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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, I1(R+1jωC)I2R=Vi ....(1) I1R+I2(2R+1jωC)I3R=0 ... (2) 0I2R+I3(2R+1jωC)=0 ...(3)
Replacing jω with s and writing equations in the matrix form, [R+1sCR0R2R+1sCR0R2R+12sC][I1I2I3]=[Vi00]
Using Cramer's rule to find out I3,
Let, D=[R+1sCR0R2R+1sCR0R2R+12sC] |D|=|R+1sCR0R2R+1sCR0R2R+12sC| =1+sRCsC[(1+2sRC)2s2C2R2]+R[R(1+2sRCsC)]+0=1+sRC(1+2sRC)2s3C3R2(1+sRC)sCR2(1+2sRC)C=(1+sRC)(1+4sRC+4s2C2R2)R2s2C2[1+2sRC+1+sRC]s3C3=1+5sRC+8s2C2R2+4s3C3R33s3R3C32s2R2C2s3C3=1+5sRC+6s2C2R2+s3C3R3s3C3 By replacing third column of matrix D with output we get(the vertical line is used for more convenience), D3=[R+1sCRViR2R+1sC00R0] |D3|=|R+1sCRViR2R+1sC00R0|=ViR2 I3=|D3||D|=ViR2s3C31+5sRC+6s2C2R2+s3C3R3 Now VO=Vf=I3R=ViR3s3C31+5sRC+6s2C2R2+s3C3R3 Since feedback factorβ=VOViβ=R3s3C31+5sRC+6s2C2R2+s3C3R3 Replacing s by jω, s2 by ω2 and s3 by jω3, β=jω3R3C31+5jωCR6ω2R2C2jω3R3C3 Deviding denominator and numerator by jω3R3C3 and using, α=1ωRC Then, β=11+6jα5α2jα3 β=1(15α2)+jα(6α2) To have phase shift of 180, the imaginary part of denominator must be zero. α(6α2)=0 α2=6 α=6 That is, 1ωRC=6 ω=1RC6 f=12πRC6 This is the frequency with which circuit oscillates.
At this frequency,β=115(6)2=129 Negative sign indicates that phase shift of 180. Thus, |β|=129 Now to have oscillations, |Aβ|29. |A||β|1 |A|1|β|1(129) |A|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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