add diode current

02_carrier_transport.tex

@@ -90,8 +90,8 @@ Which gives us the diffusion current density:
 (Defined as density times charge,
 ergo the double negative for electron diffusion.)
 \begin{align}
-    J_n^{diff} & = qD_n\frac{\mathrm{d} n}{\mathrm{d} x}  \\
-    J_p^{diff} & =- qD_p\frac{\mathrm{d} p}{\mathrm{d} x}
+    J_n^{diff} & = qD_n\frac{\mathrm{d} n}{\mathrm{d} x}    \label{label:eq:diff_current_n} \\
+    J_p^{diff} & =- qD_p\frac{\mathrm{d} p}{\mathrm{d} x}   \label{label:eq:diff_current_p}
 \end{align}
 
 

03_pn_junction_basics.tex

@@ -55,9 +55,9 @@ If the doping changes slowly with x:
 We saw in \autoref{label:sss:einstein_rel_mob_diff} the relation between mobility and diffusion coefficients.
 From this we find
 \begin{align}
-    n        & =n_{ref}e^{q(\phi-\phi_{ref})/kT} \\
-    \phi_ref & =0                                \\
-    n_{ref}  & =n_i
+    n          & =n_{ref}e^{q(\phi-\phi_{ref})/kT} \\
+    \phi_{ref} & =0                                \\
+    n_{ref}    & =n_i
 \end{align}
 And by extension
 \begin{align}
@@ -65,10 +65,10 @@ And by extension
     p & =n_ie^{-q\phi/kT}
 \end{align}
 
-Rearranging the above, we find a rule of thumb for the potential:
+Rearranging the above, we find an expression for the potential:
 \begin{align}
-    \phi & =\frac{kT}{q}\ln\frac{n}{n_i}   \\
-    \phi & = -\frac{kT}{q}\ln\frac{p}{n_i}
+    \phi & =\frac{kT}{q}\ln\frac{n}{n_i}   \label{label:eq:boltzman:phi_n} \\
+    \phi & = -\frac{kT}{q}\ln\frac{p}{n_i} \label{label:eq:boltzman:phi_p}
 \end{align}
 For Si at room temperature this is an increase of 60 mV per decade in doping.
 \begin{equation}

05_pn_junction_bias.tex

@@ -2,7 +2,7 @@
 \subsection{Model}
 
 We can see the junction as a series of resistors as follows.
-(Junction $V_{pn}>0$ is a forward bias.)
+(Junction $V_{pn}>0$ is a forward bias, QNR is a quasi-neutral region.)
 \begin{center}
     \begin{circuitikz}
         \draw (0,0) to[R, l=$R_{mp}$] ++(2,0)

06_pn_junction_diode.tex

+\section{PN junction diode}
+\subsection{Carrier concentration under bias}
+Under forward bias,  the net current is no longer zero.
+\begin{equation}
+    \left| J_{drift} \right|<\left| J_{diff} \right|
+\end{equation}
+Which causes injection of minority carriers into the QNR regions giving rise to `high' currents.
+
+
+\subsection{Diode current}
+To calculate the current, we \begin{enumerate}
+    \item Calculate concentration of minority carriers at the edges of SCR
+    \item Calculate minority carrier diffusion current in each QNR for $I_n$ and $I_p$
+    \item Sum the currents $I_n$ and $I_p$
+\end{enumerate}
+
+\subsubsection{Minority carrier conditions}
+We use the quasi-equilibrium equation to misuse equations for equilibrium.
+\begin{align}
+    \frac{n(x_1)}{n(x_2)} & \approx \exp{\frac{q(\phi(x_1)-\phi(x_2))}{kT}}  \\
+    \frac{p(x_1)}{p(x_2)} & \approx \exp{\frac{-q(\phi(x_1)-\phi(x_2))}{kT}}
+\end{align}
+
+So by using $x_n$ and $x_p$ in the above equation we have the following:
+\begin{align}
+    \frac{n(x_n)}{n(-x_p)} & \approx \exp{\frac{q(\phi_B-V)}{kT}}  \\
+    \frac{p(x_n)}{p(-x_p)} & \approx \exp{\frac{-q(\phi_B-V)}{kT}} \\
+    p(-x_p)                & =N_a                                  \\
+    n(x_n)                 & =N_d
+\end{align}
+
+And so we find what we needed:
+\begin{align}
+    n(-x_p) & \approx N_d\exp\frac{q(V-\phi_B)}{kT} \\
+    p(x_n)  & \approx N_a\exp\frac{q(V-\phi_B)}{kT}
+\end{align}
+
+
+Then by using the Boltzman relations \eqref{label:eq:boltzman:phi_n} and \eqref{label:eq:boltzman:phi_p} we find
+\begin{align}
+    \phi_B              & = \frac{kT}{q}\ln\frac{N_dN_a}{n_i^2}      \\
+    \Rightarrow n(-x_p) & \approx \frac{n_i^2}{N_a}\exp\frac{qV}{kT} \\
+    \Rightarrow p(x_n)  & \approx \frac{n_i^2}{N_d}\exp\frac{qV}{kT}
+\end{align}
+
+\subsubsection{Diffusion current in QNR}
+We assume a linear gradient between $n(-W_p)$ and $n(-x_p)$ to easily use \eqref{label:eq:diff_current_n} to find
+\begin{equation}
+    \begin{split}
+        J_n^{diff} &= qD_n\frac{n_p(-x_p)-n_p(-W_p)}{W_p-x_p}\\
+        &= qD_n \frac{\left(\frac{n_i^2}{N_a}\exp{\frac{qV}{kT}}\right)-\frac{n_i^2}{N_a}}{W_p-x_p}\\
+        &= q\frac{n_i^2}{N_a}\frac{D_n}{W_p-x_p}\left(\exp{\frac{qV}{kT}}-1\right)
+    \end{split}
+\end{equation}
+
+\subsubsection{Total diode current}
+\begin{equation}
+    \begin{split}
+        J & = J_n+J_p                                                                                             \\
+        & =q n_i^2 \left( \frac{1}{N_A}\frac{D_n}{W_p-x_p} + \frac{1}{N_D}\frac{D_p}{W_n-x_n} \right)\left(\exp
+        \frac{qV}{kT} - 1\right)
+    \end{split}
+\end{equation}
+Or so simplify
+\begin{align}
+    I   & =I_0\left(\exp\frac{qV}{kT}-1\right)                                 \\
+    I_0 & = A q n_i^2 \left( \frac{D_n}{L_n N_A} + \frac{D_p}{L_p N_D} \right)
+\end{align}
+
+
+\subsection{PN junction reverse bias}
+When applying a reverse bias, the depletion region gets wider and the electric field increases.
+There comes a point when the diode breaks down and destroys itself.
+\begin{equation}
+    W_{dep} = \sqrt{\frac{2\varepsilon}{q}\left(\frac{1}{N_A}+\frac{1}{N_D}\right)\left(V_0+V_R\right)}
+\end{equation}
\ No newline at end of file

semiconductor_summary.tex

@@ -42,5 +42,6 @@
 \include{02_carrier_transport}
 \include{03_pn_junction_basics}
 \include{04_pn_junction}
-\include{05_pn_junction_bias.tex}
+\include{05_pn_junction_bias}
+\include{06_pn_junction_diode}
 \end{document}