1 Answers
The Mott–Schottky equation relates the capacitance to the applied voltage across a semiconductor-electrolyte junction.
1 C 2 = 2 ϵ ϵ 0 A 2 e N d {\displaystyle {\frac {1}{C^{2}}}={\frac {2}{\epsilon \epsilon _{0}A^{2}eN_{d}}}}
where C {\displaystyle C} is the differential capacitance ∂ Q ∂ V {\displaystyle {\frac {\partial {Q}}{\partial {V}}}} , ϵ {\displaystyle \epsilon } is the dielectric constant of the semiconductor, ϵ 0 {\displaystyle \epsilon _{0}} is the permittivity of free space, A {\displaystyle A} is the area such that the depletion region volume is w A {\displaystyle wA} , e {\displaystyle e} is the elementary charge, N d {\displaystyle N_{d}} is the density of dopants, V {\displaystyle V} is the applied potential, V f b {\displaystyle V_{fb}} is the flat band potential, k B {\displaystyle k_{B}} is the Boltzmann constant, and T is the absolute temperature.
This theory predicts that a Mott–Schottky plot will be linear. The doping density N d {\displaystyle N_{d}} can be derived from the slope of the plot. The flatband potential can be determined as well; absent the temperature term, the plot would cross the V {\displaystyle V} -axis at the flatband potential.