- #1

EmilyRuck

- 136

- 6

The (potential) energy of an electron in a solid structure is always negative; also the [itex]E_c[/itex] and [itex]E_v[/itex] 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 [itex]\phi_i[/itex].

But I have some doubts about this.

1) First of all,

[itex]\phi_i = \displaystyle \frac{kT}{q} \log \left( \displaystyle \frac{N_D N_A}{n_i^2} \right)[/itex]

where [itex]k[/itex] is Boltzmann's constant, [itex]T[/itex] is the absolute temperature and [itex]q[/itex] is the absolute value of the electron charge. The energy step due to this built-in potential is [itex]q \phi_i[/itex], but why a positive [itex]q[/itex] is used to calculate this step? Given a voltage difference [itex]\Delta V = \phi_i[/itex], the corresponding potential energy difference should be [itex]\Delta U = q \Delta V = - |q| \Delta V[/itex] in the case of an electron.

2) Let's suppose anyway that [itex]q \phi_i > 0[/itex].

In the conduction band of the n-region, just the charge carriers with an energy [itex]E = E_c + q \phi_i[/itex] 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 [itex]V[/itex] as function of position, the n-side is at a higher potential and the p-side at [itex]V = 0[/itex]? (Consider [itex]V_A = 0[/itex], 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