--- title: "Resistance of Si cube doped with As and B" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:si-cube-arsenic-boron-billion" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-cube-arsenic-boron-billion site: BCT Engineering --- # Resistance of Si cube doped with As and B Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2082 Kartik (new course), 4 marks; 2075 Chaitra, 8 marks; 2074 Chaitra, 8 marks; 2073 Shrawan, 6 marks (asked 4 times) - Video: [Tutorial: Doping](https://www.youtube.com/watch?v=k12GMjtN8aA) — MIT OpenCourseWare - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-cube-arsenic-boron-billion - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question Calculate the resistance of pure silicon cubic crystal of 1cm³ at room temperature. What will be the resistance of the cubic when it is doped with 1 arsenic in 10⁹ silicon atoms and 1 boron atom per billion silicon atoms? Atomic concentration of silicon is 5×10²² cm⁻³, ni = 1.45×10¹⁰ cm⁻³. ## Short answer Pure Si: R = 2.39×10^5 Ω. With 1 As per 10^9 Si atoms and 1 B per 10^9 Si atoms, compensation yields net doping zero, so R returns to 2.39×10^5 Ω. ## Answer Resistance of a cube with side $L = 1$ cm: $R = \rho L/A = \rho \times \dfrac{1\ \text{cm}}{1\ \text{cm}^2}$, so $R$ in $\Omega$ equals $\rho$ in $\Omega$ cm. **Data:** $n_i = 1.45\times10^{10}$ cm$^{-3}$, $N_{Si} = 5\times10^{22}$ cm$^{-3}$, $e = 1.602\times10^{-19}$ C. Assumed room-temperature mobilities for Si: $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$ (at such low doping, impurity scattering is negligible). #### (a) Pure silicon $$ \begin{aligned} \sigma &= e n_i(\mu_e + \mu_h) \\ &= 1.602\times10^{-19} \times 1.45\times10^{10} \times 1800 \\ &= 4.18\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 2.39\times10^{5}\ \Omega \end{aligned} $$ #### (b) Doped with 1 As per $10^9$ Si atoms Arsenic is a donor: $$ \begin{aligned} N_d &= \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3} \gg n_i \\ n &\approx N_d = 5\times10^{13}\ \text{cm}^{-3} \\ p &= \frac{n_i^2}{n} = \frac{(1.45\times10^{10})^2}{5\times10^{13}} = 4.2\times10^{6}\ \text{cm}^{-3} \ (\text{negligible}) \\ \sigma &= e n\mu_e = 1.602\times10^{-19} \times 5\times10^{13} \times 1350 \\ &= 1.081\times10^{-2}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 92.5\ \Omega \end{aligned} $$ #### (c) Doped with 1 B per billion Si atoms Boron is an acceptor: $N_a = 5\times10^{13}$ cm$^{-3}$. **If boron is added together with the arsenic** (both present): $N_a = N_d$, so the acceptors capture all the electrons given by the donors (full **compensation**). The net doping $N_d - N_a = 0$, so the crystal behaves as intrinsic: $n = p = n_i$. $$ R \approx 2.39\times10^{5}\ \Omega $$ **If the crystal is doped with boron alone** (p-type): $p \approx N_a = 5\times10^{13}$ cm$^{-3}$. $$ \begin{aligned} \sigma &= e p\mu_h = 1.602\times10^{-19} \times 5\times10^{13} \times 450 \\ &= 3.60\times10^{-3}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{\sigma} = 277\ \Omega \end{aligned} $$ **Answer:** Pure Si: $R = 2.39\times10^{5}$ $\Omega$. With 1 As per $10^9$: $R = 92.5$ $\Omega$. With 1 As and 1 B per $10^9$: compensated, so $R$ returns to about $2.39\times10^{5}$ $\Omega$. (With 1 B per $10^9$ alone: $R \approx 277$ $\Omega$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 1 cm cube, how is resistance related to resistivity? A. R in ohms equals rho in ohm-cm B. R = rho x 1 cm^3 C. R = rho / 10 D. R = rho x 100 Answer: A. R = rho L/A = rho x (1 cm / 1 cm^2) = rho x 1 cm^-1, so numerically R equals rho. (Key point: Resistance of a cube) 2. Why does doping with 1 As per 10^9 Si atoms cut the resistance from about 2.39 x 10^5 to 92.5 ohm? A. Arsenic lowers the band gap of silicon B. Arsenic increases hole mobility C. Arsenic is an acceptor and adds holes D. Arsenic donors give Nd = 5 x 10^13 cm^-3 electrons, far above ni Answer: D. Nd = 5e22/1e9 = 5e13 >> ni = 1.45e10, so n is about Nd and conductivity rises sharply. (Key point: Donor doping effect) 3. In the As-doped sample, why can the hole contribution be ignored? A. p = ni^2/n is about 4 x 10^6 cm^-3, negligible compared with n B. Holes have zero mobility in silicon C. All holes are trapped by arsenic D. Holes recombine only at the cube surface Answer: A. By the mass action law p = ni^2/n = 4.2e6 cm^-3, vastly smaller than n = 5e13. (Key point: Minority holes negligible) 4. What happens when 1 boron per billion Si atoms is also added to the As-doped crystal? A. Resistance falls further because of extra carriers B. Boron acceptors cancel the donors (Na = Nd), so it behaves as intrinsic with R about 2.39 x 10^5 ohm C. The crystal becomes p-type with low resistance D. Resistance becomes zero Answer: B. Equal Na and Nd give zero net doping, so n = p = ni and the resistance returns to the intrinsic value. (Key point: Compensation) ## Close variants set in other exams - [Resistance of 8 cm³ Si cube](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-8cm3) — Electrical Engineering Material, chapter 6, asked 5× - [Si cube doped 1 As in 10⁶](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-1e6) — Electrical Engineering Material, chapter 6, asked 5× ## Related in Electrical Engineering Material - [Si cube with As and B per million](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-boron-million) — Electrical Engineering Material, chapter 6 - [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 - [Si crystal resistance with As then B doping](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-arsenic-boron-resistance) — Electrical Engineering Material, chapter 6 - [Resistance of arsenic-doped germanium wire](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Agermanium-arsenic-wire) — Electrical Engineering Material, chapter 6 - [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×