\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}