@@ -217,27 +217,19 @@ 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 valance $E_V$ and conduction $E_C$ bands connect continuously in the middle region.
On the contrary, if the bands are discontinuous, then an electric field must develop at a single point in the middle region to shift the bands in energy.
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.
On a more microscopic level, the electrons at the junction in the $n$-type semiconductor will move into the $p$-type semiconductor to **recombine** with the holes.
After the recombination, the $n$ and $p$-type semiconductors lose an electron and a hole respectively.
As a result, a single positive ionized donor dopant is not screened anymore and the $n$-type semiconductor obtains a positive overall charge.
As a result, a positive ionized donor dopant is not screened anymore and the $n$-type semiconductor obtains a positive overall charge.
Similarly, $p$-type region obtains a negative charge.
Therefore, an electric field develops across the junction.
As the recombination process continues, more unbalanced charges develop and the electric field grows until it is large enough to prevent the electrons/holes from crossing the junction.
The region where the electric field develops is called the **depletion region** because the electrons and holes are depleted in that area.
As the recombination process continues, a larger charge density $\rho$ develops and thus the electric field grows until it is large enough to prevent the electrons/holes from crossing the junction.
If the resulting potential difference becomes $\delta \varhpi \gg kT$, the density of electrons and holes drops exponentially fast.
Therefore, we refer to the region as the **depletion region**.
We now understand the rough qualitative picture of what happens at the $np$-junction.
However, can we understand things more quantitatively?
Specifically, how fast does the electrostatic potential change in the **depletion region** ?
What is the charge density at the junction if we make the chemically-defined boundary between materials very precise?
We may instead use a key bit of insight: **the density of electrons and holes drops exponentially fast as soon as potential deviates by $kT \ll \delta \varphi$ from its bulk value**.
Using this information we can expect a formation of the **depletion region** with almost no electrons or holes (but donors and acceptors won't move anywhere from that region).
So this is our expectation of the dependence of $\rho(x)$:
A schematic of a depletion region is shown below:
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@@ -252,6 +244,7 @@ Because the conductivity of the $p$-region and $n$-region is much larger than th
The number of majority carriers moving across the junction is proportional to their concentration.
Increasing the voltage bias "pushes" carriers up in energy, it depends exponentially on the voltage.
