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.
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