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\section{MOS structure}
\begin{figure}[h]
\centering
\caption{Gate structure}
\begin{tikzpicture}
\draw (0,0) rectangle ++(1,2)
rectangle ++(2,-2)
rectangle ++(2,2)
rectangle ++(3,-2)
;
\node at (0.5,1) () {$M$};
\node at (2,1) () {$SiO_2$};
\node at (4,1) () {$SCR$};
\node at (6,1) () {$p-Si$};
\end{tikzpicture}
\end{figure}
Where $M$ is the metal, $SiO_2$ is the gate oxide, $SCR$ is the space charge region and $p-Si$ is the p doped substrate.
(note that in for the SCR, there is short between the metal and the bulk.)
\subsection{Electrostatic analysis}
\begin{figure}[h]
\centering
\caption{Charge distribution at thermal equilibrium}
\begin{tikzpicture}
\draw (0,0) rectangle ++(1,2)
rectangle ++(2,-2)
rectangle ++(2,2)
rectangle ++(3,-2)
;
\node at (0.5,1) () {$M$};
\node at (2,1) () {$SiO_2$};
\node at (4,1) () {$SCR$};
\node at (6,1) () {$p-Si$};
\draw[red] (0,0.2) node[left](){$0$} -- ++(1,0)
-- ++(0,1.5)
-- ++(0,-1.5)
-- ++(2,0)
-- ++(0,-0.5)
-- ++(2,0)
-- ++(0,0.5)
-- ++(3,0)
;
\node[red] at (4,0.2) () {$Q_M$};
\end{tikzpicture}
\end{figure}
Where at the interface of the metal and the oxide, we have continuity of the electric displacement:
\begin{align}
\varepsilon_{ox} E_{ox} & = Q_M \\
\Rightarrow E_{ox} & = \frac{Q_M}{\varepsilon_{ox}} \\
D_{ox} & = D_{sc,int} \\
\Rightarrow \varepsilon{ox}E_{ox} & = \varepsilon{sc}E_{sc,int} \\
\Rightarrow \frac{E_{ox}}{E_{sc,int}} & = \frac{\varepsilon_{r,sc}}{\varepsilon_{r,ox}}\qquad\approx 3
\end{align}
And at the interface of the SCR and the substrate we have the following displacement continuity:
\begin{align}
\varepsilon_{sc} E_{sc}(x) & = -qN_A\left(x_D - x\right) \\
\Rightarrow E_{sc}(x) & = -\frac{qN_A}{\varepsilon_{sc}}\left(x-x_D\right)
\end{align}
\subsection{Electrostatic potential}
Nothing new, just integrate over the electric field.
(see \autoref{fig:mos_electrostaticpotential})
\begin{figure}[h]
\centering
\caption{Electrostatic potential of MOS}
\label{fig:mos_electrostaticpotential}
\includegraphics[width=.95\textwidth]{imgs/nmos_electrostaticpotential.png}
\end{figure}
\begin{equation}
\phi(x) = \begin{cases}
\phi_p & : x_D < x \\
\phi_p + \frac{q N_A}{2\varepsilon_{sc}}\left(x - x_D\right)^2 & : 0 < x < x_D \\
\phi_p + \frac{q N_A}{2\varepsilon_{sc}}x_D^2 - \frac{q N_A x_D}{\varepsilon_{sc}}x & : -t_{ox}<x<0 \\
\phi_{n+} & : x<t_{ox}
\end{cases}
\end{equation}
and so finally
\begin{align}
\begin{split}
\phi_B &= V_B + V_{ox}\\
&= \frac{q N_A x_D^2}{2\varepsilon_{sc}}+\frac{1 N_A x_D t_{ox}}{\varepsilon_{ox}}
\end{split} \\
\begin{split}
x_D & =\frac{\varepsilon_{sc}}{\varepsilon_{ox}}t_{ox}\left(
\sqrt{1 + \frac{2\varepsilon_{ox}^2 \phi_B}{\varepsilon_{sc} q N_A t_{ox}^2}} - 1
\right)\\
&=\frac{\varepsilon_{sc}}{C_{ox}} \left( \sqrt{1+\frac{4\phi_B}{\gamma^2}}-1 \right)
\end{split} \\
C_{ox} & = \frac{\varepsilon_{ox}}{t_{ox}} \\
\gamma & = \frac{1}{C_{ox}}\sqrt{2\varepsilon q N_A}
\end{align}
\subsection{Contact potential}