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The Mott–Schottky equation relates the capacitance to the applied voltage across a semiconductor-electrolyte junction.[1]
1 C 2 = 2 ϵ ϵ 0 A 2 e N d ( V − V f b − k B T e ) {\displaystyle {\frac {1}{C^{2}}}={\frac {2}{\epsilon \epsilon _{0}A^{2}eN_{d}}}(V-V_{fb}-{\frac {k_{B}T}{e}})}
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 (provided the area and dielectric constant are known). 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.
Under an applied potential V {\displaystyle V} , the width of the depletion region is[2]
w = ( 2 ϵ ϵ 0 e N d ( V − V f b ) ) 1 2 {\displaystyle w=({\frac {2\epsilon \epsilon _{0}}{eN_{d}}}(V-V_{fb}))^{\frac {1}{2}}}
Using the abrupt approximation,[2] all charge carriers except the ionized dopants have left the depletion region, so the charge density in the depletion region is e N d {\displaystyle eN_{d}} , and the total charge of the depletion region, compensated by opposite charge nearby in the electrolyte, is
Q = e N d A w = e N d A ( 2 ϵ ϵ 0 e N d ( V − V f b ) ) 1 2 {\displaystyle Q=eN_{d}Aw=eN_{d}A({\frac {2\epsilon \epsilon _{0}}{eN_{d}}}(V-V_{fb}))^{\frac {1}{2}}}
Thus, the differential capacitance is
C = ∂ Q ∂ V = e N d A 1 2 ( 2 ϵ ϵ 0 e N d ) 1 2 ( V − V f b ) − 1 2 = A ( e N d ϵ ϵ 0 2 ( V − V f b ) ) 1 2 {\displaystyle C={\frac {\partial {Q}}{\partial {V}}}=eN_{d}A{\frac {1}{2}}({\frac {2\epsilon \epsilon _{0}}{eN_{d}}})^{\frac {1}{2}}(V-V_{fb})^{-{\frac {1}{2}}}=A({\frac {eN_{d}\epsilon \epsilon _{0}}{2(V-V_{fb})}})^{\frac {1}{2}}}
which is equivalent to the Mott-Schottky equation, save for the temperature term. In fact the temperature term arises from a more careful analysis, which takes statistical mechanics into account by abandoning the abrupt approximation and solving the Poisson–Boltzmann equation for the charge density in the depletion region.[2]
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