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) 0−I2R+I3(2R+1jωC)=0 ...(3)
Replacing jω with s and writing equations in the matrix form, [R+1sC−R0−R2R+1sC−R0−R2R+12sC][I1I2I3]=[Vi00]
Using Cramer's rule to find out I3,
Let, D=[R+1sC−R0−R2R+1sC−R0−R2R+12sC] |D|=|R+1sC−R0−R2R+1sC−R0−R2R+12sC| =1+sRCsC[(1+2sRC)2s2C2−R2]+R[−R(1+2sRCsC)]+0=1+sRC(1+2sRC)2s3C3−R2(1+sRC)sC−R2(1+2sRC)C=(1+sRC)(1+4sRC+4s2C2R2)−R2s2C2[1+2sRC+1+sRC]s3C3=1+5sRC+8s2C2R2+4s3C3R3−3s3R3C3−2s2R2C2s3C3=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+1sC−RVi−R2R+1sC00−R0] |D3|=|R+1sC−RVi−R2R+1sC00−R0|=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+5jωCR−6ω2R2C2−jω3R3C3 Deviding denominator and numerator by jω3R3C3 and using, α=1ωRC Then, β=11+6jα−5α2−jα3 β=1(1−5α2)+jα(6−α2) To have phase shift of 180∘, the imaginary part of denominator must be zero. α(6−α2)=0 α2=6 α=√6 That is, 1ωRC=√6 ω=1RC√6 f=12πRC√6 This is the frequency with which circuit oscillates.
At this frequency,β=11−5(√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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