# Relation between intrinsic Fermi level and electric field

If I consider a semiconductor in equilibrium (type P or N, non uniformly doped, 1D situation), how can I express the electric field in terms of the intrinsic Fermi level?

I am trying to approach this problem by writing the electric field ($$E$$) as a function of the potential ($$V$$):

$$E = -\frac{\partial V(x)}{\partial x}$$

And maybe I can write $$E_{Fi}(x)=-q.V(x)$$, with $$E_{Fi}$$ the intrinsic Fermi level. I am not able to fully understand why this would be valid and under what conditions (do I need to switch the sign for P type or N type semiconductor?).

Edit : the band energy would look something like this

• If you are applying a field, why does the Fermi energy matter? If you are in equilibrium in one type of homogeneously doped material, why is there a field? Jan 20, 2019 at 20:00
• There should be a field because the material is non uniformly doped, right? (And there is no external field in this case). Am I missing something? Jan 20, 2019 at 20:16
• Assuming that your dopant levels are not reaching degenerative-doping levels (and even then...) - under normal conditions the Fermi level shifts to the dopant levels. So, where is any field coming from? Jan 20, 2019 at 23:26