
+$$E = -\frac{m_e^*}{m_e \varepsilon^2} R_E = -0.01 \text{eV (in Ge)}$$
+$r_B = 4$ nm (vs $r_B = 0.5$ Ã… in H).
+The electron is very weakly bound to the impurity! At room temperature (0.026 eV), the donor electron is easily thermally excited into the conduction band. 
+
+On the other hand, we can add a group III element to reduce the average number of electrons in the system.
+Group III elements lacks 1 electron and 1 proton and are therefore known as **acceptors**.
+We treat the absence of an electron as a hole and the lacking proton as an effective negative charge.
+As a result, we once again end up with a Hydrogen model, except this time the charges are flipped (hole circles around a negative center).
+That allows us to use the previous results and to conclude that an acceptor creates a weakly bound state above the valance band.
 
 ### Density of states with donors and acceptors
 
@@ -85,11 +98,21 @@ ax.set_xticklabels(['$E_V$', '$E_C$', '$E_A$', '$E_D$'])
 draw_classic_axes(ax, xlabeloffset=.2)
 ```
 
-All donor/acceptor states at the same energy:
-$$g_A(E) = N_A \delta(E-E_A),\quad g_D(E) = N_D \delta(E- E_D)$$
-
-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 $N_D \lesssim (1/4\textrm{nm})^3 \sim 10^{-5}\ll N_C$.
-
+In order to model **multiple** donor/acceptor states, we assume that they are all degenerate at the binding energy.
+Therefore, we model the density of states of donors/acceptors as a Dirac delta function:
+$$
+g_D(E) = N_D \delta(E- E_D), \quad  g_A(E) = N_A \delta(E-E_A), 
+$$
+where N_D and N_A are donor and acceptor concentrations respectively.
+The binding energies of the donor and acceptor are defined as $E_A$ and $E_D$.
+
+How good is this Dirac delta approximation?
+That depends on the concentrations. 
+If we keep on adding impurities, then at some point the weakly bound states will begin to overlap.
+The overlap will create an effective tight-binding model that leads to a formation of an "impurity" band which breaks our approximation.
+We must therefore prevent the overlap of impurity bound states.
+From the previous section, we know that the extent of the bound state is roughly 4 nm and thus the distance between impurity atoms cannot exceed that. 
+As a result, the impurity concentration is bounded to $N_D \lesssim (1/4\textrm{nm})^3 \sim 10^{-5}.
 
 ## Number of carriers