Derivation of the frequency of a phase shift oscillator

Consider the following oscillator circuit:
For a circuit to oscillate, it must meet the following criteria:
  1. Total loop gain is 1
  2. The total phase shift is a multiple of 2π radians
One can see that the op-amp is wired as an inverting amplifier, so it has a phase shift of π radians. Therefore, the phase shift network consisting of capacitors C1, C2, C3 and resistors R1, R2, R3 must provide the remaining π degrees of phase shift in order for the circuit to meet criterion 2.

Let \(f\) be the frequency at which the circuit oscillates. The impedance of the capacitors at frequency \(f\) is \(Z_c=\frac{1}{2\pi jfC}\), where j is the imaginary unit and \(C\) is the capacitance of all the capacitors. Let \(R\) be the resistance of all the resistors except Rf.
It is also known that given two linear devices with impedances Z1 and Z2 (either real or complex impedances):
  • their total impedance when wired in series is \(Z_1+Z_2\)
  • their total impedance when wired in parallel is \(Z_1 || Z_2 = \frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}}\ = \frac{Z_1 Z_2}{Z_1 + Z_2}\)
In an inverting amplifier, the inverting and non-inverting inputs of the op amp are at the same voltage. Since R1 and C1 are in series, their combined impedance is \(R+Z_c\). Due to the virtual ground effect, they are in parallel with R2, which is in series with C2. By continuing this pattern, we can find the total impedance of the phase shift network:
\(Z = Z_c + Z_A\), where \(Z_A = R || \left(Z_c + Z_B\right)\), where \(Z_B = R || \left(Z_c + R\right) = \frac{R\left(Z_c + R\right)}{Z_c + 2R}\)

One can see that C3 and the rest of the phase shift network form a voltage divider. Suppose the input voltage to the network has a magnitude of 1. The voltage at node A is: $$ V_A = \frac{Z_A}{Z_A + Z_c} $$ $$ V_A = \frac{\frac{R\left(Z_c + Z_B\right)}{R + Z_c + Z_B}}{\frac{R\left(Z_c + Z_B\right)}{R + Z_c + Z_B} + Z_c} $$ $$ V_A = \frac{R\left(Z_c + Z_B\right)}{R\left(Z_c + Z_B\right) + Z_c\left(R + Z_c + Z_B\right)} $$ $$ V_A = \frac{R\left(Z_c + \frac{R\left(Z_c + R\right)}{Z_c + 2R}\right)}{R\left(Z_c + \frac{R\left(Z_c + R\right)}{Z_c + 2R}\right) + Z_c\left(R + Z_c + \frac{R\left(Z_c + R\right)}{Z_c + 2R}\right)} $$ $$ V_A = \frac{R\left(Z_c\left(Z_c + 2R\right) + R\left(Z_c + R\right)\right)}{R\left(Z_c\left(Z_c + 2R\right) + R\left(Z_c + R\right)\right) + Z_c\left(R\left(Z_c + 2R\right) + Z_c\left(Z_c + 2R\right) + R\left(Z_c + R\right)\right)} $$ There is another voltage divider between VA and ground consisting of C2 and the rest of the network: $$ V_B = V_A\frac{Z_B}{Z_B + Z_c} $$ $$ V_B = V_A\frac{\frac{R\left(Z_c + R\right)}{Z_c + 2R}}{\frac{R\left(Z_c + R\right)}{Z_c + 2R} + Z_c} $$ $$ V_B = V_A\frac{R\left(Z_c + R\right)}{R\left(Z_c + R\right) + Z_c\left(Z_c + 2R\right)} $$ One can see that \(R\left(Z_c + R\right) + Z_c\left(Z_c + 2R\right)\) exists in the denominator of \(V_B\) and in the numerator of \(V_A\) $$ V_B = \frac{R^2\left(Z_c + R\right)}{R\left(Z_c\left(Z_c + 2R\right) + R\left(Z_c + R\right)\right) + Z_c\left(R\left(Z_c + 2R\right) + Z_c\left(Z_c + 2R\right) + R\left(Z_c + R\right)\right)} $$ There is another voltage divider between VB and ground consisting of C1 and R1: $$ V_C = V_B\frac{R}{R + Z_c} $$ $$ V_C = \frac{R^3}{R\left(Z_c\left(Z_c + 2R\right) + R\left(Z_c + R\right)\right) + Z_c\left(R\left(Z_c + 2R\right) + Z_c\left(Z_c + 2R\right) + R\left(Z_c + R\right)\right)} $$ $$ V_C = \frac{R^3}{R\left(R^2 + 3RZ_c + Z_c ^2\right) + Z_c\left(3R^2 + 4RZ_c + Z_c ^2\right)} $$ $$ V_C = \frac{R^3}{R^3 + 6R^2 Z_c + 5RZ_c ^2 + Z_c ^3} $$ $$ V_C = \frac{R^3 Z_c ^{-3}}{R^3 Z_c^{-3} + 6R^2 Z_c^{-2} + 5RZ_c ^{-1} + 1} $$ $$ V_C = \frac{R^3\left(2\pi jfC\right)^3}{R^3\left(2\pi jfC\right)^3 + 6R^2\left(2\pi jfC\right)^2 + 5R\left(2\pi jfC\right) + 1} $$ $$ V_C = \frac{R^3\left(2\pi fC\right)^3 j^3}{\left(1 - 6R^2 \left(2\pi fC\right)^2\right) + j\left(5R\left(2\pi fC\right) - R^3\left(2\pi fC\right)^3\right)} $$ $$ V_C = \frac{R^3\left(2\pi fC\right)^3}{j\left(1 - 6R^2\left(2\pi fC\right)^2\right) - \left(5R\left(2\pi fC\right) - R^3\left(2\pi fC\right)^3\right)} $$ If the network has a phase shift of π radians, then the imaginary part must be zero. $$ 1 - 6R^2\left(2\pi fC\right)^2 = 0 $$ $$ 6\left(2\pi fRC\right)^2 = 1 $$ $$ 2\pi fRC = \frac{1}{\sqrt{6}} $$ $$ f = \frac{1}{2\pi RC \sqrt{6}} $$ According to criterion 1, the loop gain must be 1. Since the input to the phase shift network is assumed to be 1 and the output is VC, the gain of the op-amp amplifier must be \(\frac{1}{V_C}\). Let \(\beta\) be the gain of the op-amp amplifier. $$ \beta = \frac{R^3\left(2\pi fC\right)^3 - 5R\left(2\pi fC\right)}{R^3\left(2\pi fC\right)^3} $$ Replace \(2\pi fRC\) with \(\frac{1}{\sqrt{6}}\), as shown above. $$ \beta = \frac{\left(\frac{1}{\sqrt{6}}\right)^3 - 5\left(\frac{1}{\sqrt{6}}\right)}{\left(\frac{1}{\sqrt{6}}\right)^3} $$ $$ \beta = \frac{1 - 5\left(\sqrt{6}\right)^2}{1} $$ $$ \beta = -29 $$ Since \(\beta=-R_f/R_1\), the value for Rf is \(29R_1\).

In conclusion, the frequency of a phase shift oscillator is \(\frac{1}{2\pi RC\sqrt{6}}\) and the minimum value for Rf is \(29R\).

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