# Computing the phase shift of an RC phase shift oscillator

Consider this RC phase shift oscillator, taken from here:

As described in this answer, the phase shifts shown in the image are actually incorrect: https://electronics.stackexchange.com/a/216308/5546

If we start with the RC circuit right before the opamp, we can calculate the phase shift like this:

\begin{align*} R &= 10\ k\Omega \\ C &= 1\ nF \\ f &= 6500\ Hz \\ X_c &= \frac{1}{2\pi f C} = 24485\ \Omega \\ \theta &= \arctan{\frac{X_c}{R}} = 1.18\ \text{rad} = 67.78^\circ \end{align*}

What I'm struggling with is calculating the phase shift of the other two RC stages. According to that answer, the phase shift from the middle stage should be $$\ 56.3^\circ \$$. I've simulated this circuit and confirmed that this phase shift is correct, but I'm unable to compute this value.

I'm self-taught and so it's very likely my approach is completely off, but for completeness what I've tried is to compute the impedance of the final RC stage in parallel with one resistor as $$Z = R\ \|\ (R - jX_c)$$ after which I can do $$\ \arctan{\frac{X_c}{Z}} \$$ to compute the phase shift. Is this the correct approach? If so, could someone walk through the steps to calculate Z (I would show my work, but I've tried a number of methods that are all producing different and incorrect results, and I don't think explaining each will be helpful)? If not, what should be done instead?

• Here's a link of something I wrote, earlier. In your case, don't forget that your feedback resistor loads down the prior passive part. Jul 17, 2023 at 0:31
• You have no other choice than computing the transfer function from the op-amp output node (the left connection of the first cap.) to the (-) input. This is a 3rd-order system and you can't separate the stages as they all load each other. The links given by periblepsis contain your answer. Jul 17, 2023 at 8:29

I don't see your calculations for the 2nd stage's addition, but your simulation is close. I calculate the difference between one stage ($$\\approx 67.785^\circ\$$) and two stages ($$\\approx 124.217^\circ\$$) should be $$\\approx 56.433^\circ\$$. Obviously, the 3rd stage should bring this around to a final $$\180^\circ\$$ (a remaining change of $$\\approx 55.783^\circ\$$.) But loading matters and you cannot take each stage in isolation. They combine and interact.

I've discussed your question here and here. Probably should have been able to find them, among others, with a search. But in the first of the two links, do take note of Verbal Kint's answer there, as well. Mine is just brute mathematics (his description of my work elsewhere, not there) and the approach he illustrates may provide a more useful tool than my brutish method manages.

Analysis or simulation are the usual tools. For analysis, only the impedance ratio is required. This is because a voltage divider with source impedance $$\Z_S\$$ and load impedance $$\Z_L\$$ can be usefully re-written as $$\\frac1{1+\frac{Z_S}{Z_L}}\$$. (See if you can re-arrange the usual voltage divider fraction so as to look like that.) This helps emphasize the importance of this ratio.

But do note that while $$\Z_L=R\$$ for the last of three stages, $$\Z_S\$$ won't just be the obvious single capacitor, but instead the complex combination of all the parts preceding the final resistor. So it's not so simple as I think you are attempting.

In this case, the ratio is $$\\alpha=\frac{Z_C}{R}=\frac1{R\,C\,s}\$$ and from it can be computed the source impedances, one at a time, through each stage. Complex variable $$\s\$$ can be replaced with imaginary-only $$\j\omega\$$ as there's no interest in this oscillator's failure modes (where it may damp out or rail the output.)

On the first link provided earlier, note the Pascal's triangle. From it, for three stages, find the constants to be $$\[1, 5, 6, 1]\$$. This means the resulting polynomial of interest is $$\\alpha^3+5\alpha^2+6\alpha+1\$$.

We want the ratio of the imaginary part vs the real part for performing the arctangent to get the total phase shift. The odd powers will be the imaginary part and the even powers will be the real part. So the ratio is easy enough. Therefore the phase angle for three such stages will be $$\\phi=\arctan\left(\frac{j\,\alpha\cdot\left(\alpha^2+6\right)}{5\alpha^2+1}\right)\$$. With $$\f\approx 6497.47\:\text{Hz}\$$ and therefore $$\\omega\approx 40824.83\:\frac{\text{rad}}{\text{s}}\$$ and plugging in your values, you should find the numerator to be appreciably close to zero.

Which gets to another point. You can work out the oscillation frequency from the above. All you need do is to set the imaginary part to zero and solve. So, here you just need to solve $$\j\,\alpha\cdot\left(\alpha^2+6\right)=0\$$ for $$\\omega\$$. This is just where $$\\alpha^2=-6=\frac{-1}{R^2\,C^2\,\omega^2}\$$. This quickly re-arranges into $$\\omega^2=\frac1{6\,R^2\,C^2}\$$ or $$\\omega=\frac1{\sqrt{6}\,\cdot\, R\, C}\$$. From that, find $$\f=\frac1{2\pi\,\sqrt{6}\,\cdot\, R\, C}\$$.

Just for fun, suppose four stages.

Then find $$\[1, 7, 15, 10, 1]\to \alpha^4+7\alpha^3+15\alpha^2+10\alpha+1\$$. So the fraction is now $$\\frac{j\,\alpha\cdot\left(7\alpha^2+10\right)}{\alpha^4 +15\alpha^2+1}\$$. This is just where $$\7\alpha^2=-10=\frac{-7}{R^2\,C^2\,\omega^2}\$$. Then find $$\\omega^2=\frac7{10\,R^2\,C^2}\$$ or $$\\omega=\frac{\sqrt{70}}{10\,\cdot\, R\, C}\$$. From that, find $$\f=\frac{\sqrt{70}}{20\pi\,\cdot\, R\, C}\$$.

With your values, I compute a four-stage value of $$\f\approx 13.316\:\text{kHz}\$$. With LTspice I get a simulation of $$\\approx 13.1\:\text{kHz}\$$. Which I consider close enough to not worry more about. (By the way, while the required gain must be 29 for the 3-stage case, it only needs to be about 19 for the 4-stage case. Can you suggest a reason why that trend should have been anticipated?)

Your circuit is wrong - the resistor between the inv. opamp input and ground has no effect. This resistor should be a series element BETWEEN the most right capacitor and the inv. input.