@@ -46,15 +46,15 @@ _(based on chapters 17–18 of the book)_
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.
An empty band is not going to 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.
Without a doubt, the greatest example is *semiconductors*—the bedrock of modern electronics.
In this lecture, we shall grasp the basics of semiconductors by learning how to treat bands at different levels of filling.
## Review of band structure properties
Before proceeding further, it is vital to remind ourselves some of the concepts in band structures.
Before proceeding further, let us remind ourselves of important band structure properties.
* Group velocity $v=\hbar^{-1}\partial E(k)/\partial k$.
Descibes how quickly electrons move within the lattice.
...
...
@@ -66,68 +66,69 @@ The quantity is vital in order to calculate any bulk property of the material su
In order to check that everything makes sense, we apply the concepts to the free electron model:
$$H = \hbar^2 k^2/2m$$
$$H = \frac{\hbar^2 k^2}{2m}$$
The velocity is $\hbar^{-1}\partial E(k)/\partial k = \hbar k / m \equiv p/m$.
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.
So in this simplest case the definitions match the usual expressions.
## Filled vs empty bands
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:
Despite being two opposite extreme extreme cases, filled and empty bands are very similar.
For example, both filled and empty bands carry no electric current:
$$
\begin{align}
j = 2e \frac{1}{2\pi} \int_{-\pi/a}^{\pi/a} v(k) dk = 2e \frac{1}{2\pi \hbar} \int_{-\pi/a}^{\pi/a} \frac{dE}{dk} \times dk = \\
2e \frac{1}{2\pi \hbar} [E(-\pi/a) - E(\pi/a)] = 0.
\end{align}
$$
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.
Similar results apply to many other physical quantities such as heat capacity and magnetization.
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.
As a result, rather than to consider hundreds of bands that a material contains, we neglect most of them and just focus on the handful of partially filled bands around Fermi level.
## From electrons to holes
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.
Because completely filled or completely empty bands have simple properties, we may search for a convenient way to describe a band that only has a few electrons missing or extra.
While keeping track of a few electrons has no tricks, even a few electrons missing from a band seem to require considering all the other electrons in a band.
A more efficient approach to describing a nearly filled band is motivated by the following analogy.
Let us say we have 100 boxes: 99 are occupied and 1 is empty.
To keep track which box is occupied/empty, we could write down the numbers of all 99 occupied boxes.
If, on the other hand, we only keep track which single box is empty, we solve the problem with a lot less book-keeping.
The same approach applies 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.
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.
Naturally, dealing with electrons is more convenient whenever a band is almost empty and with holes when a band is almost full.
## Properties of holes
Let us compare the properties of electrons and holes.
Let us compare the properties of an electron and a hole obtained by removing that electron.
Since removing an electron reduces the total energy of the system, the hole's energy is opposite to that of an electron $E_h = -E$.
The probability for an electron state to be occupied in equilibrium is given by $f(E)$:
$$f(E) = \frac{1}{e^{(E-E_F)/kT} + 1}.$$
On the other hand, since a hole is a missing electron, the probability for a hole state to be occupied is
Since a hole is a missing electron, the probability for a hole state to be occupied is
therefore for holes both energy $E_h= -E$ and $E_{F,h} = -E_F$.
which is the Fermi distribution of particles with 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.
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 $-e$).
Similarly, the total **charge** should be the same regardless of whether we count electrons or holes, so holes have a positive charge $+e$ (electrons having $-e$).
On the other hand, the velocity of a hole is **the same**:
On the other hand, hole's velocity is **the same** as that of an electron:
Finally, we derive the hole effective mass from the equations of motion:
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...
@@ -138,42 +139,56 @@ Comparing with
$$m_e \frac{d v}{d t} = -e (E + v\times B),$$
we get $m_h = -m_e$.
we get $m_h = -m_e$ (we could also obtain this by differentiating the hole's velocity).
## 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.
Semiconductors are materials with all bands either nearly occupied or almost empty.
Unlike in insulators, however, the band gap in semiconductors is sufficiently small, for it to be possible to create a few electrons in the lowest unoccupied band or the highest filled band.
Because in the unoccupied band the **charge carriers** (particles carrying electric current) are electrons, it is called **conduction band**, while in the almost occupied **valence band** the charge carriers are holes.
We can control the position of the Fermi level (or create additional excitations) making semiconductors conduct when needed.
!!! note "Holes in semiconductors"
Only the bottom of the conduction band has electrons and the top of the valence band has holes because the temperature is always smaller than the size of the band gap.
When introducting holes, we discussed holes obtained by removing *any* electron.
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.
As a result, only the bottom of the conduction band has electrons and the top of the valence band has holes.
Therefore we can approximate the dispersion relation of both bands as parabolic.
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 $E_h > -E_v$.
??? 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?"
The corresponding density of states of the two types of particles is
??? 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
### Part 1: Intrinsic semiconductor
### 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)
...
...
@@ -197,6 +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.
Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$
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.
Therefore Fermi-Dirac distribution is approximately similar to Boltzmann distribution.
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.
Therefore, the fourth step assumes that the Fermi level is far from both bands $E_F-E_v \gg kT$ and $E_c - E_F \gg kT$.
As a result, the Fermi-Dirac distribution is approximately similar to Boltzmann distribution:
An extra observation: regardless of where $E_F$ is located, $n_e n_h = N_C N_V e^{-E_g/kT} \equiv n_i^2$, where $E_g=E_c-E_v$ is the band gap of the semiconductor.
...
...
@@ -255,6 +286,25 @@ $n_i$ is the **intrinsic carrier concentration**, and for a pristine semiconduct
> $$n_e n_h = n_i^2$$
> is the **law of mass action**. The name is borrowed from chemistry, and describes the equilibrium concentration of two reagents in a reaction $A+B \leftrightarrow AB$. Here electrons and hole constantly split and recombine.
### Conduction
Earlier, we deduced that empty and filled bands provide no current.
We finish the analysis by considering partially filled bands of an intrinsic (pristine) semiconductor.
To calculate the current, we utilize the Drude model and sum the electron and hole contributions:
We see that despite opposite velocity signs for electrons and holes, they carry electric current in the same direction:
$$ \sigma \equiv \frac{j}{E} = \left(\frac{n_e e^2 \tau_e}{m_e}+\frac{n_h e^2 \tau_h}{m_h}\right) = n_e e \mu_e + n_h e \mu_h.$$
We know that for intrinsic semiconductors, the hole/electron densities are $n_e = n_h = n_i \propto e^{-E_G/kT}$.
Therefore, it is possible to measure the band gap of an intrinsic semiconductor by looking at the temperature dependant conductivity $E_G \approx d \ln \sigma / d [kT]^{-1}$.
Additional information can be obtained using Hall effect.
However Hall effect is much more complex in semiconductors since only the current in the direction perpendicular to the applied electric field must vanish.
This, however only means that the electron current is opposite of the hole current in that direction, not that the electrons and holes move parallel to the applied current.
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:
Most donors are ionized and most acceptors are occupied.
Then
$$n_e - n_h = N_D - N_A, n_e = n_i^2/n_h$$
When $|N_D-N_A| \gg n_i$ the semiconductor is **extrinsic**, so that if $N_D > N_A$ ($n$-doped semiconductor), $n_e \approx N_D -N_A$ and $n_h = n_i^2/(N_D-N_A)$. If $N_D < N_A$ ($p$-doped semiconductor), $n_h \approx N_A -N_D$ and $n_e = n_i^2/(N_A-N_D)$.
| $n_e$ | 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.
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.
We determine the electron/hole occupation of the donor/acceptor states by applying Fermi-Dirac statistics to their simple Dirac delta density of states:
Here we refer to $n_D$($n_A$) as the electron(hole) concentration inside donor(acceptor) bound state.
With this, the charge balance equation reads:
$$n_e - n_h + n_D - n_A = N_D - N_A.$$
The equation is not an easy one to solve: all of the terms on the lhs depend non-trivially on $E_F$.
In order to solve it, we require several approximations:
* Firstly, we assume that the Fermi level is far from both bands $E_F−E_v \gg kT$ and $E_c−E_F \gg kT$. The approximation allows us to use the law of mass action from the previous lecture:
$$
n_e n_h = N_C N_V e^{-E_g/kT} \equiv n_i^2.
$$
* Secondly, we determined that electrons/holes are weakly bound to the impurities. Therefore, at ambient temperatures, we assume that all the impurities are fully ionized and therefore $n_D = n_A = 0$.
The approximations allow us to simplify the charge balance equation:
$$
n_e - n_i^2/n_e = N_D - N_A,
$$
which is just the quadratic equation for $n_e$.
When $|N_D-N_A| \gg n_i$ the semiconductor is **extrinsic**, so that if $N_D > N_A$ ($n$-doped semiconductor), $n_e \approx N_D -N_A$ and $n_h = n_i^2/(N_D-N_A)$.
If $N_D < N_A$ ($p$-doped semiconductor), $n_h \approx N_A -N_D$ and $n_e = n_i^2/(N_A-N_D)$.
We can now easily find the Fermi level.
From the first approximation, we know that the simplified relation between $n_{e/h}$ and $E_F$ is:
$$
n_e \approx N_C e^{-(E_c - E_F)/kT},
$$
$$
n_h \approx N_V e^{E_v-E_F/kT}.
$$
We express the lhs with the quadratic equation solution and solve for Fermi level:
## Temperature dependence of the carrier density and Fermi level
It is instructive to consider how $E_F$, $n_e$ and $n_h$ depend on carrier concentrations.
In this case, we consider an n-doped semiconductor, however, the same logic applies to p-doped semiconductors.
Several noteworthy features:
There are several relevant temperature limits:
* At high temperature $n_e = n_h$ and $E_F$ has an upturn (if holes are heavier than electrons)
* Once the temperature is sufficiently low, we expect the electrons to "freeze away" from the conduction band to the donor band, so that the donor band starts playing a role of the new valence band at $kT \ll E_G - E_D$.
* At zero temperature $E_F$ should match the donor band since it has partially occupied states. If there are no acceptors, $E_F$ would be halfway between $E_D$ and $E_G$, and if there was no doping at all it would be at $E_G/2$.
***Intrinsic limit** . If the temperature is sufficiently large, then $n_i \gg |N_D-N_A|$ and therefore $n_e = n_h = n_i$. Additionally, if holes are heavier than electrons, then $E_F$ has an upturn in this limit.
***Extrinsic limit**. If 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.
!!! check "Exercise"
check that you can reproduce all the relevant limits in a calculation.
## Measuring band gaps
### Conduction
The change of $n_i$ is extremely rapid due to the factor $e^{-E_G/kT}$. Therefore the simplest way of determining the band gap is the temperature dependence of conductance
Combining the two we see that despite electron and hole velocities have opposite signs, they carry electric current in the same direction.
$$ \sigma \equiv \frac{j}{E} = \left(\frac{n_e e^2 \tau_e}{m_e}+\frac{n_h e^2 \tau_h}{m_h}\right) = n_e e \mu_e + n_h e \mu_h.$$
Since $n_e = n_h = n_i \propto e^{-E_G/kT}$, $E_G \approx d \log \sigma / d [kT]^{-1}$.
Additional information can be obtained using Hall effect. However Hall effect is much more complex in semiconductors since only the current in the direction perpendicular to the applied electric field must vanish. This, however only means that the electron current is opposite of the hole current in that direction, not that the electrons and holes move parallel to the applied current.
### Light absorption
See [previous lecture](12_band_structures_in_higher_dimensions.md#light-adsorption)
## Combining semiconductors: $pn$-junction
Main idea: what happens if we bring two differently doped semiconductors together (one of $p$-type, one of $n$-type)?
What happens if we bring two differently doped semiconductors together (one of $p$-type, one of $n$-type)?
