Frequency of Oscillation of RC Phase Shift Oscillator

Derivation of Frequency of Oscillation

We have to find out the transfer function of RC feedback network.

Feedback Circuit of RC Phase Shift Oscillator

Applying KVL to various loops on the figure, we get, $$I_1 \left(R+\frac{1}{j \omega C }\right) -I_2R=V_i \text{ ....(1)}$$ $$-I_1R+I_2\left (2R+\frac {1}{j\omega C}\right)-I_3R=0\text{ ... (2)}$$ $$0-I_2R+I_3\left(2R+ \frac{1}{j\omega C}\right)=0\text{ ...(3)}$$
Replacing \(j\omega\) with \(s\) and writing equations in the matrix form, $$\begin{bmatrix}R+\frac{1}{sC} & -R & 0 \\-R & 2R+\frac{1}{sC} & -R \\0 & -R & 2R+\frac{1}{2sC} \end{bmatrix}\begin{bmatrix}I_1\\I_2\\I_3\end{bmatrix}=\begin{bmatrix}V_i\\0\\0\end{bmatrix}$$
Using Cramer's rule to find out \(I_3\),
$$\text{Let, }D=\begin{bmatrix}R+\frac{1}{sC} & -R & 0 \\-R & 2R+\frac{1}{sC} & -R \\0 & -R & 2R+\frac{1}{2sC} \end{bmatrix}$$ \(|D|=\begin{vmatrix}R+\frac{1}{sC} & -R & 0 \\-R & 2R+\frac{1}{sC} & -R \\0 & -R & 2R+\frac{1}{2sC} \end{vmatrix}\\ \begin{align} \text{ } &=\frac{1+sRC}{sC}\left[\frac{(1+2sRC)^2}{s^2C^2}-R^2\right]+R\left[-R\left(\frac{1+2sRC}{sC}\right) \right]+0\\ \\ & =\frac{1+sRC(1+2sRC)^2}{s^3C^3}-\frac{R^2(1+sRC)}{sC}-\frac{R^2(1+2sRC)}{C}\\ \\ & = \frac{(1+sRC)(1+4sRC+4s^2C^2R^2)-R^2s^2C^2[1+2sRC+1+sRC]}{s^3C^3}\\ \\ &= \frac{1+5sRC+8s^2C^2R^2+4s^3C^3R^3-3s^3R^3C^3-2s^2R^2C^2}{s^3C^3}\\ \\ &=\frac{1+5sRC+6s^2C^2R^2+s^3C^3R^3}{s^3C^3} \end{align}\) By replacing third column of matrix D with output we get(the vertical line is used for more convenience), $$D_3=\left [ \begin{array}{cc|c}R+\frac{1}{sC} & -R & V_i \\-R & 2R+\frac{1}{sC} & 0 \\ 0 & -R & 0 \end{array} \right] \\ $$ $$|D_3|=\begin{vmatrix}R+\frac{1}{sC} & -R & V_i \\-R & 2R+\frac{1}{sC} & 0 \\ 0 & -R & 0 \end{vmatrix}=V_iR^2$$ $$I_3=\frac{|D_3|}{|D|}=\frac{V_iR^2s^3C^3}{1+5sRC+6s^2C^2R^2+s^3C^3R^3}$$ Now $$V_O=V_f=I_3R=\frac{V_iR^3s^3C^3}{1+5sRC+6s^2C^2R^2+s^3C^3R^3}$$ Since feedback factor$$\beta=\frac{V_O}{V_i} \\ \beta =\frac{R^3s^3C^3}{1+5sRC+6s^2C^2R^2+s^3C^3R^3}$$ Replacing s by \(j\omega\), s² by \(-\omega^2\) and s³ by \(j\omega^3\), $$\beta =\frac{-j \omega^3 R^3C^3}{1+5j\omega CR-6\omega^2R^2C^2-j\omega^3R^3C^3}$$ Deviding denominator and numerator by \(j\omega^3R^3C^3\) and using, $$\color{red} {\alpha =\frac{1}{\omega RC}}$$ Then, $$\beta= \frac{1}{1+6j\alpha -5\alpha^2-j\alpha^3}$$ $$\beta= \frac{1}{(1-5\alpha^2)+j\alpha(6-\alpha^2)}$$ To have phase shift of \(180^\circ\), the imaginary part of denominator must be zero. $$\alpha(6-\alpha^2)=0$$ $$\alpha^2=6$$ $$\alpha=\sqrt 6$$ That is, $$\frac{1}{\omega RC}=\sqrt 6$$ $$\omega =\frac{1}{RC\sqrt 6}$$ $$f= \frac{1}{2\pi RC\sqrt 6}$$ This is the frequency with which circuit oscillates.
At this frequency,$$\beta= \frac{1}{1-5(\sqrt 6)^2}=-\frac{1}{29}$$ Negative sign indicates that phase shift of \(180^\circ \). Thus, $$\color{red}{|\beta |=\frac{1}{29}}$$ Now to have oscillations, \(|A\beta | \geq 29 \). $$|A||\beta| \geq 1$$ $$|A|\geq \frac{1}{|\beta |}\geq \frac{1}{\left( \frac{1}{29} \right)}$$ $$\color{red}{|A| \geq 29}$$ Thus we can conclude that gain of amplifier should be at least 29 for getting oscillations from a RC Phase shift oscillator.