docs/13_semiconductors.md

@@ -254,7 +254,7 @@ n_e = \int_{E_c}^\infty f(E)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi
 \end{gather}
 
 Note that whenever calculating the hole dependent quantities, we replace all the relevant physical quantities with their hole equivalents.
-Since the hole energy is opposite $E_h = -E$, we replace the Fermi energy $E_F \to -E_F$ and the bottom of the valance band by $E_v \to -E_v$ in the integration limits.
+Since the hole energy is opposite $E_h = -E$, we replace the Fermi energy $E_F \to -E_F$ and the bottom of the valence band by $E_v \to -E_v$ in the integration limits.
 
 In the third step, we need to solve the equation under charge balance $n_e = n_h$.
 The equation is not a pleasant one and cannot be solved analytically unless an approximation is made.

docs/14_doping_and_devices.md

@@ -50,8 +50,8 @@ In today's lecture, we will take a look at how doping allows the fine control of
 
 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. 
+In both cases, there are 4 valence 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 valence 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.
@@ -87,7 +87,7 @@ On the other hand, we can add a group III element to reduce the average number o
 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.
+That allows us to use the previous results and to conclude that an acceptor creates a weakly bound state above the valence band.
 
 ### Density of states with donors and acceptors
 
@@ -134,20 +134,22 @@ As a result, the impurity concentration is bounded to $N_D \lesssim (1/4\textrm{
 
 ## Number of carriers
 
-| Symbol | Meaning |
-| - | - |
-| $n_e$ | Concentration of electrons in the conduction band |
-| $n_h$ | Concentration of holes in the valance band |
-| $n_D$ | Concentration of electrons in the donor bound state|
-| $n_A$ | Concentration of holes in the acceptor bound state|
-| $N_D$ | Concentration of donor impurities| 
-| $N_A$ | Concentration of acceptor impurities|
-
-We now have the necessary tools to determine how the Fermi level changes with doping. 
+We now have the necessary tools to determine how the Fermi level changes with doping.
 The algorithm to determine the Fermi level of a semiconductor was outlined in the previous lecture and we continue to use it here.
-The process is the same up until the third step - charge conservation. 
-The semiconductor now contains impurities that become charged through ionization.
-For example, if the donor impurity bound state loses an electron - it becomes positively charged.
+The process is the same up until the third step—charge conservation. 
+The semiconductor now contains dopant atoms that become ionized, and therefore charged.
+For example, if a donor impurity bound state loses an electron, it becomes positively charged.
+To solve the problem let us introduce all the necessary concentrations:
+
+| Symbol | Concentration of |
+| - | - |
+| $n_e$ | electrons in the conduction band |
+| $n_h$ | holes in the valence band |
+| $n_D$ | electrons in the donor bound states |
+| $n_A$ | holes in the acceptor bound states |
+| $N_D$ | donor impurities |
+| $N_A$ | acceptor impurities |
+
 We determine the electron/hole occupation of the donor/acceptor states by applying Fermi-Dirac statistics to their simple Dirac delta density of states:
 
 $$
@@ -255,7 +257,7 @@ As we go from highest to lowest temperature, we observe several regimes:
 * **Intrinsic limit** . If the temperature is sufficiently large, then $n_i \gg |N_D-N_A|$ and therefore $n_e \approx n_h \approx n_i$. Additionally, if holes are heavier than electrons, then $E_F$ grows with temperature in this limit.
 * **Extrinsic limit**. As we decrease the temperature, we decrease the number of intrinsic carriers to the point where most of the charge carriers come form the fully ionized donors. As a result, the number of carriers stays approximately constant in this temperature range.
 * **Freeze-out limit**. Once the temperature is sufficiently low $kT \ll E_G - E_D$, we expect the electrons to "freeze away" from the conduction band to the donor band. The charge carriers still come from the donors, however, not all donors are ionized now.
-* **Zero temperature**. There are no charge carriers in neither conduction nor valance bands. The highest energy electrons are in the donor band and therefore $E_F$ should match the donor band. 
+* **Zero temperature**. There are no charge carriers in neither conduction nor valence bands. The highest energy electrons are in the donor band and therefore $E_F$ should match the donor band.
 
 !!! check "Exercise"
     check that you can reproduce all the relevant limits in a calculation.
@@ -603,7 +605,7 @@ However, a question arises: what happens at the junction?
 
 We can understand the junction with a simple picture. 
 In physics, most of the time we expect things to change *continuously*.
-Therefore, we expect that the valance $E_V$ and conduction $E_C$ bands connect continuously in the middle region as shown in the "Band Bending" tab.
+Therefore, we expect that the valence $E_V$ and conduction $E_C$ bands connect continuously in the middle region as shown in the "Band Bending" tab.
 On the contrary, if the bands were to be discontinuous, then an electric field must develop at a single point in the middle region to shift the bands in energy.
 However, we do not expect such point-like electric fields to develop because electrons can move freely in semiconductors.