--- title: "pn junction from 5 Ω cm resistivity" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:pn-resistivity-5-ohm-cm" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-resistivity-5-ohm-cm site: BCT Engineering --- # pn junction from 5 Ω cm resistivity Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2075 Chaitra, 6 marks; 2071 Shrawan, 8 marks (asked 2 times) - Video: [Conductivity of Semiconductors Numerical (Part 1)](https://www.youtube.com/watch?v=WbDvUitRKi8) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-resistivity-5-ohm-cm - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question A pn junction semiconductor has resistivity of 5Ω cm. If mobility of holes is 450 cm²/Vs, and electron mobility is three times the mobility of holes at room temperature, find i) Built in potential ii) Depletion width that lies in n-region and p-region respectively iii) Built in electric field at x=0. ## Short answer Find doping from resistivity: Na = 1/(5×1.6e-19×450) = 2.77×10^15 cm^-3, Nd = 1/(5×1.6e-19×1350) = 9.25×10^14 cm^-3. V0 = 0.02586 ln(NaNd/ni²) = 0.601 V. Total W = 1.07 μm (Wn = 0.80 μm, Wp = 0.27 μm). Maximum field |E0| = 1.13×10^4 V/cm. ## Answer **Assumptions:** silicon at 300 K; both the p-side and the n-side have resistivity 5 $\Omega$ cm; $n_i = 1.45\times10^{10}$ cm$^{-3}$, $\varepsilon_r = 11.9$, $kT/e = 0.02586$ V; abrupt junction, full ionisation. **Given:** $\mu_h = 450$ cm$^2$/Vs, $\mu_e = 3\mu_h = 1350$ cm$^2$/Vs. #### Doping levels from resistivity Majority carriers dominate, so $\rho = 1/(eN\mu)$: $$ \begin{aligned} N_a &= \frac{1}{\rho e\mu_h} = \frac{1}{5 \times 1.602\times10^{-19} \times 450} = 2.77\times10^{15}\ \text{cm}^{-3} \\ N_d &= \frac{1}{\rho e\mu_e} = \frac{1}{5 \times 1.602\times10^{-19} \times 1350} = 9.25\times10^{14}\ \text{cm}^{-3} \end{aligned} $$ #### (i) Built-in potential $$ \begin{aligned} V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2} = 0.02586 \ln\frac{2.77\times10^{15} \times 9.25\times10^{14}}{(1.45\times10^{10})^2} \\ &= 0.02586 \times 23.22 = 0.601\ \text{V} \end{aligned} $$ #### (ii) Depletion widths With $\varepsilon = 11.9 \times 8.854\times10^{-14} = 1.054\times10^{-12}$ F/cm: $$ \begin{aligned} W &= \sqrt{\frac{2\varepsilon(N_a+N_d)V_0}{eN_aN_d}} \\ &= \sqrt{\frac{2 \times 1.054\times10^{-12} \times 3.70\times10^{15} \times 0.601}{1.602\times10^{-19} \times 2.77\times10^{15} \times 9.25\times10^{14}}} \\ &= 1.067\times10^{-4}\ \text{cm} = 1.067\ \mu\text{m} \\ W_n &= \frac{N_a}{N_a+N_d}W = 0.75 \times 1.067 = 0.800\ \mu\text{m} \\ W_p &= \frac{N_d}{N_a+N_d}W = 0.25 \times 1.067 = 0.267\ \mu\text{m} \end{aligned} $$ #### (iii) Built-in field at $x = 0$ $$ \begin{aligned} |E_0| &= \frac{eN_dW_n}{\varepsilon} = \frac{1.602\times10^{-19} \times 9.25\times10^{14} \times 8.00\times10^{-5}}{1.054\times10^{-12}} \\ &= 1.13\times10^{4}\ \text{V/cm} \end{aligned} $$ It is directed from n to p. Check: $\frac{1}{2}|E_0|W = 0.5 \times 1.126\times10^4 \times 1.067\times10^{-4} = 0.601$ V. **Answer:** $V_0 = 0.601$ V; $W_n = 0.800$ $\mu$m, $W_p = 0.267$ $\mu$m ($W = 1.07$ $\mu$m); $|E_0| = 1.13\times10^4$ V/cm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are Na and Nd found from rho = 5 ohm-cm? A. N = rho e mu B. N = 1/(rho e mu) using the majority mobility C. N = rho / (e mu) D. N = e mu / rho Answer: B. rho = 1/(e N mu) for majority carriers, so Na uses mu_h = 450 and Nd uses mu_e = 1350. (Key point: Doping from resistivity) 2. Which is true for Na and Nd here? A. Nd = 2.77 x 10^15 is larger than Na B. Na = Nd C. Na = 2.77 x 10^15 cm^-3 is larger than Nd = 9.25 x 10^14 cm^-3 because holes are less mobile D. Both equal 10^16 Answer: C. Same resistivity with lower hole mobility needs 3 times more acceptors than donors. (Key point: Which side is more doped) 3. Which side holds more of the depletion width? A. The p-side, because Na is larger B. The n-side (0.800 um), because it is lightly doped C. Both equal D. Neither Answer: B. Wn = Na W/(Na+Nd) = 0.75 W, larger than Wp = 0.25 W. (Key point: Depletion split) 4. How is the built-in field at x = 0 computed? A. |E0| = e Nd Wn / epsilon B. |E0| = V0 / Wn C. |E0| = epsilon Nd / (e Wn) D. |E0| = e Na Wn / epsilon Answer: A. Gauss's law across the n-side gives |E0| = e Nd Wn/epsilon, about 1.13 x 10^4 V/cm. 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