3 Ways of treating an external Potential in a PN-Junction

Question

From my understanding of the PN-Junction in a semiconductor (without applied voltage), the chemical Potential $\mu(x)$ is required to be the same at every position (in equilibrium). We divide the semiconductor into suitably small Volumes at position x, and assume every volume to behave like a grandcanonical ensemble, where the Energy of one electron is given by a hamiltonian $H_0$. Thermodynamical equilibrium is reached when

$\mu(x) = constant$.

In the neutral regions, $\mu(x)$ is linked to the Energies $E_v$ and $E_c$, so one way of satisfying $\mu(x)$ being constant is raising those Energies by applying a potential $\Phi(x)$. With this potential existing $\mu$ is constant, and $\mu$ is still linked to the Energies in the n-doped or p-doped neutral regions (of course the potential is created by charge densities at the junction, so there $\mu$ is not anymore linked to $E_V$ or $E_c$). So far so good. Now my Question:

What exactly happens, when you apply an external Potential? My Model for that is simply adding an external Potential $\Phi_{ext}(x)$ to the existing Potential, but up from there you have several possibilities:

1.You once again assume that the junction will again reach an equilibrium state, which means: $\tilde{\mu}$, the chemical Potential, is constant. I wrote it with a ~ because this is the chemical Potential for another System (many Volumes at position x with the hamiltonian $H_0 + \Phi_{ext}(x)$

2.Aequivalent to that, you assume that the only effect of $\Phi_{ext}$ is changing your chemical potential $\mu$: Your system still constists of many Volumes at position x, the hamiltonian for each volume still being $H_0$, but now with your chemical Potential $\mu$ no longer being constant. Instead $\mu$ now has to satisfy the condition: \begin{align} \mu(x) + \Phi_{ext}(x) = constant \end{align} Both treatments are aequivalent since the partition function of a grandcanonical ensemble with 1particle-hamiltonian $H_0 + V$ and chemical Potential $\tilde{\mu}$ is aequivalent to a grandcanonical ensemble with 1particle-hamiltonian $H_0$ and chemical Potential $\mu - V$.

3. You don't assume a new equilibrium of the system, but instead $\Phi_ext(x)$ changes the equilibrium distribution of the unbiased pn-junction intwo a nonequilibrium distribution, following the boltzmann-equations for the electron / hole density $f(x,p)$, $p$ being the momentum, and thus creating a current.

Option 1 and 2 will give you the same carrier densities, but will they also give you the same current? How are these options linked to option 3? Is option 3 applicable here?

Answer 1

When you apply an external potential on a pn-junction in equilibrium you will not have an equilibrium of the system anymore and you don't have a constant chemical potential or sum of chemical potential with electrical potential throughout the junction. Therefore you cannot use your solutions 1. and 2. As you point out in 3. you have a non-equilibrium situation with electron and hole currents and, in principle, you have to find the non-equilibrium concentrations of electrons and holes and their currents by solving transport equations, like, e.g., the drift-diffusion equations for electrons and holes and the current continuity equations with recombination/generation terms and Poisson equation for the potential. In the classical approach to get approximations for the carrier distributions and currents often separate quasi-chemical potentials (i.e. quasi-fermi levels) for electrons and holes are introduced that are separated by the applied voltage at the contacts and assumed to be approximately constant over respective regions of the junction. The non-equilibrium carrier currents due to diffusion and drift are then calculated with recombination and generation terms. This approach is described in most textbooks on semiconductor devices.