--- title: "Fermi level at middle of band gap" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:fermi-level-middle-band-gap" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-level-middle-band-gap site: BCT Engineering --- # Fermi level at middle of band gap Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2071 Chaitra, 6 marks - Video: [Fermi level in semiconductors](https://www.youtube.com/watch?v=eSTu7G-0pgg) — Engineering Physics by Sanjiv - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-level-middle-band-gap - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question Prove that the position of Fermi level is near the middle of band gap in pure silicon semiconductor. ## Short answer For intrinsic Si, setting n=p gives EFi=(Ec+Ev)/2 + (kT/2) ln(Nv/Nc). With mh*/me*≈0.56/1.08, the correction is about −0.013 eV, so EFi is practically at mid-gap. ## Answer In an intrinsic (pure) semiconductor the number of electrons in the conduction band equals the number of holes in the valence band. Equating their expressions shows that the Fermi level lies at (or very near) the centre of the gap. #### Carrier concentrations Using the density of states and the Fermi–Dirac function (Boltzmann approximation): $$ n = N_c\exp\left[-\frac{E_c-E_F}{kT}\right],\qquad N_c = 2\left(\frac{2\pi m_e^*kT}{h^2}\right)^{3/2} $$ $$ p = N_v\exp\left[-\frac{E_F-E_v}{kT}\right],\qquad N_v = 2\left(\frac{2\pi m_h^*kT}{h^2}\right)^{3/2} $$ #### Equate n and p (intrinsic, $E_F = E_{Fi}$) $$ N_c\exp\left[-\frac{E_c-E_{Fi}}{kT}\right] = N_v\exp\left[-\frac{E_{Fi}-E_v}{kT}\right] $$ Take logarithms: $$ \begin{aligned} \frac{2E_{Fi}-E_c-E_v}{kT} &= \ln\frac{N_v}{N_c}\\ E_{Fi} &= \frac{E_c+E_v}{2} + \frac{kT}{2}\ln\frac{N_v}{N_c} \end{aligned} $$ Since $N_v/N_c = (m_h^*/m_e^*)^{3/2}$: $$ E_{Fi} = E_v + \frac{E_g}{2} + \frac{3}{4}kT\ln\frac{m_h^*}{m_e^*} $$ #### Why it is near the middle for Si - The first term $E_v + E_g/2$ is exactly mid-gap. - The second term is a small correction. For Si, $m_e^* = 1.08m_e$, $m_h^* = 0.56m_e$, $kT = 0.0259$ eV at 300 K: $$ \frac{3}{4}(0.0259)\ln\frac{0.56}{1.08} = -0.0127\ \text{eV} $$ This is only about 1% of $E_g = 1.1$ eV. So $E_{Fi}$ lies about 0.013 eV below the exact middle, i.e. **practically at mid-gap**. If $m_e^* = m_h^*$ (or at T = 0 K), it is exactly at the middle. ```text Ec ------------------------- 0.55 eV mid ......................... Eg/2 EFi - - - - - - (~0.013 eV below mid) 0.55 eV Ev ------------------------- ``` Physically, every electron excited to the CB leaves one hole in the VB; with similar densities of states on both sides, the level of 50% occupancy must sit midway between them. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition is equated to locate E_Fi? A. N_c = N_v B. n = p C. E_c = E_v D. n = N_d Answer: B. In pure semiconductor every electron in the CB leaves a hole in the VB, so n = p. (Key point: Intrinsic condition) 2. Which is the result for E_Fi? A. (E_c + E_v)/2 + (kT/2) ln(N_v/N_c) B. (E_c + E_v)/2 + kT ln(N_v N_c) C. E_c - E_g D. (E_c - E_v) + (kT/2) ln(N_c/N_v) Answer: A. Taking logs of N_c exp(-(E_c-E_F)/kT) = N_v exp(-(E_F-E_v)/kT) gives this form. (Key point: General expression) 3. What does the correction term (3/4) kT ln(m_h*/m_e*) depend on? A. The doping level B. The ratio of hole to electron effective masses C. The applied voltage D. The lattice constant Answer: B. N_v/N_c = (m_h*/m_e*)^(3/2), so the offset comes from unequal density of states. (Key point: Effective mass term) 4. For Si (m_e* = 1.08, m_h* = 0.56) the offset from mid-gap is about? A. 0.55 eV above B. 0.013 eV below, about 1% of Eg C. 0.5 eV below D. 0.013 eV above Answer: B. 0.75 x 0.0259 x ln(0.56/1.08) = -0.0127 eV, so E_Fi is practically at mid-gap. 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