# Energy Band of pn junction in thermal equilibrium (Zero bias)

Consider the following diagrams:

Question:

1. Why in thermal equilibrium—the Fermi energy level is constant throughout the entire system? I know the following explanation A gradient in the Fermi level is the driving force for carrier motion:

$$F_n = \frac{D_n}{k_BT} n \frac{\mathrm{d}E_F}{\mathrm{d}x}$$

In equilibrium (zero bias), $$F_n = 0$$ and therefore $$E_F=\text{const}$$. but don't get the derivation due to following question.

1. In First diagram why the $$E_v$$ elevates? Is that mean that the highest energy state in valence band of p-type semiconductor is increase?if yes How? Consider that I know the explanation that you have to make that fermi level in equillibrium condition constant throughout the entire system.
2. See the second figure $$E_{Fi}$$ in both the figure are equal(atleast in figure) As I joint them How $$E_{Fi}$$ get increased in p type ,I mean it's condition for intrinsic.

3.They get the expression $$V_{bi}=|\phi_{Fp}|+|\phi_{Fn}|.$$(Is that in figure the $$E_{Fi}$$ of p-side is equal to $$E_c$$ of n side if yes How?)

• You say , "An electron at the top of the valence band on the the left side will have more energy than one on the right hand side." I understand but $E_v$ is the energy of highest occupied state not the energy of electron,So how state energy elevate? Commented Mar 20, 2020 at 12:39
• because the energy of a state is the sum of the energy in the absence of an external electric potential and $-e V$, the energy of an electron in an external electric potential V. the potential is lower on the left so the energy of the state at the top of the valence band is higher than on the right. Commented Mar 21, 2020 at 14:22