--- title: "Band bending and Einstein relationship" subject: "Electrical Engineering Material" chapter: 6 key: "semester_3/electrical_engineering_material:band-bending-einstein" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/band-bending-einstein site: BCT Engineering --- # Band bending and Einstein relationship Electrical Engineering Material · Chapter 6: Semiconductors - Asked in: 2075 Chaitra, 10 marks; 2072 Chaitra, 10 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/band-bending-einstein - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6 ## Question How band bending occurs in semiconductors? Derive Einstein relationship. ## Short answer Band bending results from internal electric fields due to nonuniform doping or junctions, while the Fermi level stays constant. Electrons diffuse, leaving ionised donors that create a field; bands shift by -eV(x). In equilibrium, zero net current gives D/μ = kT/e, the Einstein relation linking drift and diffusion. ## Answer #### Band bending **Band bending** is the change of the energy band edges ($E_c$, $E_v$) with position inside a semiconductor. It happens whenever there is an electric field or a space charge, for example in a non-uniformly doped sample, at a pn junction, or at a metal-semiconductor contact or surface. How it occurs: 1. In thermal equilibrium the **Fermi level is constant** throughout the material. 2. If doping varies with position (say $N_d$ falls with $x$), the gap $E_c - E_F$ must be smaller where doping is high and larger where it is low, because $n = N_c\exp[-(E_c - E_F)/kT]$. 3. Electrons diffuse from high to low concentration, leaving positive donor ions. This space charge sets up an internal field $E$ and a potential $V(x)$. 4. The electron potential energy is $-eV(x)$, so all bands shift by $-eV(x)$: they **bend**. The slope gives the field: $$ E = \frac{1}{e}\frac{dE_c}{dx} $$ ```text high Nd low Nd _________ Ec ________/ Ec ____/ EF ------------------------ (flat) _________ Ev ________/ Ev ____/ ---> e- diffuse <--- e- drift (built-in E) ``` Electrons roll "downhill" in energy; holes float "uphill". At a pn junction the total bending equals $eV_0$. #### Einstein relation In the sample above, in equilibrium, the electron current is zero: $$ J_e = en\mu_eE + eD_e\frac{dn}{dx} = 0 \quad (1) $$ With constant $E_F$: $$ \frac{dn}{dx} = -\frac{n}{kT}\frac{dE_c}{dx} = -\frac{neE}{kT} \quad (2) $$ Substituting (2) in (1): $$ \begin{aligned} en\mu_eE - eD_e\frac{neE}{kT} &= 0 \\ \frac{D_e}{\mu_e} &= \frac{kT}{e} \end{aligned} $$ Similarly for holes, $D_h/\mu_h = kT/e$. This is the **Einstein relation**. At 300 K, $kT/e = 25.9$ mV, so for Si electrons ($\mu_e = 1350$ cm$^2$/Vs), $D_e \approx 35$ cm$^2$/s. It shows that drift and diffusion are linked: both depend on the same random thermal motion and scattering of carriers. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do the bands bend in a sample with non-uniform doping in equilibrium? A. The Fermi level tilts with position B. Electrons diffuse leaving ionised donors, creating an internal field and potential V(x) that shifts the bands while EF stays constant C. The band gap changes with position D. Holes are absent Answer: B. EF must be constant in equilibrium; the space charge field produces a potential energy -eV(x) that bends Ec and Ev. (Key point: Cause of band bending) 2. How is the electric field related to the band bending? A. E = e dEc/dx B. E = (1/e) dEc/dx C. E = Eg / e D. E = EF / x Answer: B. The slope of the band edge gives the field: E = (1/e) dEc/dx. (Key point: Field from bending) 3. What condition is used to derive the Einstein relation? A. Electron drift current plus diffusion current is zero in equilibrium B. Drift current equals recombination current C. Hole and electron currents add D. Diffusion current is zero Answer: A. J = e n mu E + e D dn/dx = 0, with dn/dx = -(n e E)/(kT) from constant EF. (Key point: Zero-current condition) 4. What does the Einstein relation state? A. D x mu = kT/e B. D/mu = e/kT C. D/mu = kT/e (about 25.9 mV at 300 K) D. D/mu = kT Answer: C. Substituting gives D_e/mu_e = kT/e, linking diffusion and drift through the same thermal motion and scattering. 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