--- title: "Fermi level with 10¹⁰ Sb then 2×10¹⁵ B" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:antimony-1e10-boron-2e15" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/antimony-1e10-boron-2e15 site: BCT Engineering --- # Fermi level with 10¹⁰ Sb then 2×10¹⁵ B Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2069 Chaitra, 6 marks - Video: [Fermi Energy Level in Intrinsic and Extrinsic Semiconductors](https://www.youtube.com/watch?v=SxLX65vVMtg) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/antimony-1e10-boron-2e15 - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question An n-type silicon wafer is uniformly doped with 10¹⁰ antimony atoms per cm³. Where will be the Fermi level compared to its intrinsic Fermi level? Where will the Fermi level be shifted if the sample is further doped with 2×10¹⁵ boron atoms per cm³? ## Short answer With 10^10 cm⁻³ Sb (Nd < ni), solving n = (Nd/2)+√[(Nd/2)²+ni²] gives n≈2.03×10^10 cm⁻³, so EF − Ei = 0.0087 eV. After adding 2×10^15 cm⁻³ B, net p ≈ 2×10^15 cm⁻³ (p-type), and Ei − EF = 0.306 eV. ## Answer Take $T = 300$ K, $kT = 0.02586$ eV and $n_i = 1.45\times10^{10}$ cm⁻³ for Si. Fermi level positions are measured from the intrinsic Fermi level $E_{Fi}$. #### Part 1: Only $N_d = 10^{10}$ cm⁻³ antimony Here $N_d$ is **smaller than** $n_i$, so we cannot use $n \approx N_d$. Use charge neutrality $n = N_d + p$ with $np = n_i^2$: $$ \begin{aligned} n^2 - N_d n - n_i^2 &= 0 \\ n &= \frac{N_d}{2} + \sqrt{\left(\frac{N_d}{2}\right)^2 + n_i^2} \\ &= 0.5\times10^{10} + \sqrt{(0.5\times10^{10})^2 + (1.45\times10^{10})^2} \\ &= 2.03\times10^{10}\ \text{cm}^{-3} \end{aligned} $$ Since $n = n_i \exp[(E_F - E_{Fi})/kT]$: $$ E_F - E_{Fi} = kT\ln\frac{n}{n_i} = 0.02586\ln\frac{2.03\times10^{10}}{1.45\times10^{10}} = 0.0087\ \text{eV} $$ The Fermi level is only about **8.7 meV above** $E_{Fi}$; such light doping makes the wafer almost intrinsic. #### Part 2: Further doping with $N_a = 2\times10^{15}$ cm⁻³ boron Now $N_a > N_d$, so the sample becomes **p-type** (compensated). Net acceptor concentration: $$ N_a - N_d = 2\times10^{15} - 10^{10} \approx 2\times10^{15}\ \text{cm}^{-3} \gg n_i $$ So $p \approx 2\times10^{15}$ cm⁻³ and $n = n_i^2/p \approx 1.05\times10^{5}$ cm⁻³. $$ \begin{aligned} E_{Fi} - E_F &= kT\ln\frac{p}{n_i} = 0.02586 \ln\frac{2\times10^{15}}{1.45\times10^{10}} \\ &= 0.02586 \times 11.83 = 0.306\ \text{eV} \end{aligned} $$ ```text Ec --------------------------- EF (part 1) 0.0087 eV above Ei Ei - - - - - - - - - - - - - - | 0.306 eV EF (part 2) ----+------------- Ev --------------------------- ``` **Answer:** With $10^{10}$ cm⁻³ Sb, $E_F$ is 0.0087 eV (8.7 meV) above $E_{Fi}$. After adding $2\times10^{15}$ cm⁻³ B, the sample is p-type and $E_F$ shifts to 0.306 eV **below** $E_{Fi}$, a total shift of about 0.315 eV. Note: if the textbook version of this problem ($10^{16}$ Sb, then $2\times10^{17}$ B) is intended, the same method gives $E_F - E_{Fi} = 0.348$ eV, then $E_{Fi} - E_F = 0.424$ eV. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can n = Nd not be used for Nd = 10^10 cm^-3 Si? A. Antimony is not a donor B. Nd is too large C. Si is degenerate D. Nd is smaller than ni, so intrinsic carriers matter Answer: D. Nd < ni (1.45e10), so the full quadratic must be used. (Key point: Why n is not Nd) 2. Where is EF after doping with 10^10 Sb? A. About 8.7 meV above Efi B. About 0.35 eV above Efi C. At Efi exactly D. Below Efi Answer: A. n = 2.03e10 gives EF - Efi = kT ln(n/ni) = 0.0087 eV. (Key point: Fermi level, part 1) 3. What is the type after adding 2e15 boron? A. n-type B. p-type C. Intrinsic D. Degenerate n-type Answer: B. Na exceeds Nd, so the sample is compensated and p-type. (Key point: Type after boron) 4. Where is EF after boron doping? A. About 0.306 eV above Efi B. At mid-gap C. About 0.306 eV below Efi D. At Ev Answer: C. EFi - EF = kT ln(p/ni) = 0.02586 x ln(2e15/1.45e10) = 0.306 eV. (Key point: Fermi level, part 2) ## Close variants set in other exams - [Fermi level with 10¹⁶ Sb then 2×10¹⁷ B](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e16-boron-2e17) — Electrical Engineering Material, chapter 6, asked 6× - [Fermi level with 10¹⁷ Sb then further doping](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e17-fermi-shift) — Electrical Engineering Material, chapter 6, asked 6× - [Fermi level with 10¹⁶ B then 10¹⁷ Sb](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aboron-1e16-antimony-1e17) — Electrical Engineering Material, chapter 6, asked 5× - [Fermi level of Si with 10¹⁶ Sb](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e16-fermi) — Electrical Engineering Material, chapter 6, asked 5× ## Related in Electrical Engineering Material - [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 - [Fermi level shift with As then B](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aarsenic-1e15-fermi-shift) — Electrical Engineering Material, chapter 6, asked 2× - [Si with 10¹⁵ As: resistance and Fermi level](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aarsenic-1e15-resistance-fermi) — Electrical Engineering Material, chapter 6, asked 2× - [Si cube with As then 10¹⁴ boron](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-boron-1e14) — Electrical Engineering Material, chapter 6 - [Resistance of Si cube doped with Sb](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-antimony) — Electrical Engineering Material, chapter 6