--- title: "Si with 10¹⁵ As: resistance and Fermi level" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:arsenic-1e15-resistance-fermi" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/arsenic-1e15-resistance-fermi site: BCT Engineering --- # Si with 10¹⁵ As: resistance and Fermi level Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2081 Baisakh, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/arsenic-1e15-resistance-fermi - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question An n-type Si wafer has been doped uniformly with 10¹⁵ Arsenic atoms per cm³. Calculate its resistance and the position of the Fermi energy with respect to the intrinsic Fermi energy level EFi at 27°C. If this sample is further doped with 10²² Boron atoms per cm³, what will be change in its resistance? ## Short answer 10¹⁵ As/cm³: σ = 0.216 Ω⁻¹cm⁻¹, R = 4.62 Ω, EF – EFi = 0.288 eV. Adding 10²² B/cm³ makes it p-type with p ≈ 10²² cm⁻³, σ = 7.21×10⁵ Ω⁻¹cm⁻¹, R = 1.39×10⁻⁶ Ω. ## Answer **Assume:** Si at 300 K ($27^\circ$C), $kT = 0.02586$ eV, $n_i = 1.45\times10^{10}$ cm$^{-3}$, $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$. Sample size is not given, so take a 1 cm cube: $R$ ($\Omega$) $= \rho$ ($\Omega$ cm). #### Resistance with $10^{15}$ As cm$^{-3}$ $$ \begin{aligned} n &\approx N_d = 10^{15}\ \text{cm}^{-3}, \quad p = \frac{n_i^2}{n} = 2.1\times10^{5}\ \text{cm}^{-3}\ (\text{negligible}) \\ \sigma &= eN_d\mu_e = 1.602\times10^{-19}\times10^{15}\times1350 = 0.216\ \Omega^{-1}\text{cm}^{-1} \\ \rho &= 4.62\ \Omega\,\text{cm} \;\Rightarrow\; R = 4.62\ \Omega \end{aligned} $$ #### Fermi level $$ \begin{aligned} E_F - E_{Fi} &= kT\ln\frac{N_d}{n_i} = 0.02586\ln\frac{10^{15}}{1.45\times10^{10}} \\ &= 0.02586\times11.14 = 0.288\ \text{eV} \end{aligned} $$ $E_F$ is 0.288 eV above $E_{Fi}$. #### Further doped with $10^{22}$ B cm$^{-3}$ $N_a \gg N_d$, so the sample becomes strongly p-type: $p = N_a - N_d \approx 10^{22}$ cm$^{-3}$. $$ \begin{aligned} \sigma &= ep\mu_h = 1.602\times10^{-19}\times10^{22}\times450 = 7.21\times10^{5}\ \Omega^{-1}\text{cm}^{-1} \\ \rho &= 1.39\times10^{-6}\ \Omega\,\text{cm} \;\Rightarrow\; R = 1.39\times10^{-6}\ \Omega \end{aligned} $$ The resistance falls by a factor of $4.62/1.39\times10^{-6} \approx 3.3\times10^{6}$. Note: $10^{22}$ cm$^{-3}$ is 20% of all Si atoms, so the material would be heavily degenerate and the real hole mobility far below 450 cm$^2$/Vs; the figure above follows the simple model as asked. (If the intended value is $10^{17}$ cm$^{-3}$, then $p = 9.9\times10^{16}$ cm$^{-3}$ and $R = 0.140$ $\Omega$.) **Answer:** $R = 4.62$ $\Omega$, $E_F - E_{Fi} = 0.288$ eV; after boron doping, p-type with $R \approx 1.39\times10^{-6}$ $\Omega$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In n-type Si with 10^15 As per cm^3, what are n and p? A. n about 10^15 and p = ni^2/n about 2x10^5 cm^-3 B. n = p = 10^15 C. n about 2x10^5 and p about 10^15 D. n = 1.45x10^10 and p = 10^15 Answer: A. All donors ionise, n = Nd; the minority p follows from np = ni^2. (Key point: Majority carrier) 2. What is E_F - E_Fi for this sample? A. kT ln(Nd/ni) = 0.288 eV below E_Fi B. kT ln(Nd/ni) = 0.288 eV above E_Fi C. kT ln(ni/Nd) = 0.288 eV above E_Fi D. 0.55 eV above E_Fi Answer: B. n-type puts E_F above the intrinsic level by kT ln(n/ni). (Key point: Fermi level position) 3. How is the resistance of the n-type sample found (1 cm cube)? A. sigma = e Nd mu_h, so R = 13.9 ohm B. sigma = e ni (mu_e + mu_h), so R = 2.4x10^5 ohm C. sigma = e Nd mu_e = 0.216 S/cm, so R = 4.62 ohm D. R = 1/(Nd mu_e) = 4.6x10^-12 ohm Answer: C. Electrons dominate, so sigma = e n mu_e; R = 1/sigma = 4.62 ohm. (Key point: Resistance method) 4. After adding 10^22 B per cm^3 (much more than Nd), what is the type and carrier density? A. Still n-type with n = 10^15 B. Intrinsic with n = p C. n-type with n = 10^22 D. p-type with p = Na - Nd about 10^22 cm^-3 Answer: D. Na >> Nd, so the compensated sample is strongly p-type and p = Na - Nd. (Key point: After boron) ## Close variants set in other exams - [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× ## 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 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 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× - [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 10¹⁰ Sb then 2×10¹⁵ B](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e10-boron-2e15) — 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× ## Related in other subjects - [Sound energy flow per square cm](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asound-energy-flow-frequency-amplitude) — Engineering Physics, chapter 3