# Fermi Energy for Semiconductor

I usually post question in bulleted form for convenience that does not mean I am asking more than one question.

Although Fermi energy and fermi Level is discussed many times on this website. Still, I have some doubt. What I Understand is

Fermi Energy is the difference highest occupied state and lowest occupied state at 0k.

Fermi Level is the state for which there is a 50% probability of occupation.

1. This definition of Fermi energy is valid only for the system in which electrons are free (metals or superconductor), or any system.

For the intrinsic semiconductor at 0k,

1. Is fermi energy and Fermi level both are equal? (I don't know)

2. If Yes, then the value of this fermi energy is the same as derived for free electrons system (metal) as $$\frac{\hbar^2}{2m}(\frac{3\pi^2N}{V})^{2/3}$$ and

3. In one of the answers (here) it is written that the Fermi energy falls in the band gap If it is a highest occupied state, then how it can fall on bandgap. The energy states in the bandgap are not permissible.

4. If No, then what is the significance of Fermi energy for semiconductors? I think it should lie in the valence band?

5. Some formula we use in solid state semiconductors like $$E_{\mathrm{F}} == \frac{E_{\mathrm{CB}} - E_{\mathrm{VB}}}{2} + \frac{3}{4} k_\mathrm{B} T \ln(\frac{m_{\mathrm{CB}}^*}{m_{\mathrm{VB}}^*})$$

Here the$$E_f$$ is fermi energy or fermi level?

1. In kittel's, book It is has been mentioned that at 0 k the chemical potential equals fermi energy, here, is this fermi energy equals one that derived for metals. Also, I did not understand the significance of this line.

I am seriously confused. I will edit my question later after getting answer, I will make it more precise for future help.

Edit: This question discusses fermi level, And In my question, doubt is solely related to fermi energy.

• Does this answer your question? Understanding the Fermi level and the Fermi-Dirac distribution Commented Mar 11, 2021 at 17:55
• The question is asked wrt to fermi level and band gap, I have doubt with fermi energy Commented Mar 11, 2021 at 17:57
• yes, but the answer addresses your question as well, doesn't it? Commented Mar 11, 2021 at 18:21

Fermi energy and chemical potential are two terms that are very frequently confused.

1. The chemical defined in the Fermi-Dirac distribution. $$\tag{1} f(E) = \frac{1}{1 + \exp(\frac{E-\mu}{K T})}.$$ Chemical potential defined at where the probability equals to 1/2. The width of the region where the probability drops from approximately $$1$$ to nearly $$0$$ is in order of a few $$K T$$. The position of chemical potential depends on the temperature $$T$$ and total particle number $$N$$ by the equation: $$\tag{2} N = V \int_0^\infty g(E) \frac{1}{1 + \exp(\frac{E-\mu}{K T})} dE.$$ Where $$N$$ is the total number of particles, and $$V$$ the volume. Define the density of particle $$n = N/V$$.

2. Fermi energy is the chemical potential at $$0$$ kelvin; $$E_F = \mu(0)$$. At temperature $$0$$K, the Fermi-Dirac distribution becomes a step function equals to $$1$$ for $$E and $$0$$ for $$E> E_F$$. $$N = V \int_0^{E_f} g(E) dE.$$ For a parabolic band, $$E_f = \frac{\hbar^2}{2m} (3\pi n^2)^{2/3}$$. Where $$k_F = (3\pi n^2)^{1/3}$$ is the Fermi wave vector.

3. Typically for a fixd $$n$$, the chemical potential decreases as temperature increases from $$0$$K: $$\tag{3} \mu(T) = E_F \left\{ 1 - C_1 (\frac{K T}{E_F})^2 - ...\right\}.$$ Image when $$T\to\infty$$, the $$\mu \to -\infty$$ to make all energy levels have equal probability of filling. The Fermi level is a term to incdicate where the Fermi energy is located, mostly an indication in a figure.

4. For metallic material, the particle density $$n$$ is great, also the $$E_F \gg K T$$ is larger. The chemical potential is approximately equal to the Fermi level from Eq.(3). This is called the degenerate case in band filling, when the chemical potential can be approximated by Fermi energy.

5. For intrinsic semiconductor, the Ok Fermi energy is located in the gap, a location depends on the curvatures of both conduction and valance band. It will be right at the mid-gap if effective masses $$m_c^* = m_v^*$$. The position of chemical potential is calculated to ensure that the number of vacance in the valence band is equal to the occupied number in the conduction band: $$N_h = N_c$$ (because no impurties): $$\int_0^\infty g(E_c) \frac{1}{1 + \exp(\frac{E_c-\mu}{K T})} dE_c = \int_{-\infty}^0 g(E_v) \left\{1- \frac{1}{1 + \exp(\frac{E_v-\mu}{K T})} \right\} dE_v.$$ The chemical potential can be viewed as the supporting points to balance the weights of conduction band and valance band (weighted by each effective mass).

6. For lightly n-type doped semiconductor, the chemical potential is near and under the conduction band (near the donor imupurity energy level in the gap), typically by an amount 5 meV.

7. For lightly p-type doped semiconductor, the chemical potential is near and above the conduction band (near the acceptor energy level).

Look at the picture below. The half-circular shapes (I do not see why they are drawn in this way though) represent bands of available electron energies. is represents the electron energy. White means non-occupied while black means totally occupied. For a metal, you can see clearly that electrons can move higher to states of energies within the band they occupy. So without jumping a gap. For semi-conductors, you can see that a part of the gaps with energies above the Fermi energy is filled (except for the p-type). For an insulator, all electrons occupy states with an energy below the energies that are "present" in the band above it. The electrons have to be excited to very high energies before they can jump the gap.
Note that the Fermi energy can lie between bands.

3. I don't see how $$\frac{h^2}{2m}(\frac{3n}{8\pi})^{2/3}$$ represents an energy. Just look at the units. O course, this formula should be somewhat changed: multiply the formula by inversed volume and taking the sqareroot of the number density gives you the right formula.
6. In the formula $$E_f$$ refers to the Fermi level. There is a temperature dependence and the Fermi energy certainly is not temperature-dependent.