PHT.301 Physics of Semiconductor Devices  

MOS Capacitor  Depletion approximationThe charge density, electric field, and electrostatic potential can be calculated for a MOS capacitor in the depletion approximation. This approximation is valid in depletion but not in accumulation or inversion. It is useful for estimating the threshold voltage and for estimating the concentration of charge carriers in the channel in the subthreshold regime. Consider a MOS capacitor with a psubstrate. We will ground the substrate and apply a voltage $V_g$ to the metal gate. Initially the gate voltage is set to the flatband voltage $V_{fb}$ so that the hole concentration is constant in the semiconductor all the way to the oxide. The flatband voltage is determined by the work functions of the materials as well as trapped charge in the oxide or at the semiconductoroxide interface. To bring the MOS capacitor into depletion, the voltage is made more positive than the flatband voltage to push the holes back exposing the negatively charged acceptors. The basic assumption of the depletion approximation is to assume that the charge density in the depletion region has the form, $$\rho(x) = \begin{cases} 0 & \mbox{for } t_{\text{ox}} \lt x\lt 0, \\ eN_A & \mbox{for } 0\lt x\lt x_p, \\ 0 & \mbox{for } x \gt x_p.\end{cases}$$The electric field is the integral of the charge density, $$E(x) = \begin{cases} \frac{eN_Ax_p}{\epsilon_{\text{ox}}} & \mbox{for } t_{\text{ox}} \lt x\lt 0, \\ \frac{eN_A}{\epsilon_s}(x_px) & \mbox{for } 0\lt x\lt x_p, \\ 0 & \mbox{for } x \gt x_p.\end{cases}$$The electric field is constant in the oxide and there is an abrupt jump in the field at the interface between the semiconductor and the oxide. The condition at this interface is that $\epsilon_{\text{ox}}E(0) = \epsilon_sE(0+)$ where $E(0)$ is the electric field next to the interface on the oxide side and and $\epsilon_{\text{ox}}$ is the dielectric constant in the oxide while $E(0+)$ is the electric field next to the interface on the semiconductor side and $\epsilon_s$ is the dielectric constant in the semiconductor. The electric field can be integrated to obtain the electrostatic potential, $$\phi(x) = \begin{cases} \frac{eN_Ax_px}{\epsilon_{\text{ox}}} + \frac{eN_A}{2\epsilon_s}x_p^2 & \mbox{for } t_{\text{ox}} \lt x\lt 0, \\ \frac{eN_A}{2\epsilon_s}(x_p  x)^2 & \mbox{for } 0\lt x\lt x_p, \\ 0 & \mbox{for } x \gt x_p.\end{cases}$$The voltage at the gate is the potential at $x = t_{\text{ox}}$, $$V_g  V_{fb} = \frac{eN_Ax_pt_{\text{ox}}}{\epsilon_{\text{ox}}} + \frac{eN_A}{2\epsilon_s}x_p^2.$$This can be solved for the depletion width, $$x_p = \frac{\epsilon_s}{\epsilon_{\text{ox}}}t_{\text{ox}} + \sqrt{\left( \frac{\epsilon_s}{\epsilon_{\text{ox}}}t_{\text{ox}} \right)^2 + \frac{2\epsilon_s}{eN_A}(V_g  V_{fb})}$$The depletion width goes to zero at $V_g = V_{fb}$. The electron concentration and the hole concentration are, $$n(x) = N_c\exp\left(\frac{E_F  E_c(x)}{k_BT}\right), \qquad p(x) = N_v\exp\left(\frac{E_v(x)  E_F}{k_BT}\right),$$where $$E_F  E_c(x) = \begin{cases} k_BT\ln\left(\frac{n_i^2}{N_cN_A}\right) + \frac{e^2N_A}{2\epsilon_s}(x_p  x)^2 & \mbox{for } 0 \lt x\lt x_p, \\ k_BT\ln\left(\frac{n_i^2}{N_cN_A}\right) & \mbox{for } x\gt x_p,\end{cases},$$and $$E_v(x)  E_F = \begin{cases} k_BT\ln\left(\frac{N_A}{N_v}\right)  \frac{e^2N_A}{2\epsilon_s}(x_p  x)^2 & \mbox{for } 0 \lt x\lt x_p, \\ k_BT\ln\left(\frac{N_A}{N_v}\right) & \mbox{for } x\gt x_p,\end{cases}$$therefore, $$n(x) = \begin{cases}\frac{n_i^2}{N_A}\exp\left(\frac{e^2N_A(x_p  x)^2}{2\epsilon_sk_BT}\right) & \mbox{for } 0 \lt x\lt x_p, \\ \frac{n_i^2}{N_A} & \mbox{for } x\gt x_p,\end{cases}$$and $$p(x) = \begin{cases}N_A\exp\left(\frac{e^2N_A(x_p  x)^2}{2\epsilon_sk_BT}\right) & \mbox{for } 0 \lt x\lt x_p, \\ N_A & \mbox{for } x\gt x_p.\end{cases}$$The form below will plot the bands, the charge density, and the electric field in the depletion approximation.
