However, we have to remember that the above equations are valid for a hydrogen atom in free space. In our case, the extra electron moves in the semiconductor.
Therefore, there are a couple of differences from the Hydrogen model.
One difference is that the electron's mass is conduction band's effective mass.
One difference is that the electron's mass is determined by the conduction band's effective mass.
Another difference is that the interaction between the electron and the nucleus is screened by the lattice.
As a result, we need to introduce the following substitutions: $m_e \to m_e^*$, $\epsilon_0 \to \epsilon\epsilon_0$.
This turns out to yield the following estimation of the energy of the bound state associated with the impurity:
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.
***Extrinsic limit**. As we decrease the temperature, we decrease the number of intrinsic carriers to the point where most of the electrons in the conduction band (the charge carriers) originate from 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 $kT \ll E_C - E_D$, we expect the electrons to "freeze away" from the conduction band to the donor band. The charge carriers still originate from the donors, however, not all donors are ionized now.
***Zero temperature**. There are no charge carriers in neither conduction nor valence bands. If there are no acceptors, then the Fermi level goes to halfway between the donor and the conduction bands. Otherwise, the highest energy electrons are in the donor band and therefore $E_F$ should match the donor band.
!!! check "Exercise"
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@@ -273,7 +273,7 @@ What happens if we bring two differently doped semiconductors together (one of $
### Band diagram
Previously we dealt with homogeneous materials, now the position coordinate (let's call it $x$) starts playing a role.
Until now we dealt with spatially homogeneous materials. From now on the position coordinate (let's call it $x$) starts playing a role.
We represent the properties of inhomogeneous materials using the **band diagram**.
The main idea is to plot the dependence of various energies ($E_F$, bottom of conduction band $E_C$, top of the valence band $E_V$) as a function of position.
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@@ -604,13 +604,13 @@ fig1
The main difference between $n$-type and $p$-type semiconductors is the location of the Fermi level $E_F$ (see "n and p" tab above).
The Fermi level of an $n$-type semiconductor is close to the donor states.
On the other hand, the $p$-type semiconductor has its Fermi level near the acceptor states.
At equilibrium (no external fields), we do not expect to see any currents in the system and therefore the **Fermi level $E_F$ must be constant** across the system (see "Equilibrium" tab).
To achieve a homogenous Fermi level, we could bring up in energy the $p$-type region or bring down the $n$-type region until the Fermi levels are aligned.
At equilibrium (i.e., in the absence of an external electric field), we do not expect to see any currents in the system. Therefore, the **Fermi level $E_F$ must be constant** across the system (see "Equilibrium" tab).
To achieve a constant Fermi level, we bring up in energy the $p$-type region or bring down the $n$-type region until the Fermi levels are aligned.
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 valence $E_V$ 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.
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@@ -623,14 +623,13 @@ As the recombination process continues, a larger charge density $\rho$ develops
Inside the region, energy deviates by $\delta \varphi \gg kT$ from the bulk value and thus the density of electrons/holes drops exponentially fast.
Therefore, we refer to the region as the **depletion region**.
The charge density $\rho$ distribution inside of a depletion region is shown below:
The charge density $\rho$ distribution inside the depletion region is shown below:
The typical values of $w_n+w_p$ are $\sim 1 \mu \textrm{m}$ at $N_A,\,N_D \sim 10^{16} \textrm{cm}^{-3}$, and $\sim 0.1 \mu \textrm{m}$ at $N_A,\,N_D \sim 10^{18} \textrm{cm}^{-3}$, so it may be much larger than the distance between the dopant atoms.
