# Built-in potential in pn junction

• I
Hello!
The (potential) energy of an electron in a solid structure is always negative; also the $E_c$ and $E_v$ levels (conduction band and valence band limits) are negative, in the band diagram of a pn junction.
When the junction is built and thermal equilibrium is reached, the depletion region creates an electric field and a potential barrier $\phi_i$.
1) First of all,

$\phi_i = \displaystyle \frac{kT}{q} \log \left( \displaystyle \frac{N_D N_A}{n_i^2} \right)$

where $k$ is Boltzmann's constant, $T$ is the absolute temperature and $q$ is the absolute value of the electron charge. The energy step due to this built-in potential is $q \phi_i$, but why a positive $q$ is used to calculate this step? Given a voltage difference $\Delta V = \phi_i$, the corresponding potential energy difference should be $\Delta U = q \Delta V = - |q| \Delta V$ in the case of an electron.
2) Let's suppose anyway that $q \phi_i > 0$.
In the conduction band of the n-region, just the charge carriers with an energy $E = E_c + q \phi_i$ can cross the junction and diffuse to the p-region. So, I would say that the p-region, with respect to the n-region, is at a higher potential.
In order to have a reference image, let's consider this document, page 11. Why in the last plot, which shows $V$ as function of position, the n-side is at a higher potential and the p-side at $V = 0$? (Consider $V_A = 0$, that is the case of thermal equilibrium)
3) What about the vacuum level? It is never depicted in the band diagram of a pn junction; but how is it affected by the built-in voltage?
Thank you anyway for having read.

Emily

From the point of view of electron what you say is right p-side is at higher potential. But it is a convention to consider from the point of view of a positive test charge. What do you mean by vacuum level you mean at absolute zero. At absolute zero all the bonds will be in tact and there will not be any free charge to create potential difference or depletion region.

EmilyRuck
From the point of view of electron what you say is right p-side is at higher potential. But it is a convention to consider from the point of view of a positive test charge.
Ok, thank you.

What do you mean by vacuum level you mean at absolute zero. At absolute zero all the bonds will be in tact and there will not be any free charge to create potential difference or depletion region.
By "vacuum level" I mean the energy of a free electron outside the crystal, as stated here, par. 2.3.3.2. Anyway, maybe the vacuum level is bent like the energy bands across the junction due to the electric field.