From f55a3c49a3b0acffb58f59feae7b753a7d03afb4 Mon Sep 17 00:00:00 2001
From: Anton Akhmerov <anton.akhmerov@gmail.com>
Date: Mon, 29 Mar 2021 10:19:53 +0200
Subject: [PATCH] expand the semiconductor E_F calculation
---
src/13_semiconductors.md | 35 ++++++++++++++++++++++-------------
1 file changed, 22 insertions(+), 13 deletions(-)
diff --git a/src/13_semiconductors.md b/src/13_semiconductors.md
index 0547cad4..74bd4076 100644
--- a/src/13_semiconductors.md
+++ b/src/13_semiconductors.md
@@ -150,7 +150,7 @@ Because in the unoccupied band the **charge carriers** (particles carrying elect
!!! note "Holes in semiconductors"
When introducting holes, we discussed holes obtained by removing *any* electron.
- From this point on we will only speak of holes in valence band and electrons in conduction band.
+ From this point on, similar to most other discussions of semiconductors in the literature, we will only speak of holes in valence band and electrons in conduction band.
The occupation of the two bands is dictated by the Fermi distribution.
Furthermore, the Fermi level of a semiconductor lies between the conduction and the valence bands, and the band gap $E_G \gg k_B T$ in most materials.
@@ -161,29 +161,35 @@ Therefore we can approximate the dispersion relation of both bands as parabolic.
Or in other words
-$$E_e = E_c + \frac{\hbar^2k^2}{2m_e}$$
-$$E_h = E_{v,h} + \frac{\hbar^2k^2}{2m_h} = -E_{v} + \frac{\hbar^2k^2}{2m_h},$$
-
-with the corresponding density of states
-
-$$ g(E) = (2m_e)^{3/2}\sqrt{E-E_c}/2\pi^2\hbar^3$$
-$$ g_h(E_h) = (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3.$$
+$$
+\begin{align}
+E_e &= E_c + \frac{\hbar^2k^2}{2m_e},\\
+E_h &= E_{v,h} + \frac{\hbar^2k^2}{2m_h} = -E_{v} + \frac{\hbar^2k^2}{2m_h}.
+\end{align}
+$$
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.
Observe that because we are describing particles in the valence band as holes, $m_h > 0$ and $E_h > -E_v$.
-Lastly, it is important to point out the notation that we adopt here.
-In the above plot, $m_h \neq -m_e$ because it refers to two **different** bands: valance band for holes and conduction band for electrons.
-In most literature, the band indices are neglected such that $m_{h,v} \to m_{h}$ and $m_{e,c} \to m_e$.
+The corresponding density of states of the two types of particles is
+$$
+\begin{align}
+g(E) &= (2m_e)^{3/2}\sqrt{E-E_c}/2\pi^2\hbar^3,\\
+g_h(E_h) &= (2m_h)^{3/2}\sqrt{E_h+E_v}/2\pi^2\hbar^3.
+\end{align}
+$$
-??? question "a photon gives a single electron enough energy to move from the valence band to the conduction band. How many particles does this process create?"
+??? question "A photon gives a single electron enough energy to move from the valence band to the conduction band. How many particles does this process create?"
Two: one electron and one hole.
## Semiconductor density of states and Fermi level
### Intrinsic semiconductor
+Our next task is to figure out *how many* electrons and holes there are, and for that we need to find where the Fermi level $E_F$ is located.
+Let us plot the density of states, the Fermi distribution function, and the density of particles at each energy in the same plot:
+
```python
E = np.linspace(-3, 3, 1000)
fig, ax = pyplot.subplots()
@@ -206,6 +212,9 @@ ax.legend()
draw_classic_axes(ax, xlabeloffset=.2)
```
+We know that by itself, the semiconductor should have no charge, and therefore the total numbers of electrons and holes must be equal.
+Since increasing the Fermi level increases the number of electrons and reduces the number of holes, we will use the charge neutrality condition to determine where the Fermi level is situated.
+
**The key algorithm of describing the state of a semiconductor:**
1. Compute the density of states of all types of particles.
@@ -229,7 +238,7 @@ Therefore, the fourth step assumes that the Fermi level is far from both bands $
As a result, the Fermi-Dirac distribution is approximately similar to Boltzmann distribution:
$$
-f(E)_{e/h} \approx e^{-(E_{e/h}\pm E_F)/kT}.
+f(E)_{e/h} \approx \exp\left[-(E_{e/h}\pm E_F)/kT\right].
$$
Now we can move to the last step and calculate $n_e$ and $n_h$:
--
GitLab