A varactor - variable reactor - is a diode with a pn-junction in reverse bias, where the capacitance can be varied by changing the applied voltage. The capacitance changes because the depletion width changes with voltage. Varactors are typically used as voltage controlled capacitors and have largely replaced rotary capacitors. They are widely used in amplifiers, for frequency selection in radio receivers, for frequency modulation, in frequency multipliers, in filters and for oscillator circuits.
For an abrupt pn junction the relationship between the voltage and the depletion width is, $W \propto \sqrt{V_{bi}-V)}$, while for a linearly graded pn junction, $W \propto (V_{bi}-V))^{1/3}$. More generally it can be shown that if the charge density in the depletion region has the form $\rho = e\alpha \,\mathrm{sgn}(x)\mathrm{abs}(x)^m$, then the depletion width $W \propto (V_{bi}-V)^{\frac{1}{m+2}}$. For an abrupt pn-junction, $m=0$ and for a linearly graded junction $m=1$. For $m= -1$ the depletion width is linear in $(V_{bi}-V)$ while for $m=-3/2$, $W \propto (V_{bi}-V)^{2}$.
Consider a doping profile such that the charge density has the form,
\begin{equation*} N_D(x) - N_A(x) = \alpha \mathcal{H}\Big(x+\frac{W}{2}\Big) \mathcal{H}\Big(\frac{W}{2} - x\Big) \mathrm{sgn}(x) \mathrm{abs}(x)^{m} = \frac{\rho(x)}{e} \end{equation*}with $\alpha$ being the generalized doping distribution and $\mathcal{H}$ the Heaviside function. Inside the depletion width, $-\frac{W}{2} < x < \frac{W}{2}$, the charge density $\rho$ can be simplified to
\begin{equation*} \rho = e \alpha \cdot \mathrm{sgn}(x) \cdot \mathrm{abs}(x)^{m} \end{equation*}In the following, the $\mathrm{sgn}(x)$ is omitted for simplicity and the integrals are done for $0 < x < W/2$ and are symmetric. One has to keep in mind that the space charge is negative on the p-side.
By integrating the charge density, the electric field $E$ can be determined.
\begin{equation*} E(x) = \int \dfrac{e \alpha x^m}{\epsilon} dx = \dfrac{e \alpha x^{m+1}}{\epsilon (m+1)} + C \end{equation*}Solving for the integration constant $C$ with the condition $E(x=W/2) = 0$ gives
\begin{equation*} E(x=W/2) = \dfrac{e \alpha (W/2)^{m+1}}{\epsilon (m+1)} + C = 0 \end{equation*} \begin{equation*} C = - \dfrac{e \alpha (W/2)^{m+1}}{\epsilon (m+1)} \end{equation*} \begin{equation*} E(x) = \dfrac{e \alpha x^{m+1}}{\epsilon (m+1)} - \dfrac{e \alpha (W/2)^{m+1}}{\epsilon (m+1)} \end{equation*}Integrating over the electric field gives the electrostatic potential as follows
\begin{equation*} \phi(x) = - \int E(x) dx = - \int \dfrac{e \alpha}{\epsilon (m+1)} \Big[ x^{m+1} - (W/2)^{m+1} \Big] dx \end{equation*} \begin{equation*} \phi(x) = \dfrac{e \alpha x}{\epsilon} \dfrac{\Big( (m+2) (W/2)^{m+1} - x^{m+1} \Big)}{(m+1)(m+2)} + C \end{equation*}With the conditions $\phi(x=0) = 0$ and $\phi(x=W/2) = (V_{bi} - V)/2$ the depletion width $W$ as a function of the built-in voltage $V_{bi}$ and the applied bias voltage $V$ can be determined as follows
\begin{equation*} \phi(x=W/2) - \phi(x=0) = \dfrac{e\alpha (W/2)}{\epsilon} \dfrac{\Big( (m+2) (W/2)^{m+1} - (W/2)^{m+1} \Big)}{(m+1)(m+2)} \end{equation*} \begin{equation*} \dfrac{V_{bi} - V}{2} = \dfrac{e\alpha}{\epsilon} \dfrac{(W/2)^{m+2}}{(m+2)} \end{equation*} \begin{equation*} W = 2\Bigg( \dfrac{\epsilon (m+2) (V_{bi} - V)}{2 e \alpha} \Bigg)^{\tfrac{1}{m+2}} \propto \Big(V_{bi} - V \Big)^{n} \end{equation*}with the voltage sensitivity $n = \dfrac{1}{m+2}$.
The capacitance per area of the junction $C_j$ is determined with the relation
\begin{equation*} C_j = \dfrac{\epsilon}{W} \end{equation*} \begin{equation*} C_j = \dfrac{1}{2}\Bigg( \dfrac{2 e \alpha \epsilon^{m+1}}{(m+2) (V_{bi} - V)} \Bigg)^{n} \propto \Big(V_{bi} - V\Big) ^{-n} \end{equation*}