--- title: "Minority suppression and donor-defined conduction" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:minority-suppression-impurity-donor" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/minority-suppression-impurity-donor site: BCT Engineering --- # Minority suppression and donor-defined conduction Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2071 Shrawan, 8 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/minority-suppression-impurity-donor - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question What is minority carrier suppression? Prove electron concentration and conduction in n-type semiconductor is defined by impurity donor. ## Short answer Minority carrier suppression is the reduction of minority carriers below n_i due to doping, from np = n_i². In n-type, charge neutrality n = p+N_d and N_d >> n_i yields n ≈ N_d, p = n_i²/N_d. Conductivity σ ≈ e N_d μ_e, controlled by donors. ## Answer **Minority carrier suppression** is the reduction of the minority carrier concentration below the intrinsic value $n_i$ when a semiconductor is doped. In n-type material the holes are suppressed; in p-type the electrons are suppressed. It follows from the mass action law $np = n_i^2$: if doping raises one carrier type, the other must fall in the same ratio. #### Proof for n-type semiconductor Consider Si doped with $N_d$ donors per cm³ (P, As, Sb). Each donor level lies about 0.05 eV below $E_c$, so at room temperature practically all donors are ionised: $N_d^+ = N_d$. ```text Ec ------------------------- Ed - * - * - * - (donors, ~0.05 eV) EFn ........................ EFi - - - - - - - - - - - - - Ev ------------------------- ``` **1. Charge neutrality:** $$ n = p + N_d $$ **2. Mass action law** (thermal equilibrium): $$ np = n_i^2,\quad n_i^2 = N_cN_ve^{-E_g/kT} $$ **3. Solve:** substitute $p = n_i^2/n$: $$ n^2 - N_dn - n_i^2 = 0 \;\Rightarrow\; n = \frac{1}{2}\left[N_d + \sqrt{N_d^2+4n_i^2}\right] $$ **4. Normal doping, $N_d \gg n_i$:** $$ n \approx N_d,\qquad p = \frac{n_i^2}{N_d} \ll n_i $$ So the electron concentration is fixed by the donor concentration only, not by temperature or band gap, and the holes are suppressed by the factor $n_i/N_d$. #### Conduction is defined by the donors $$ \sigma = e(n\mu_e + p\mu_h) = e\left(N_d\mu_e + \frac{n_i^2}{N_d}\mu_h\right) $$ The ratio of the hole term to the electron term is $\dfrac{n_i^2\mu_h}{N_d^2\mu_e}$, which is tiny. Hence $$ \sigma \approx eN_d\mu_e $$ **Example:** Si with $N_d = 10^{16}$: $n = 10^{16}$, $p = 2.1\times10^{4}$ cm⁻³; hole/electron current ratio $\approx 7\times10^{-13}$. $\sigma = 1.6\times10^{-19}\times10^{16}\times1350 = 2.16$ S cm⁻¹. #### Conclusions - Electron concentration $\approx N_d$ (donor defined). - Conductivity $\approx eN_d\mu_e$ (donor defined), and it can be set accurately by controlling the doping. - This holds over the **extrinsic range** of temperature; at very high T, $n_i$ becomes comparable to $N_d$ and the material turns intrinsic. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is minority carrier suppression? A. Minority carriers rise above ni on doping B. Minority carriers vanish at all temperatures C. Doping with equal donors and acceptors D. The minority carrier density drops below ni when doping raises the majority density Answer: D. By np = ni^2, raising n lowers p to ni^2/n, below ni. (Key point: Minority carrier suppression) 2. Which pair of conditions is used to find n in n-type material? A. Charge neutrality n = p + Nd and np = ni^2 B. n = p and np = Nd C. n + p = Nd and n = ni D. p = Nd and n = ni^2 Answer: A. Substituting p = ni^2/n into n = p + Nd gives a quadratic for n. (Key point: Governing equations) 3. When Nd >> ni, what are n and p? A. n = ni, p = Nd B. n ~ Nd, p = ni^2/Nd C. n = Nd/2, p = Nd/2 D. n ~ Nd, p ~ Nd Answer: B. The quadratic root reduces to n ~ Nd and p = ni^2/Nd << ni. (Key point: Result for Nd >> ni) 4. Why is sigma ~ e Nd mu_e in n-type material? A. Holes have zero mobility B. Donors carry the current C. The hole term e (ni^2/Nd) mu_h is negligible D. Mobility cancels Answer: C. The ratio of hole to electron terms is tiny, so only electrons matter. 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