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

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 $\mu - V$.

1. 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?

• I am wondering why the standard treatments of pn junctions in semiconductor physics books are not helpful to you? Commented Oct 6, 2016 at 12:43
• Because when it comes to applying voltage, they mostly become somewhat "handweaving", and either don't exactly say what happens (to the chemical potential etc ...), or they use the argument that the chemical Potential is bent by the potential, whithout saying why that is allowed. Option 3 is never mentioned in those books, hence they don't give me an answer to why it is not used, and how it is linked to the other two. But maybe this is the more complicated question how the boltzmann equation is linked to a statistical treatment of a kinetic gas of particles. Commented Oct 6, 2016 at 22:47
• Why would applying a potential across a device not change the chemical potential of the situation? Commented Oct 7, 2016 at 0:37
• I thought there could also be an equilibrium situation with a potential applied. This situation is: Electron/Hole densities don't change anymore (and a current flows). I thought, describing this situation would again mean the chemical potential being constant. Can you point out the exact mechanism that changes the chemical potential? Is the chemical potential then still well defined? Commented Oct 7, 2016 at 9:41
• If current is flowing you are not in equilibrium. And, that current is from electrons and holes meeting and annihilating each other. The exact mechanism that changes the potential is that an external voltage source is applied to one end. It feeds electrons at this elevated voltage (potential) in to the device, and the device has to figure out how to rearrange itself in to a steady state situation (with current flow). We will avoid, for the moment, the entire 'is the Fermi energy the chemical potential' question, which is not a straightforward one. Commented Oct 7, 2016 at 12:40