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?

  • $\begingroup$ I am wondering why the standard treatments of pn junctions in semiconductor physics books are not helpful to you? $\endgroup$
    – Jon Custer
    Commented Oct 6, 2016 at 12:43
  • $\begingroup$ 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. $\endgroup$ Commented Oct 6, 2016 at 22:47
  • $\begingroup$ Why would applying a potential across a device not change the chemical potential of the situation? $\endgroup$
    – Jon Custer
    Commented Oct 7, 2016 at 0:37
  • $\begingroup$ 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? $\endgroup$ Commented Oct 7, 2016 at 9:41
  • $\begingroup$ 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. $\endgroup$
    – Jon Custer
    Commented Oct 7, 2016 at 12:40

1 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.

  • $\begingroup$ Isn't the introduction of quasi-chemical potenials and treating any electron current as diffusion process exactly the case of a thermodamic equilibrium with potential +quasi- chemical potential being constant throughout the device? $\endgroup$ Commented Oct 6, 2016 at 22:51
  • $\begingroup$ No, in thermodynamic equilibrium there are by definition no electron or hole currents. Also, the quasi-chemical potentials for electrons and holes are, in general, different and only approximately constant in parts of the device. If you assume a local quasi-equilibrium for either electrons or holes then currents are due to the gradient of the quasi-chemical poetential. In a current carrying device the quasi/chemical potentials cannot be exactly constant, they have to have at least a small gradient. $\endgroup$
    – freecharly
    Commented Oct 6, 2016 at 23:00
  • $\begingroup$ I never said the quasi-chemical potential has to be constant, but instead it + the potential together should be constant. That is the way a potential bends the chemical potential in the equilibrium situation. Why does equilibrium mean that there are no currents? To me equilibrium just means that the distribution doesn't change. The term equilibrium may be not sharply definded thouth, since we're looking at many statistic systems linked together here, not just one. $\endgroup$ Commented Oct 6, 2016 at 23:13
  • $\begingroup$ Also the chemical potential + the electrical potential, which is called electrochemical potential, is not constant in a a pn-junction with an applied voltage. It is constant in a device in equilibrium without an applied voltage. In equilibrium there is no flow of energy into the device, which you would have when you have a current I and an applied voltage V. This would give you a power P=V·I that is dissipated and thus supplied to the device. $\endgroup$
    – freecharly
    Commented Oct 6, 2016 at 23:24
  • $\begingroup$ The term "chemical potential" you used has to be understood in the sense of "total chemical potential" which is synonymous to "electrochemical potential" which includes electric potential. $\endgroup$
    – freecharly
    Commented Oct 6, 2016 at 23:26

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