Oscillation Frequency for RC Phase-Shift Oscillator¶
In the pahse-shift oscillator the feedback circuit is three cascaded RC secrions. For simple design the capacitors and resistors in each section have the same value R and C.
Calculate the Oscillation Frequency¶
At each capacitor and resistor the currencies are
$
\begin{aligned}
i_1 &= j\omega C\,(v_1-v_a)\\
i_2 &= \cfrac{v_a}{R}\\
i_3 &= j\omega C\,(v_a-v_b)\\
i_4 &= \cfrac{v_b}{R}\\
i_5 &= j\omega C\,(v_b-v_2)\\
i_6 &= \cfrac{v_2}{R}\\
i_7 &= \cfrac{v_2}{R_i}
\end{aligned}
$
Kirchhoff's Current Law gives
$
\begin{aligned}
i_1 &= i_2+i_3\kern 2em&\dots(1)\\
i_3 &= i_4+i_5&\dots(2)\\
i_5 &= i_6+i_7&\dots(3)\\
\end{aligned}
$
Eq.(1) becomes
$
\begin{aligned}
&j\omega C\,(v_1-v_a)=\cfrac{v_a}{R}+j\omega C\,(v_a-v_b)\\
&(2j\omega C+\cfrac{1}{R})\,v_a-j\omega C\,v_b=j\omega C\,v_1 \kern 4ex\dots(1)'\\
\end{aligned}
$
Eq.(2) becomes
$
\begin{aligned}
&j\omega C\,(v_a-v_b)=\cfrac{v_b}{R}+j\omega C\,(v_b-v_2)\\
&j\omega C\,v_a-(2j\omega C+\cfrac{1}{R})\,v_b+j\omega C\,v_2=0 \kern 4ex\dots(2)'\\
\end{aligned}
$
Eq.(3) becomes
$
\begin{aligned}
&j\omega C\,(v_b-v_2)=\cfrac{v_2}{R}+\cfrac{v_2}{R_i}\\
&j\omega C\,v_b-(j\omega C+\cfrac{1}{R}+\cfrac{1}{R_i})\,v_2=0 \kern 4ex\dots(3)'\\
\end{aligned}
$
Set up a simultaneous equation from these three equations.
$$ \begin{aligned} {(2j\omega C+\cfrac{1}{R})\,v_a}& &{-j\omega C\,v_b}& &{}& = {j\omega C\,v_1} &\kern 4ex\dots(1)'\\ {j\omega C\,v_a}& &{-(2j\omega C+\cfrac{1}{R})\,v_b}& &{+j\omega C\,v_2}& = {0} &\kern 4ex\dots(2)'\\ {}& &{j\omega C\,v_b}& &{-(j\omega C+\cfrac{1}{R}+\cfrac{1}{R_i})\,v_2}& = {0} &\kern 4ex\dots(3)'\\ \end{aligned} $$Shown in matix.
$$ \begin{pmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}&{0}\\ {j\omega C}&{-(2j\omega C+\cfrac{1}{R})}&{j\omega C}\\ {0}&{j\omega C}&{-(j\omega C+\cfrac{1}{R}+\cfrac{1}{R_i})} \end{pmatrix} \begin{pmatrix} {v_a}\\ {v_b}\\ {v_2} \end{pmatrix} =\begin{pmatrix} {j\omega C\,v_1}\\ {0}\\ {0} \end{pmatrix} $$Solve for $\,v_2\,$ using Cramers's rule.
$$ v_2=\cfrac{1}{\Delta} \begin{vmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}&{j\omega C\,v_1}\\ {j\omega C}&{-(2j\omega C+\cfrac{1}{R})}&{0}\\ {0}&{j\omega C}&{0} \end{vmatrix} \qquad\dots(4) $$Here $\,\Delta\,$ is
$$ \Delta = \begin{vmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}&{0}\\ {j\omega C}&{-(2j\omega C+\cfrac{1}{R})}&{j\omega C}\\ {0}&{j\omega C}&{-(j\omega C+\cfrac{1}{R}+\cfrac{1}{R_i})} \end{vmatrix} $$The numerator in the right hand side in Eq.(4)
$$ = \begin{vmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}&{j\omega C\,v_1}\\ {j\omega C}&{-(2j\omega C+\cfrac{1}{R})}&{0}\\ {0}&{j\omega C}&{0} \end{vmatrix} $$and calculate this determinant with cofactor expansion
$$ = \begin{vmatrix} {j\omega C}&{-(2j\omega C+\cfrac{1}{R})}\\ {0}&{j\omega C} \end{vmatrix} {j\omega C\,v_1} $$The fractin in the right hand side in Eq(4) is $\,\Delta\,$.
$$ \Delta = \begin{vmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}&{0}\\ {j\omega C}&{-(2j\omega C+\cfrac{1}{R})}&{j\omega C}\\ {0}&{j\omega C}&{-(j\omega C+\cfrac{1}{R}+\cfrac{1}{R_i})} \end{vmatrix} $$Calculate $\Delta\,$ with cofactor expansion
$$ \Delta= -\begin{vmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}\\ {0}&{j\omega C} \end{vmatrix} {j\omega C} -\begin{vmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}\\ {j\omega C}&{-(2j\omega C+\cfrac{1}{R})} \end{vmatrix} {(j\omega C+\cfrac{1}{R})} -\begin{vmatrix} {(2j\omega C+\cfrac{1}{R})}&{-j\omega C}\\ {j\omega C}&{-(2j\omega C+\cfrac{1}{R})} \end{vmatrix} {\cfrac{1}{R_i}} $$Sum up $\,\Delta\,$
$$ \begin{aligned} \Delta=& -2{(j\omega C)^3}-{\cfrac{1}{R}(j\omega C)^2} +{\{{4(j\omega C)^2}+{\cfrac{4}{R}(j\omega C)} +{\cfrac{1}{R^2}}\}(j\omega C+\cfrac{1}{R})} -{(j\omega C)^3}-{\cfrac{1}{R}(j\omega C)^2} \\&+{\{{4(j\omega C)^2}+{\cfrac{4}{R}(j\omega C)} +{\cfrac{1}{R^2}}\}\cfrac{1}{R_i}} -{(j\omega C)^2}{\cfrac{1}{R_i}} \\=& -2{(j\omega C)^3}-{\cfrac{1}{R}(j\omega C)^2} +4{(j\omega C)^3}+{\cfrac{4}{R}(j\omega C)^2}+{\cfrac{1}{R^2}(j\omega C)} +{\cfrac{4}{R}(j\omega C)^2}+{\cfrac{4}{R^2}(j\omega C)}+{\cfrac{1}{R^3}} -{(j\omega C)^3}-{\cfrac{1}{R}(j\omega C)^2} \\&+{{\cfrac{4}{R_i}(j\omega C)^2}+{\cfrac{4}{RR_i}(j\omega C)} +{\cfrac{1}{R^2R_i}}} -{(j\omega C)^2}{\cfrac{1}{R_i}} \\=& \ {(j\omega C)^3}+{\cfrac{6}{R}(j\omega C)^2}+{\cfrac{5}{R^2}(j\omega C)}+{\cfrac{1}{R^3}} \\&+{{\cfrac{3}{R_i}(j\omega C)^2}+{\cfrac{4}{RR_i}(j\omega C)} +{\cfrac{1}{R^2R_i}}} \end{aligned} $$Here $\,j^2=-1\,$ then
$$ \begin{aligned} \Delta=&-{j(\omega C)^3}-{\cfrac{6}{R}(\omega C)^2}+{j\cfrac{5}{R^2}(\omega C)}+{\cfrac{1}{R^3}} \\&-{{\cfrac{3}{R_i}(\omega C)^2}+{\cfrac{4}{RR_i}(j\omega C)} +{\cfrac{1}{R^2R_i}}}\\ =&\ {j\omega C\{\cfrac{5}{R^2}-(\omega C)^2+\cfrac{4}{RR_i}\}} +{\cfrac{1}{R}\{\cfrac{1}{R^2}-6(\omega C)^2-\cfrac{3R}{R_i}(\omega C)^2+{\cfrac{1}{RR_i}}\}} \end{aligned} $$Wiht the above Eq.(4) becomes
$$ v_2=\cfrac{(j\omega C)^3\,v_1}{{j\omega C\{\cfrac{5}{R^2}-(\omega C)^2+\cfrac{4}{RR_i}\}} +{\cfrac{1}{R}\{\cfrac{1}{R^2}-6(\omega C)^2-\cfrac{3R}{R_i}(\omega C)^2+{\cfrac{1}{RR_i}}\}}} $$Multiply the numerator and fraction by $\,R^3\,$.
$$ v_2=\cfrac{(j\omega CR)^3\,v_1} {{j\{5\omega CR-(\omega CR)^3+{4\omega C\cfrac{R^2}{R_i}}\}} +{\{1-6(\omega CR)^2-{3(\omega C)^2\cfrac{R^3}{R_i}}+{\cfrac{R}{R_i}}\}}} $$Multiply by $\,-j\,$ futhermore.
$$ v_2=\cfrac{(\omega CR)^3\,v_1} {{\{(\omega CR)^3-5\omega CR-{4\omega C\cfrac{R^2}{R_i}}\}} -{j\{(6+\cfrac{3R}{R_i})(\omega CR)^2-1-\cfrac{R}{R_i}\}}} $$The feedback factor is then given by $\,\beta=\cfrac{v_2}{v_1}\,$
$$ \cfrac{v_2}{v_1}=\cfrac{(\omega CR)^3} {{\{(\omega CR)^3-5\omega CR-{4\omega C\cfrac{R^2}{R_i}}\}} -{j\{(6+\cfrac{3R}{R_i})(\omega CR)^2-1-\cfrac{R}{R_i}\}}} \qquad\dots(5) $$When the imaginary part of the feedback factor is zero, the oscillation frequency is given by
$$ (6+\cfrac{3R}{R_i})(\omega CR)^2-1-\cfrac{R}{R_i}=0\qquad\dots(6) $$Here $\,\omega=2\pi f\,$, and Eq.(6) becomes
$$
(6+\cfrac{3R}{R_i})(2\pi fCR)^2-1-\cfrac{R}{R_i}=0
$$
The oscillation frequency $\,f\,$ is
$$
(\cfrac{6R_i+3R}{R_i})f^2(2\pi CR)^2=\cfrac{R_i+R}{R_i}
$$
$$
f=\cfrac{1}{2\pi CR}\sqrt{\cfrac{R_i+R}{6R_i+3R}}
$$
import numpy as np
C = 0.01e-6
R = 6.8e3
Ri = 1e10
f = 1/(2*np.pi*C*R)*np.sqrt((Ri+R)/(6*Ri+3*R))
f
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