@@ -133,14 +133,14 @@ 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 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
The approximations allow us to simplify the charge balance equation:
$$
n_e - n_i^2/n_e = N_D - N_A,
$$
...
...
@@ -151,16 +151,15 @@ If $N_D < N_A$ ($p$-doped semiconductor), $n_h \approx N_A -N_D$ and $n_e = n_i^
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:
$$
n_e = N_C e^{-(E_c - E_F)/kT},
n_e \approx N_C e^{-(E_c - E_F)/kT},
$$
$$
n_h \approx N_V e^{E_v-E_F/kT}.
$$
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.
## Combining semiconductors: $pn$-junction
What happens if we bring two differently doped semiconductors together (one of $p$-type, one of $n$-type)?
