Lecture_13

src/13_semiconductors.md

@@ -43,13 +43,28 @@ _(based on chapters 17–18 of the book)_
 
     <iframe width="100%" height="315" src="https://www.youtube-nocookie.com/embed/IxDp_JAtBQs" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
 
+
+Up until this point, we focused on calculating and understanding the band structures.
+However, the dispersion of a band is only part of the story.
+An empty band is not gonna lead to any interesting physical properties no matter how sophisticated it is.
+Therefore, it is also important *how* bands are filled by the particles.
+By carefully controlling the distribution of particles in the bands, we are able to engineer material properties that we require.
+Without a doubt, the greatest example is *semiconductors* - the bedrock of all electronics. 
+In this lecture, we shall grasp the basics of semiconductors by learning how to treat bands at different filling.
+
 ## Review of band structure properties
 
-* Group velocity $v=\hbar^{-1}\partial E(k)/\partial k$
-* Effective mass $m_{eff} = \hbar^2\left(d^2 E(k)/dk^2\right)^{-1}$
-* Density of states $g(E) = \sum_{\textrm{FS}} (dn/dk) \times (dk/dE)$ the amount of states per infinitesimal interval of energy at each energy.
+Before proceeding further, it is vital to remind ourselves some of the concepts in band structures. 
+
+* Group velocity $v=\hbar^{-1}\partial E(k)/\partial k$. 
+Descibes how quickly electrons move within the lattice. 
+* Effective mass $m_{eff} = \hbar^2\left(d^2 E(k)/dk^2\right)^{-1}$. 
+Tells us how hard it is to *accelerate* the particles and is related to the curvature of the band.
+* Density of states $g(E) = \sum_{\textrm{FS}} (dn/dk) \times (dk/dE)$. 
+The amount of states per infinitesimal interval of energy at given energy. 
+The quantity is vital in order to calculate any bulk property of the material such as conductivity, heat capacity, etc.
 
-A simple check that everything is correct is the free electron model:
+In order to check that everything makes sense, we apply the concepts to the free electron model:
 
 $$H = \hbar^2 k^2/2m$$
 
@@ -58,70 +73,78 @@ The effective mass is $m_{eff} = \hbar^2\left(d^2 E(k)/dk^2\right)^{-1} = m$.
 
 So in the simplest case the definitions match the usual expressions.
 
-#### Why are these properties important?
-
-* Group velocity descibes how quickly electrons interacting with the lattice move.
-* Effective mass tells us how hard it is to *accelerate* the particles and it enters $g(E)$ for a parabolic band
-* Density of states is required to determine the Fermi level, heat capacity, etc.
-
 ## Filled vs empty bands
 
-A completely filled band is very similar to a completely empty band.
-
-In a filled band $n(E)=1$ because $|E - E_F| \gg kT$. In an empty band $n(E)=0$.
-
-Heat capacity $C_v = \tfrac{d}{dT}\int_{-\infty}^\infty E\times g(E) \times dE\times n_F(E, T) = 0$.
-
-A completely filled band carries no electric current:
+We distinguish three different band filling types: filled, empty and partially filled. 
+Let us start with the most extreme cases - filled and empty bands.
+We treat these two cases together, because a completely filled band is very similar to a completely empty band.
+For example, both filled and empty bands lead to zero current:
 $$
 \begin{align}
 j = 2e \frac{1}{2\pi} \int_{-\pi/a}^{\pi/a} v(k) dk = 2e \frac{1}{2\pi \hbar} \int_{-\pi/a}^{\pi/a} \frac{dE}{dk} \times dk = \\
-2e \frac{1}{2\pi \hbar} [E(-\pi/a) - E(\pi/a)] = 0
+2e \frac{1}{2\pi \hbar} [E(-\pi/a) - E(\pi/a)] = 0.
 \end{align}
 $$
+An empty band has no electrons and thus no current.
+On the other hand, a filled band has an equal number of electrons going forwards and backwards which thus cancel and lead to zero current. 
+Similar results apply to many other physical quantities such as heat capacity and magnetisation. 
+Therefore, filled and empty bands do not affect most physical properties and can be disregarded.
+As a result, rather than to consider thousands of bands that a material contains, we neglect most of them and just focus on partially filled bands around Fermi level.
 
 ## From electrons to holes
 
-A completely filled band $\approx$ completely empty band. The idea of introducing **holes** is to transform an *almost* completely filled band $\Rightarrow$ almost empty one. Instead of describing a lot of electrons that are present, we focus on those that are absent.
-
-Definition: a **hole** is a state of a completely filled band with one particle missing.
+In order to understand partial filling, we begin with a simple analogy.
+Let's say we have 100 chairs: 99 are occupied and 1 is empty.
+To keep track which chair is occupied/empty, we could write down the occupation of each and every chair.
+However, that would require a lot of unnecessary writing!
+Instead, we only need to write down which chair is empty, because we know the rest wil be occupied. 
+The same philosophy is applied to band filling. 
+Instead of describing a lot of electrons that are present in an almost filled band, we focus on those that are absent.
+The absence of an electron is called a **hole**: a state of a completely filled band with one particle missing.
 
 ![](figures/holes.svg)
 
-In this schematic we can either say that 8×2 electron states are occupied (the system has 8×2 electrons counting spin), or 10×2 hole states are occupied. A useful analogy to remember: glass half-full or glass half-empty.
+In this schematic we can either say that 8×2 electron states are occupied (the system has 8×2 electrons counting spin), or 10×2 hole states are occupied.
+A useful analogy to remember: glass half-full or glass half-empty.
+Electron and hole pictures correspond to two different, but equivalent ways of describing the occupation of a band.
+In practice, we choose to deal with electrons whenever a band is almost empty and holes when a band is almost full.
 
 ## Properties of holes
 
-The probability for an electron state to be occupied in equilibrium is $f(E)$:
+Let us compare the properties of electrons and holes.
+The probability for an electron state to be occupied in equilibrium is given by $f(E)$:
 
-$$f(E) = \frac{1}{e^{(E-E_F)/kT} + 1}$$
+$$f(E) = \frac{1}{e^{(E-E_F)/kT} + 1}.$$
 
-The probability for a hole state to be occupied:
+On the other hand, since a hole is a missing electron, the probability for a hole state to be occupied is
 
-$$1 - f(E) = 1 - \frac{1}{e^{(E-E_F)/kT} + 1} = \frac{1}{e^{(-E+E_F)/kT} + 1}$$
+$$f_h(E) = 1 - f(E) = 1 - \frac{1}{e^{(E-E_F)/kT} + 1} = \frac{1}{e^{(-E+E_F)/kT} + 1},$$
 
 therefore for holes both energy $E_h= -E$ and $E_{F,h} = -E_F$.
 
-The **momentum** $p_h$ of a hole should give the correct total momentum of a partially filled band if one sums momenta of all holes. Therefore $p_h = -\hbar k$, where $k$ is the wave vector of the electron.
+The **momentum** $p_h$ of a hole should give the correct total momentum of a partially filled band if one sums momenta of all holes.
+Therefore $p_h = -\hbar k$, where $k$ is the wave vector of the electron.
 
 Similarly, the total **charge** should be the same regardless of whether we count electrons or holes, so holes have a positive charge $+e$ (electrons have $-e$).
 
 On the other hand, the velocity of a hole is **the same**:
-$$\frac{dE_h}{dp_h} = \frac{-dE}{-d\hbar k} = \frac{dE}{dp}$$
+$$\frac{dE_h}{dp_h} = \frac{-dE}{-d\hbar k} = \frac{dE}{dp}.$$
 
 Finally, we derive the hole effective mass from the equations of motion:
 
-$$m_h \frac{d v}{d t} = +e (E + v\times B)$$
+$$m_h \frac{d v}{d t} = +e (E + v\times B).$$
 
 Comparing with
 
-$$m_e \frac{d v}{d t} = -e (E + v\times B)$$
+$$m_e \frac{d v}{d t} = -e (E + v\times B),$$
 
 we get $m_h = -m_e$.
 
 ## Semiconductors: materials with two bands.
 
-In semiconductors the Fermi level is between two bands. The unoccupied band is the **conduction band**, the occupied one is the **valence band**. In the conduction band the **charge carriers** (particles carrying electric current) are electrons, in the valence band they are holes.
+In semiconductors the Fermi level is between two bands.
+The unoccupied band is the **conduction band**, the occupied one is the **valence band**. 
+In the conduction band the **charge carriers** (particles carrying electric current) are electrons, in the valence band they are holes.
 
 We can control the position of the Fermi level (or create additional excitations) making semiconductors conduct when needed.
 
@@ -133,8 +156,13 @@ Therefore we can approximate the dispersion relation of both bands as parabolic.
 
 Or in other words
 
-$$E_e = E_c + \frac{\hbar^2k^2}{2m_e},$$
-$$E_h = E_{v,h} + \frac{\hbar^2k^2}{2m_h} = -E_{v} + \frac{\hbar^2k^2}{2m_h}.$$
+$$E_e = E_c + \frac{\hbar^2k^2}{2m_e}$$
+$$E_h = E_{v,h} + \frac{\hbar^2k^2}{2m_h} = -E_{v} + \frac{\hbar^2k^2}{2m_h},$$
+
+with the corresponding density of states
+
+$$ g(E) = (2m_e)^{3/2}\sqrt{E-E_c}/2\pi^2\hbar^3$$
+$$ g_h(E_h) = (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3.$$
 
 Here $E_c$ is the energy of an electron at the bottom of the conduction band and $E_v$ is the energy of an electron at the top of the valence band.
 
@@ -169,16 +197,6 @@ ax.legend()
 draw_classic_axes(ax, xlabeloffset=.2)
 ```
 
-Electrons
-$$ E = E_c + {p^2}/{2m_e}$$
-
-$$ g(E) = (2m_e)^{3/2}\sqrt{E-E_c}/2\pi^2\hbar^3$$
-
-Holes
-$$ E_h = {p^2}/{2m_h}$$
-
-$$ g(E_h) = (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3$$
-
 **The key algorithm of describing the state of a semiconductor:**
 
 1. Compute the density of states of all types of particles.
@@ -189,9 +207,9 @@ $$ g(E_h) = (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3$$
 
 Applying the algorithm:
 
-$$n_h = \int_{-E_v}^\infty f(E+E_F)g_h(E)dE = \int_{-E_v}^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}\sqrt{E+E_v}\frac{1}{e^{(E+E_F)/kT}+1}dE$$
+$$n_h = \int_{-E_v}^\infty f_h(E_h) g_h(E_h) dE_h = \int_{-E_v}^\infty\frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}\sqrt{E_h+E_v}\frac{1}{e^{(E_h+E_F)/kT}+1}dE_h$$
 
-$$n_e = \int_{E_c}^\infty f(E-E_F)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_c}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
+$$n_e = \int_{E_c}^\infty f(E)g_e(E)dE = \int_{E_c}^\infty\frac{(2m_e)^{3/2}}{2\pi^2\hbar^3} \sqrt{E-E_c}\frac{1}{e^{(E-E_F)/kT}+1}dE$$
 
 We need to solve $n_e = n_h$
 
@@ -200,11 +218,11 @@ Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$
 
 Therefore Fermi-Dirac distribution is approximately similar to Boltzmann distribution.
 
-$$f(E\pm E_F) \approx e^{-(E\pm E_F)/kT}$$
+$$f(E)_{e/h} \approx e^{-(E_{e/h}\pm E_F)/kT}$$
 
 Now we can calculate $n_e$ and $n_h$:
 
-$$n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_{-E_v}^\infty\sqrt{E+E_v}e^{-E/kT}dE =
+$$n_h \approx \frac{(2m_h)^{3/2}}{2\pi^2\hbar^3}e^{-E_F/kT} \int_{-E_v}^\infty\sqrt{E_h+E_v}e^{-E_h/kT}dE_h =
 N_V e^{E_v-E_F/kT},$$
 
 where we used $\int_0^\infty \sqrt{x}e^{-x}dx=\sqrt{\pi}/2$ and we defined