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\section{Bipolar junction transistor (BJT)}
\begin{figure}[h]
\centering
\caption{BJT}
\includegraphics[width=.75\textwidth]{imgs/bjt_terminals_and_functioning.png}
\end{figure}
But what's going on?
If $V_{BE}>0$ injection of electrons from E to B, of holes from B to E.
If $V_{BC}<0$ extraction of electrons from B to C, of holes from C to B.
\subsection{BJT characteristics}
\begin{align}
I_E & = -I_C-I_B \\
\begin{split}
\beta &= \frac{I_C}{I_B}
=\frac{n_{pB_0}\frac{D_n}{W_B}}{p_{nE_0}\frac{D_p}{W_E}}\\
&= \frac{N_{dE} D_n W_E}{N_{aB} D_p W_B}
\end{split}
\end{align}
Collector current,
focus on electron diffusion in base:
\begin{align}
n_{pB}(0) & =n_{pB_0}e^{\frac{qV_{BE}}{kT}} \\
n_{pB}(x) & =n_{pB}(0)(1-\frac{x}{W_B}) \\[1em]
\begin{split}
J_{nB} &= qD_n\frac{\mathrm{d} n_{pB}}{\mathrm{d}x}\\
&= -qD_n\frac{n_{pB}(0)}{W_B}
\end{split} \\
\begin{split}
I_C &=-J_{nB}A_E\\
&=qA_E\frac{E_n}{W_B}n_{pB_0}e^{\frac{qV_{BE}}{kT}}
\end{split} \\
I_C & = I_Se^{\frac{qV_{BE}}{kT}}
\end{align}
Base current,
focus on hole injection and recombination in emitter:
\begin{align}
p_{nE}(-x_{BE}) & =p_{nE_0}e^{-\frac{qV_{BE}}{kT}} \\
p_{nE}(-W_E-x_{BE}) & =p_{nE_0} \\
p_{nE}(x) & =\left[ p_{nE}(-x_{BE}-p_{nE_0}) \right]\left( 1+\frac{x+x_{BE}}{W_E} \right)+P_{nE_0} & \leftarrow \text{Hole Profile} \\[1em]
\begin{split}
J_{pE}&=-qD_p\frac{\mathrm{d}p_{nE}}{\mathrm{d}x}\\
&=-qD_p\frac{p_{nE(-x_{BE})-p_{nE_0}}}{W_E}
\end{split} \\
\begin{split}
I_B&=-J_{pE}A_E\\
&=qA_E\frac{D_p}{W_E}p_{nE_0}\left( e^{\frac{qV_{VE}}{kT}} -1 \right)
\end{split} \\
I_B & =\frac{I_S}{\beta}\left(e^{\frac{qV_{BE}}{kT}}-1\right) \\
I_B\approx\frac{I_C}{\beta}
\end{align}
\subsubsection{`Good' transistor}
We want collector and emitter current to be identical and so we define $\alpha$ as measurement of how close we are:
\begin{align}
I_C & =-\alpha I_E \\
& =\alpha\left(I_B+I_C\right) \\
& =\frac{\alpha}{1-\alpha}I_B \\
& =\beta I_B \\
\beta & =\frac{\alpha}{1-\alpha}
\end{align}
\subsection{Summary forward active}
\begin{align}
I_C & = I_Se^{\frac{qV_{BE}}{kT}} \\
I_B & = \frac{I_S}{\beta}\left(e^{\frac{qV_{BE}}{kT}}-1\right) \\
I_E & = -I_C-I_B
\end{align}
For reverse, it is the same but $\beta_R\approx [0.1,5]\ll\beta$.
\subsection{Summary cut-off}
\begin{alignat}{2}
I_{B1} & = -\frac{I_S}{\beta} & & =-I_E \\
I_{B2} & =-\frac{I_S}{\beta_R} & & =-I_C
\end{alignat}
\subsection{Summary saturation}
\begin{align}
I_C & =I_S\left(e^{\frac{qV_{BE}}{kT}} - e^{\frac{qV_{BC}}{kT}}\right)-\frac{I_S}{\beta_R}\left( e^\frac{qV_{BC}}{kT} - 1 \right) \\
I_B & =\frac{I_S}{\beta}\left( e^{\frac{qV_{BE}}{kT}}-1 \right)+\frac{I_S}{\beta_R}\left( e^{\frac{qV_{BC}}{kT}} -1 \right) \\
I_E & =\frac{I_S}{\beta}\left(e^{\frac{qV_{BE}}{kT}} - 1\right) - I_S\left( e^{\frac{qV_{BE}}{kT}} -e^{\frac{qV_{BC}}{kT}} \right)
\end{align}
\subsection{Ebers-Moll model}
\begin{center}
\begin{circuitikz}
\draw (0,0) node[left] {B} to [short,*-] ++(1,0)
to [Do,l=$\frac{I_S}{\beta_R}\left( e^{\frac{qV_{BC}}{kT}} -1 \right)$] ++(0,2)
to [short] ++(2,0)
to [I,l=$I_S\left( e^{\frac{qV_{BE}}{kT}} - e^{\frac{qV_{BC}}{kT}} \right)$,i=$$] ++(0,-4)
to [short] ++(-1,0);
\draw (1,0) to [Do,l_=$\frac{I_S}{\beta}\left( e^{\frac{qV_{BE}}{kT}}-1 \right)$] ++(0,-2)
to [short] ++(1,0)
to [short,-*] ++(0,-1) node [below] {E};
\draw (2,2) to [short,-*] ++(0,1) node[above] {C};
\end{circuitikz}
\end{center}
\subsection{Early effect}
With increasing $V_{CE}$, the depletion region inceases.
To not have to deal with that, we introduce a correction factor
\begin{equation}
I_C = I_S e^{\frac{V_{BE}}{V_{th}}}\left(1+\frac{V_{CE}}{V_A}\right)
\end{equation}
\subsection{Transfer characteristics}
We evaluate the transistor at its operating point ($OP$ or $Q=(V_{BE},V_{CE})$) to find the transconductance $g_m$.
\begin{equation}
\label{label:eq:bjt_transconductance}
g_m = \left. \frac{\partial i_C}{\partial V_{BE}} \right|_{OP} = \frac{qI_C}{kT}
\end{equation}