--- title: "Compensated semiconductor concentrations and conductivity" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:compensated-conductivity" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/compensated-conductivity site: BCT Engineering --- # Compensated semiconductor concentrations and conductivity Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2073 Chaitra, 6 marks - Video: [Understanding Compensated Semiconductors: GATE Electronic Devices Numerical 1](https://www.youtube.com/watch?v=gkF8_ruL_Ng) — Ekeeda GATE & ESE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/compensated-conductivity - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question An n-type semiconductor doped with 10¹⁶cm⁻³ phosphorus atoms has been doped with 10¹⁷cm⁻³ boron atoms. Calculate the electron and hole concentrations and conductivity. [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 Compensated p-type, majority hole concentration p = N_a−N_d = 9×10^16 cm^−3, n = n_i²/p ≈ 2.3×10^3 cm^−3. Conductivity σ ≈ e p μ_h ≈ 4.75 S·cm^−1 (using μ_h ≈ 330 cm²/V·s). ## Answer Phosphorus is a donor and boron an acceptor. Since $N_a > N_d$, the sample is converted to **p-type** (compensated). **Given:** $N_d = 10^{16}$, $N_a = 10^{17}\ \text{cm}^{-3}$. **Data used:** $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$ (Si, 300 K), $e = 1.6\times10^{-19}$ C. #### Carrier concentrations $$ \begin{aligned} p &= N_a - N_d = 10^{17} - 10^{16} = 9\times10^{16}\ \text{cm}^{-3}\\ n &= \frac{n_i^2}{p} = \frac{(1.45\times10^{10})^2}{9\times10^{16}} = 2.34\times10^{3}\ \text{cm}^{-3} \end{aligned} $$ #### Mobility Mobility is limited by scattering from **all** ionised impurities: $N_a + N_d = 1.1\times10^{17}\ \text{cm}^{-3}$. From the graph (electron curve for $10^{17}$ at 300 K) $\mu_e \approx 800\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$. Holes have about 0.4 of that mobility in Si; the standard value for $\sim10^{17}\ \text{cm}^{-3}$ doping is $\mu_h \approx 330\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ (assumed, not on the graph). #### Conductivity The electron term is negligible ($n \ll p$): $$ \begin{aligned} \sigma &= e(n\mu_e + p\mu_h) \approx ep\mu_h\\ &= 1.6\times10^{-19}\times9\times10^{16}\times330\\ &= 4.75\ \text{S cm}^{-1}\\ \rho &= 1/\sigma = 0.210\ \Omega\,\text{cm} \end{aligned} $$ **Answer: $p = 9\times10^{16}\ \text{cm}^{-3}$, $n \approx 2.3\times10^{3}\ \text{cm}^{-3}$, $\sigma \approx 4.75$ S cm⁻¹ ($\rho \approx 0.21\ \Omega$ cm).** (If the lightly-doped value $\mu_h = 450$ is used, $\sigma = 6.48$ S cm⁻¹. The conductivity depends on the hole mobility chosen, but in either case the sample is p-type and conduction is by holes.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Silicon has Nd = 10^16 cm^-3 phosphorus and Na = 10^17 cm^-3 boron. What type is it? A. p-type, since Na > Nd B. n-type, since phosphorus is present C. Intrinsic, since both dopants are present D. n-type with n = Nd + Na Answer: A. Boron is an acceptor and Na exceeds Nd, so the sample is compensated and p-type. (Key point: Type after compensation) 2. What is the hole concentration in this compensated sample? A. 1.1 x 10^17 cm^-3 B. 9 x 10^16 cm^-3 C. 10^16 cm^-3 D. 10^17 cm^-3 Answer: B. p = Na - Nd = 10^17 - 10^16 = 9 x 10^16 cm^-3; donors cancel an equal number of acceptors. (Key point: Majority carrier concentration) 3. How is the electron concentration found, and about what is it? A. n = Nd = 10^16 cm^-3 B. n = p, about 9 x 10^16 cm^-3 C. n = ni^2/p, about 2.3 x 10^3 cm^-3 D. n = ni, 1.45 x 10^10 cm^-3 Answer: C. By the mass action law n = ni^2/p = (1.45e10)^2/(9e16), about 2.3 x 10^3 cm^-3. (Key point: Minority carrier concentration) 4. Why can the conductivity be written as sigma ~ e p mu_h? A. Because holes have higher mobility than electrons B. Because mobility is independent of doping C. Because electrons are trapped by boron D. Because n is far smaller than p, so the electron term is negligible Answer: D. n << p, so the electron term e n mu_e is negligible and holes carry the current. 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