# How to calculate the frequency of oscillation of this low-pass phase shift oscillator?

In the following phase-shift oscillator a low pass network is used instead of a high-pass network.

simulate this circuit – Schematic created using CircuitLab

Usually the frequency of oscillation for a high pass is given by:

$$f_{oscillation} = \frac{1}{2\pi RC\sqrt{6}}$$

However, I'm I don't really know how this is derived, I just use it. For the this circuit, how would one calculate the frequency of oscillation?

With the given values of R= 2.2Meg and C = 1uF, LTspice gave a frequency of about 0.174Hz.

(In the actual simulation I amplified Vsin)

Thank you.

• Your circuit is flawed - C3 is ineffective because it connects between a virtual earth and actual circuit earth i.e. has zero voltas across it and therefore does not contribute to shifting phase. Try and find a valid circuit that uses low pass filters. Apr 28, 2018 at 19:03
• @Andyaka I got the circuit from electronicdesign.com/analog/…. Is it wrong? Apr 28, 2018 at 19:12
• What part of my original comment did you not understand? Apr 28, 2018 at 19:22
• o_o what ? I think whoever drew this schematic interchanged resistors and capacitors. C3 is unwanted too. Apr 28, 2018 at 19:54
• I think your equation would be far closer to correct if the $\sqrt{6}$ were moved from the denominator to the numerator. The reason is that in the ideal case, $2\pi\:f\approx \operatorname{tan}\left(60^\circ\right)$. But the case isn't ideal since the RC sections do load each other in practice, so I'd expect the frequency to be a bit higher than predicted from the ideal.
– jonk
Apr 28, 2018 at 22:37

Your schematic uses a low pass filter approach. One way to mentally consider a solution is to look at the following schematic:

simulate this circuit – Schematic created using CircuitLab

Above, you can see that $$\R_3\$$ and $$\Z_3\$$ form a voltage divider that divides $$\V_Y\$$ into $$\V_\text{OUT}\$$. Also, $$\R_2\$$ and $$\Z_2\$$ form a voltage divider that divides $$\V_X\$$ into $$\V_Y\$$. Finally, $$\R_1\$$ and $$\Z_1\$$ form a voltage divider that divides $$\V_\text{IN}\$$ into $$\V_X\$$. It follows that:

\begin{align*} \text{Each Stage} \left\{ \begin{array}{rl} V_\text{OUT} = V_Y\,\frac{1}{1+\frac{R_3}{Z_3}}&&Z_3 = Z_{C_3}\mid\mid \infty=Z_{C_3}\\\\ V_Y = V_X\,\frac{1}{1+\frac{R_2}{Z_2}}&&Z_2 = Z_{C_2}\mid\mid \left(Z_3 + R_3\right)\\\\ V_X = V_\text{IN}\,\frac{1}{1+\frac{R_1}{Z_1}}&&Z_1 = Z_{C_1}\mid\mid \left(Z_2 + R_2\right) \end{array} \right. \end{align*} $$\therefore \frac{V_\text{OUT}}{V_\text{IN}}=\frac{1}{1+\frac{R_1}{Z_1}}\cdot\frac{1}{1+\frac{R_2}{Z_2}}\cdot\frac{1}{1+\frac{R_3}{Z_3}}$$

Assuming $$\R=R_1=R_2=R_3\$$ and $$\C=C_1=C_2=C_3\$$ and setting $$\\tau=R\:C\$$ I get the following answer:

$$H\left(j\omega\right)=\frac{1}{1 - 5\:\left(\omega\:\tau\right)^2 + j\:\left[6\:\omega\:\tau-\left(\omega\:\tau\right)^3\right]}$$

For a phase shift of $$\180^\circ\$$, the imaginary part in the denominator goes to zero. So:

\begin{align*}6\:\omega\:\tau-\left(\omega\:\tau\right)^3&=0\\6\:\omega\:\tau&=\left(\omega\:\tau\right)^3\\6&=\left(\omega\:\tau\right)^2\\\omega&=\frac{\sqrt{6}}{\tau}\\\therefore\\f&=\frac{\sqrt{6}}{2\pi\:R\:C}\end{align*}

Note that this differs from what you wrote.

I just tried out a simulation on LTspice using a decent R2R opamp (it can't handle more than about $$\12.5\:\text{V}\$$ between its rails.) Here's the results:

The period seems to be about $$\5.8\:\text{s}\$$. Which is close to the prediction.

## Note

Per Tony's request, it's not difficult to work out the peak or peak-to-peak voltage for the resulting sine wave at the output.

Since the imaginary component of $$\H\left(j\omega\right)\$$ is 0, the magnitude is just $$\\mid\:H\left(j\omega\right)\mid\:=\frac{1}{29}\$$ (just plug in $$\\omega=\frac{\sqrt{6}}{\tau}\$$.) The RMS of the square wave at $$\V_\text{IN}\$$ is just $$\6\:\text{V}\$$ (as it is always either $$\+6\:\text{V}\$$ or else $$\-6\:\text{V}\$$.) So the output will be:

$$V_{\text{OUT}_\text{RMS}}=\frac{1}{29}\cdot \text{V}_{\text{IN}_\text{RMS}}=\frac{1}{29}\cdot 6\:{\text{V}_\text{RMS}}\approx 207\:{\text{mV}_\text{RMS}}$$

Since the output is a sine wave, the peak should be about $$\\sqrt{2}\$$ larger, or $$\\approx 290\:\text{mV}_\text{P}\$$.

As you can see, the image I included above shows an output peak that is close to this prediction, as well.

• i like your style of answers +1 and never tried Sage/Sympy Apr 29, 2018 at 12:25
• Bonus points however might also show the sine amplitude Vpp is 26 dB down from Vcc-Vee Apr 29, 2018 at 12:32
• @TonyStewartEEsince1975 Um. Okay. I'll add a note since the added work is essentially zero.
– jonk
Apr 30, 2018 at 1:08
• A more general formula for the frequency, the long version that does not require all the R to be the same or all the C to be the same: $$f_\text{osc}=\frac{1}{2\pi}\sqrt{\frac{\left(R_1+R_2+R_3\right)C_1+\left(R_2+R_3\right)C_2+R_3C_3}{R_1R_2R_3C_1C_2C_3}}$$ Nov 23, 2019 at 3:23
• @ElEctric If your point is that one can use varying values for the resistors and capacitors and still get a useful oscillator then you have no argument from me. Did I say something that made you feel as though I felt all the R values had to be the same and that all the C values also had to be? If so, please let me know. (Just simple parasitics alone would mean none of these would ever work, if that were the case.) It's nice to add a formula to handle varying values -- interesting even -- but beyond the scope of my answer here. Perhaps you'd like to write a fuller exposition? I'd upvote it.
– jonk
Nov 23, 2019 at 3:28

The shown circuit is a bad one - why? Because the opamp is driven into deep saturation. As a consequence, the filtered output is not a "good" sinus signal. More than that, you need an additional ouput buffer for further processing of the filtered oscillation signal.

One little modification - and the circuit is much better: Connect the capacitor C3 not to ground but, instead to the output of the opamp.

Thus, you have an inverting integrator (phase sift +90 deg). Together with a phase shift (-90 deg) of the two remaining lowpass RC sections you can fulfill the oscillation condition as far as the phase is concerned. For a unity loop gain (amplitude condition) the value of C3 must be somewhat smaller than C/12. (C1=C2=C)

The frequency of oscillation is wo=SQRT(3)/RC (R1=R2=R; C1=C2=C)

As another advantage: A good-quality oscillation signal is available at a low-resistive opamp output (no additional buffer needed).

If you want to improve the quality of the signal, a soft-limiting technique can be incoroprated: Use a series connection of another (small) capacitor C4 and two anti-parallel diodes. This series combination is connected in parallel to C3 (select the parallel combination C3+C4>C/12).

• The Circuit shown is a good one and can also be done with 3 stage HPF but demands a linear gain. Here the Op Amp is simply a Comparator but the input is a good sine wave, 26.7 dB down if outputs are Rail-Rail to Vcc-Vee Apr 29, 2018 at 12:53
• But - you need a buffer for further processing. More than that, a soft-limiting path (a small non-linearity) is ALWAYS better than hard-limiting at the power rails. So - I don`t think that the original circuit (with an opamp without internal feedback) is a "good one". The signal quality is certainly not as good if compared with the classical integrator version.
– LvW
Apr 29, 2018 at 13:40
• It is actually very clean with fundamental down 27dB and 3rd harmonic down -6*3*2=36dB from fundamental or 63dB down from supply voltage so all you need is 20dB Gain or so. This is a comparator design essentially not a linear op Amp like the HPF phase shift Osc Apr 29, 2018 at 15:31
• sorry -6db/octave *3rd order Apr 29, 2018 at 15:42
• So let me ask the other way round: Where are the advantages of the shown circuit (C3 grounded) if compared with "my" proposal (invertiing integrator, with C3 connected to the opamp output)?
– LvW
Apr 29, 2018 at 19:34

The proper way to derive the oscillation frequency from this oscillator is to go back to Barkhausen's oscillation criteria. First find the transmittance of the 3 RC blocks and the one of your amplifier. Then apply Barkhausen's criteria for phase shift: the sum of the phase shifts from the two transmittance must be equal to zero for an oscillation to exist.

Since your amplifier has a real negative transmittance, its phase shift is equal to -180°. Hence the transmittance of the RC blocks must be a real negative number, which implies its imaginary value must be zero. From this condition, you may find out that $\omega RC = \frac{1}{\sqrt{6}}$, ending with $f = \frac{1}{2\pi RC \sqrt{6}}$. Calculation of the RC block transmittance is quite harsh though.

Edit: this would be the same method for your low-pass filter version.

This is how the experts do it: