---
title: "Intrinsic Fermi level of pure semiconductor"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:intrinsic-fermi-level"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/intrinsic-fermi-level
site: BCT Engineering
---

# Intrinsic Fermi level of pure semiconductor

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2068 Chaitra, 2+4 marks
- Video: [Fermi Energy Level in Intrinsic Semiconductor | Electronics Devices and Circuits - EDC](https://www.youtube.com/watch?v=xvVnBN1HMMU) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/intrinsic-fermi-level
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Explain about intrinsic Fermi level of a pure semiconductor and derive a relationship of the intrinsic Fermi level assuming that intrinsic carrier concentration is known.

## Short answer

Intrinsic Fermi level E_Fi lies near mid-gap where n=p=n_i. From n_i=N_c exp[−(E_c−E_Fi)/kT] we get E_Fi = E_c − kT ln(N_c/n_i); also E_Fi = E_v + (E_g/2) + (3/4)kT ln(m_h*/m_e*). It is the reference for doping type and carrier concentration.

## Answer

#### Intrinsic Fermi level

The **intrinsic Fermi level** $E_{Fi}$ is the Fermi level of a pure (undoped) semiconductor, where $n = p = n_i$. It is the energy at which the probability of occupation is 1/2. Because each electron excited into the conduction band leaves one hole in the valence band, $E_{Fi}$ lies almost exactly at the **middle of the band gap**, shifted slightly towards the band with the smaller density of states. It serves as the reference level from which the Fermi level of doped material is measured ($E_F > E_{Fi}$ for n-type, $E_F < E_{Fi}$ for p-type).

#### Derivation (given $n_i$)

In any non-degenerate semiconductor

$$
n = N_c\exp\left[-\frac{E_c-E_F}{kT}\right],\qquad p = N_v\exp\left[-\frac{E_F-E_v}{kT}\right]
$$

For intrinsic material $n = n_i$ and $E_F = E_{Fi}$:

$$
n_i = N_c\exp\left[-\frac{E_c-E_{Fi}}{kT}\right] \;\Rightarrow\; E_{Fi} = E_c - kT\ln\frac{N_c}{n_i}
$$

Similarly from $p = n_i$:

$$
E_{Fi} = E_v + kT\ln\frac{N_v}{n_i}
$$

Adding the two forms and dividing by 2:

$$
E_{Fi} = \frac{E_c+E_v}{2} + \frac{kT}{2}\ln\frac{N_v}{N_c} = E_v + \frac{E_g}{2} + \frac{3}{4}kT\ln\frac{m_h^*}{m_e^*}
$$

#### Doped material in terms of $n_i$

Dividing $n = N_ce^{-(E_c-E_F)/kT}$ by $n_i = N_ce^{-(E_c-E_{Fi})/kT}$:

$$
n = n_i\exp\left(\frac{E_F-E_{Fi}}{kT}\right),\qquad p = n_i\exp\left(\frac{E_{Fi}-E_F}{kT}\right)
$$

**Example (Si, 300 K):** $N_c = 2.8\times10^{19}$, $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$: $E_c - E_{Fi} = 0.02585\ln(1.93\times10^9) = 0.553$ eV, i.e. about half of $E_g = 1.1$ eV.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the intrinsic Fermi level?
   A. Level of the conduction band edge
   B. The Fermi level of undoped semiconductor where n = p = n_i
   C. The donor level
   D. The level of the top of the valence band
   Answer: B. E_Fi is the 50% occupancy energy of pure semiconductor, close to mid-gap. (Key point: Definition)

2. Given n_i, which expression follows from n_i = N_c exp(-(E_c-E_Fi)/kT)?
   A. E_Fi = E_c + kT ln(N_c/n_i)
   B. E_Fi = E_c - kT ln(n_i/N_c)
   C. E_Fi = E_c kT ln(N_c/n_i)
   D. E_Fi = E_c - kT ln(N_c/n_i)
   Answer: D. Taking logs gives E_c - E_Fi = kT ln(N_c/n_i). (Key point: Derivation from n_i)

3. Toward which band is E_Fi displaced from mid-gap?
   A. The band with the smaller density of states
   B. The band with the larger density of states
   C. Always the valence band
   D. It is never displaced
   Answer: A. E_Fi = mid-gap + (kT/2) ln(N_v/N_c), so it shifts toward the band with smaller density of states. (Key point: Shift from mid-gap)

4. How is n written for a doped sample using n_i and E_Fi?
   A. n = n_i exp((E_F - E_Fi)/kT)
   B. n = n_i exp((E_Fi - E_F)/kT)
   C. n = n_i^2 exp((E_F-E_Fi)/kT)
   D. n = N_c exp((E_F-E_Fi)/kT)
   Answer: A. Dividing n by n_i cancels N_c, giving n = n_i exp((E_F - E_Fi)/kT). (Key point: Doped material in terms of n_i)

## Related in Electrical Engineering Material

- [Intrinsic concentration and its temperature dependence](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aintrinsic-concentration-derivation) — Electrical Engineering Material, chapter 6
- [Temperature effect on carrier concentration](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Atemperature-carrier-concentration) — Electrical Engineering Material, chapter 6, asked 2×
- [n-type carrier concentration versus temperature](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Antype-concentration-temperature) — Electrical Engineering Material, chapter 6, asked 2×
- [Doping to raise Fermi level by 0.7 eV](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Araise-fermi-0-7ev) — Electrical Engineering Material, chapter 6, asked 2×
- [Fermi level at middle of band gap](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Afermi-level-middle-band-gap) — Electrical Engineering Material, chapter 6
- [Si with 10¹⁵ Sb: carriers, resistance, Fermi level](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e15-resistance-fermi) — Electrical Engineering Material, chapter 6

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