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\begin{document}

\underline{\textbf{BJT Phase Shift Oscillator design}}

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 Let take \smallskip $ V_{CC} = 5\,\text{V}, \quad \beta = 110, \quad f = 4096\,\text{Hz}, \quad I_C = 2.5\,\text{mA}, \quad C = 5 \times 10^{-9}\,\text{F}  \text{(feedback)}$ \smallskip \smallskip 
 
 Since the quiescent point is at the center of the load line, 
\begin{equation*}
V_{CE} = 0.5 V_{CC} = \SI{2.5}{\volt}, \quad 
V_{RE} = 0.1 V_{CC} = \SI{0.5}{\volt}
\end{equation*}

\textbf{Collector Resistor (\(R_C\))}
\smallskip
\begin{align*}
V_{RC} &= V_{CC} - V_{CE} - V_{RE} = 5 - 2.5 - 0.5 = \SI{2}{\volt} \\[0.5em] % Adds half of an 'M' width of space
R_C &= \frac{V_{RC}}{I_C} = \frac{2}{0.0025} = \SI{800}{\ohm}
\end{align*}


\textbf{Emitter Resistor (\(R_E\))}
\begin{equation*}
R_E = \frac{V_{RE}}{I_C} = \frac{0.5}{0.0025} = \SI{200}{\ohm}
\end{equation*}

\textbf{Base Current (\(I_B\))}
\begin{align*}
I_B &= \frac{I_C}{\beta} = \frac{2.5 \times 10^{-3}}{110} \\[0.5em] 
    &= \SI{0.0227}{\milli\ampere} = \SI{22.7}{\micro\ampere}
\end{align*}

\textbf{Bias Resistors (\(R_1, R_2\))}
\begin{align*}
I_2 &= 10 I_B = \SI{0.227}{\milli\ampere}, \\[0.5em]  
I_1 &= I_2 + I_B = \SI{0.25}{\milli\ampere} \\[0.5em] 
V_B &= V_{RE} + 0.7 = \SI{1.2}{\volt}
\end{align*}

\begin{align*}
R_2 &= \frac{V_B}{I_2} = \frac{1.2}{0.000227} = \SI{5286}{\ohm} \approx \SI{5.3}{\kilo\ohm}  \\[0.5em] 
R_1 &= \frac{V_{CC} - V_B}{I_1} = \frac{5 - 1.2}{0.00025} = \SI{15200}{\ohm} \approx \SI{15.2}{\kilo\ohm} 
\end{align*}

\textbf{Input Coupling Capacitor (\(C_{C1}\))}

\smallskip\smallskip
Capacitor impedance should be less than the input impedance of the amplifier $(R_{in})$.
\begin{align*}
r_e &= \frac{26}{I_E} = \frac{26}{2.5} = \SI{10.4}{\ohm} \\[0.5em] 
h_{ie} &= \beta r_e = 110 \times 10.4 = \SI{1144}{\ohm} \\[0.5em] 
R_{in} &= R_1 \parallel R_2 \parallel h_{ie} = \SI{15}{\kilo\ohm} \parallel \SI{5.6}{\kilo\ohm} \parallel \SI{1.144}{\kilo\ohm} = \SI{1.05}{\kilo\ohm} \\[0.5em] 
X_{C1} &\leq \frac{R_{in}}{10} = \SI{105}{\ohm} 
\end{align*}

Calculate for a lower cut off frequency of 200 Hz
\begin{align*}
C_{C1} &= \frac{1}{2 \pi f X_{C1}} = \frac{1}{2\pi \times 200 \times 105} \\[0.5em] 
       &= \SI{7.57}{\micro\farad}
\end{align*}


\textbf{Bypass Capacitor (\(C_E\))}
\begin{align*}
X_{CE} &\leq \frac{R_E}{10} = \SI{20}{\ohm} 
\end{align*}

Calculate for a lower cut off frequency of 500 Hz
\begin{align*}
C_E &= \frac{1}{2\pi f X_{CE}} = \frac{1}{2\pi \times 500 \times 20} \\[0.5em] 
    &= \SI{15.9}{\micro\farad}
\end{align*}

\textbf{Feedback Resistor (\(R\))}
\begin{align*}
f &= \frac{1}{2\pi R C \sqrt{6}} \implies \\[0.5em] 
R &= \frac{1}{2\pi \times 4096 \times 5 \times 10^{-9} \times \sqrt{6}} \\[0.5em] 
  &= \SI{3173}{\ohm} \approx \SI{3.17}{\kilo\ohm}
\end{align*}


\textbf{Final Component Values}

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Component values are changed little bit by trial and error method for proper working.

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Parallel combination of components used for non-standard values of components.
\begin{table}[h!]
    \centering
    \begin{tabular}{lll}
        \toprule % toprule is better than \hline
        \textbf{Component} & \textbf{Calculated Value} & \textbf{Used Value} \\
        \midrule % midrule is better than \hline
        $R_C$ & \SI{800}{\ohm} & \SI{825}{\ohm} \\
        $R_E$ & \SI{200}{\ohm} & \SI{224}{\ohm} \\
        $R_1$ & \SI{15.2}{\kilo\ohm} & \SI{23.5}{\kilo\ohm} \\
        $R_2$ & \SI{5.3}{\kilo\ohm} & \SI{5}{\kilo\ohm} \\
        $R$ & \SI{3.17}{\kilo\ohm} & \SI{3.195}{\kilo\ohm} \\
        $C_{C1}$ & \SI{7.57}{\micro\farad} & \SI{10}{\micro\farad} \\
        $C_E$ & \SI{15.9}{\micro\farad} & \SI{22}{\micro\farad} \\
        \bottomrule % bottomrule is better than \hline
    \end{tabular}
\end{table}

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