--- title: "Si with 10¹⁵ Sb: carriers, resistance, Fermi level" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:antimony-1e15-resistance-fermi" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/antimony-1e15-resistance-fermi site: BCT Engineering --- # Si with 10¹⁵ Sb: carriers, resistance, Fermi level Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2074 Asoj, 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/antimony-1e15-resistance-fermi - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question Sample of silicon wafer is doped with 10¹⁵ Antimony atoms/cm³. Find the carrier concentrations, its resistance and the shift in Fermi level from its intrinsic Fermi level at 27°C. If this sample is further doped with 10²² Boron atoms/cm³, what will be the change in its resistance. [Graph attached with the paper: log-log plot of electron drift mobility μₑ (cm² V⁻¹ s⁻¹, 50 to 10⁴) versus temperature T (100 to 1000 K) for n-type Si with curves for Nd = 10¹⁴, 10¹⁶, 10¹⁷, 10¹⁸, 10¹⁹ cm⁻³; inset shows ln(μₑ) vs ln(T) rising as T^(3/2) for impurity scattering and falling as T^(−3/2) for lattice scattering.] ## Short answer Initial n-type: n = 10^15 cm^-3, p = 2.1×10^5 cm^-3; for 1 cm cube, R = 1/(e N_d μ_e) = 4.63 Ω, Fermi level 0.288 eV above intrinsic. Adding 10^22 cm^-3 boron makes p-type, R ≈ 1.25×10^-5 Ω. ## Answer Antimony is a donor. The sample size is not given, so the resistance is found for a **1 cm × 1 cm × 1 cm cube** (R in Ω = ρ in Ω cm). **Data used:** $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT = 0.02585$ eV at 300 K, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$. From the attached graph, at 300 K and $N_d = 10^{15}$ (between the $10^{14}$ and $10^{16}$ curves) $\mu_e \approx 1350\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$. #### Carrier concentrations $$ n \approx N_d = 10^{15}\ \text{cm}^{-3},\qquad p = \frac{n_i^2}{n} = \frac{(1.45\times10^{10})^2}{10^{15}} = 2.1\times10^{5}\ \text{cm}^{-3} $$ #### Resistance $$ \begin{aligned} \sigma &= e(n\mu_e + p\mu_h) \approx eN_d\mu_e\\ &= 1.6\times10^{-19}\times10^{15}\times1350 = 0.216\ \text{S cm}^{-1}\\ \rho &= 4.63\ \Omega\,\text{cm} \;\Rightarrow\; R = 4.63\ \Omega\ (\text{1 cm cube}) \end{aligned} $$ #### Fermi level shift $$ \begin{aligned} E_{Fn}-E_{Fi} &= kT\ln\frac{N_d}{n_i} = 0.02585\ln\frac{10^{15}}{1.45\times10^{10}}\\ &= 0.288\ \text{eV above } E_{Fi} \end{aligned} $$ #### Further doped with $10^{22}$ boron cm⁻³ Now $N_a \gg N_d$, so the sample becomes strongly **p-type**: $$ p = N_a - N_d \approx 10^{22}\ \text{cm}^{-3},\qquad n = n_i^2/p \approx 0.02\ \text{cm}^{-3} $$ At such heavy doping impurity scattering dominates; the hole mobility falls to about its minimum value, taken as $\mu_h \approx 50\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (assumed; the graph gives only electron mobility). $$ \begin{aligned} \sigma &= ep\mu_h = 1.6\times10^{-19}\times10^{22}\times50 = 8\times10^{4}\ \text{S cm}^{-1}\\ \rho &= 1.25\times10^{-5}\ \Omega\,\text{cm} \;\Rightarrow\; R = 1.25\times10^{-5}\ \Omega \end{aligned} $$ **Answer: $n = 10^{15}$, $p = 2.1\times10^5\ \text{cm}^{-3}$; $R \approx 4.63\ \Omega$ (1 cm cube); $E_F$ is 0.288 eV above $E_{Fi}$. After boron doping, $R \approx 1.25\times10^{-5}\ \Omega$, so the resistance falls by a factor of about $3.7\times10^{5}$.** Note: $10^{22}$ cm⁻³ is 20% of the Si atom density; such a sample is degenerate and almost metallic, so this result is only an estimate. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For Si doped with 10^15 Sb at 300 K, what are n and p? A. n = 10^15, p = 2.1 x 10^5 cm^-3 B. n = 10^15, p = 10^15 C. n = 1.45 x 10^10, p = 10^15 D. n = 10^15, p = 1.45 x 10^10 Answer: A. n is about N_d and p = n_i^2/n = 2.1 x 10^5 cm^-3. (Key point: Majority and minority carriers) 2. What is the Fermi level shift for this sample? A. 0.288 eV below E_Fi B. 0.288 eV above E_Fi C. 0.348 eV above E_Fi D. 0.02585 eV above E_Fi Answer: B. E_Fn - E_Fi = kT ln(N_d/n_i) = 0.288 eV, above because it is n-type. (Key point: Fermi shift) 3. Where does the electron mobility used for the resistance come from? A. Hole mobility of the sample B. The intrinsic carrier concentration C. The graph of mu_e versus T for the given N_d at 300 K D. The built-in potential Answer: C. The graph gives mu_e of about 1350 cm^2/Vs for N_d = 10^15 at 300 K. (Key point: Mobility source) 4. After doping with 10^22 boron, what happens to the sample and its resistance? A. It stays n-type and resistance rises B. It becomes intrinsic C. It becomes p-type and resistance rises D. It becomes p-type and resistance falls by about 10^5 times Answer: D. N_a >> N_d gives p of about 10^22, so sigma is huge and R falls from 4.63 ohm to about 1.25 x 10^-5 ohm. 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