--- title: "Built-in potential of a pn junction" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:pn-builtin-potential-derivation" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-builtin-potential-derivation site: BCT Engineering --- # Built-in potential of a pn junction Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2081 Chaitra (new course), 5 marks; 2069 Chaitra, 6 marks (asked 2 times) - 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-builtin-potential-derivation - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question Derive the relation for finding built in potential of PN junction taking necessary assumptions. ## Short answer The built-in potential V0 is derived by equating Fermi levels across the junction. For an abrupt, fully ionized step junction under equilibrium, V0 = (kT/e) ln(Na Nd / ni²). Key assumptions: non-degenerate doping, constant Fermi level, depletion region free of mobile carriers. ## Answer The **built-in potential** $V_0$ is the potential difference across the depletion region of a pn junction in equilibrium. It balances the diffusion of carriers. #### Assumptions 1. Abrupt (step) junction: $N_a$ on the p-side, $N_d$ on the n-side, uniform. 2. All dopants ionised: $p_{p0} = N_a$, $n_{n0} = N_d$. 3. Non-degenerate doping, so Boltzmann statistics and $np = n_i^2$ hold. 4. Thermal equilibrium, no applied voltage: $E_F$ is constant through the device. 5. Depletion region free of mobile carriers. #### Derivation (using the Fermi level) Before contact, measured from the intrinsic level: $$ E_{Fn} - E_{Fi} = kT\ln\frac{N_d}{n_i}, \qquad E_{Fi} - E_{Fp} = kT\ln\frac{N_a}{n_i} $$ ```text before contact after contact p n p |W| n --Ec --Ec --Ec__ ..EFn \___ Ec - Ei - - Ei - EF ............... ..EFp Ev__ eV0 --Ev --Ev \___ Ev ``` On contact, electrons flow from n to p (and holes from p to n) until the Fermi levels line up. The bands on the n-side drop by $eV_0$ relative to the p-side, where $$ \begin{aligned} eV_0 &= E_{Fn} - E_{Fp} = kT\ln\frac{N_d}{n_i} + kT\ln\frac{N_a}{n_i} \\ V_0 &= \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2} \end{aligned} $$ #### Check using the Boltzmann relation Hole concentrations on the two sides differ by the barrier $eV_0$: $$ \frac{p_{p0}}{p_{n0}} = \exp\left(\frac{eV_0}{kT}\right), \quad p_{n0} = \frac{n_i^2}{N_d} $$ $$ \frac{N_aN_d}{n_i^2} = \exp\left(\frac{eV_0}{kT}\right) \;\Rightarrow\; V_0 = \frac{kT}{e}\ln\frac{N_aN_d}{n_i^2} $$ Example: Si, $N_a = 10^{17}$, $N_d = 10^{16}$ cm$^{-3}$, $n_i = 1.45\times10^{10}$ cm$^{-3}$: $V_0 = 0.02586\ln(4.76\times10^{12}) = 0.755$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens on contact that creates the built-in potential in the derivation? A. Carriers diffuse until the Fermi levels on both sides line up, bending the bands by eV0 B. The Fermi level of the n-side rises above that of the p-side C. An external voltage shifts the n-side bands up D. Donors and acceptors move across the junction Answer: A. Diffusion continues until E_F is constant throughout, and the band offset left is eV0. (Key point: Fermi levels align) 2. Which expression for the built-in potential results? A. V0 = (kT/e) ln(Na/Nd) B. V0 = (kT/e) ln(NaNd/ni^2) C. V0 = (kT/e) ln(ni^2/(NaNd)) D. V0 = kT ln(NaNd) / ni Answer: B. eV0 = E_Fn - E_Fp = kT ln(Nd/ni) + kT ln(Na/ni), which combines to ln(NaNd/ni^2). (Key point: Result formula) 3. Which is one of the assumptions used in the derivation? A. A large forward bias is applied B. Doping is so heavy that the Fermi level lies inside the bands C. Thermal equilibrium with np = ni^2 and non-degenerate (Boltzmann) doping D. The depletion region is full of mobile carriers Answer: C. The result needs equilibrium, Boltzmann statistics and a depletion region free of mobile carriers. (Key point: Assumptions) 4. In the Boltzmann check, p_p0/p_n0 = exp(eV0/kT) with p_n0 = ni^2/Nd gives what? A. NaNd/ni^2 = exp(eV0/kT), the same V0 as before B. Na/Nd = exp(eV0/kT) C. NaNd = exp(eV0/kT) with no ni D. A V0 that depends on applied bias Answer: A. Substituting p_p0 = Na and p_n0 = ni^2/Nd gives NaNd/ni^2, so the same expression for V0. 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