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 $n_e = n_i^2/(N_A-N_D)$.
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 $n_D = n_A = 0$.
We can now easily find the Fermi level:
The two approximations allow us to to express the charge balance equation purely in terms of $n_h$ (n-doped) or $n_e$ (p-doped).
For an n-doped semiconductor, the approximations reduce the charge balance equation to
$$
n_e - n_i^2/n_e = N_D - N_A,
$$
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:
$$
n_e = N_C e^{-(E_c - E_F)/kT},
$$
$$
n_h \approx N_V e^{E_v-E_F/kT}.
$$
We express the lhs with the quadratic equation solution and solve for Fermi level:
$$E_F = kT\log[N_V/(N_A-N_D)], \textrm{ for } N_A > N_D$$
...
...
@@ -158,33 +180,6 @@ Several noteworthy features:
!!! 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)?
