My aim is to find the oscillation frequency of a Phase Shift Oscillator.

enter image description here

I start by finding transfer function of the cascaded RC network.

$$ V_o(s) = \dfrac{\dfrac{1}{C_3s}}{R_3+\dfrac{1}{C_3s}} V_2(s) = \dfrac{1}{1+R_3C_3s} V_2(s). $$


$$ V_2(s) = \dfrac{1}{1+R_2C_2s} V_1(s) \quad\text{and}\quad V_1(s) = \dfrac{1}{1+R_1C_1s} V_i(s). $$

Then the transfer function is:

$$ \begin{array}{lcl} H(s) = \dfrac{V_o(s)}{V_i(s)} &=& \dfrac{1}{(1+R_1C_1s)(1+R_2C_2s)(1+R_3C_3s)} \\ &=& \dfrac{1}{R_1R_2R_3C_1C_2C_3s^3 + \dots} \cdots \\ && \dfrac{}{(R_1R_2C_1C_2 + R_2R_3C_2C_3 + R_1R_3C_1C_3)s^2 + \dots} \cdots \\ && \dfrac{}{(R_1C_1 + R_2C_2 + R_3C_3)s + 1} \end{array} $$

So the frequency response is:

$$ \begin{array}{lcl} H(j\omega) &=& \dfrac{1}{j\omega \left[ (R_1C_1 + R_2C_2 + R_3C_3) - R_1R_2R_3C_1C_2C_3\omega^2 \right] + \dots} \cdots \\ && \dfrac{}{\left[ 1 - (R_1R_2C_1C_2 + R_2R_3C_2C_3 + R_1R_3C_1C_3)\omega^2 \right]} \end{array} $$

Now, we are looking for a special \$\omega\$ value, \$\omega_0\$, for which the argument of \$H(wj)\$ will be \$\pm180^o\$. Clearly, it happens when

$$ R_1C_1 + R_2C_2 + R_3C_3 = R_1R_2R_3C_1C_2C_3\omega^2 \Big|_{\omega=\omega_0}. $$

Hence we find the oscillation frequency as

$$ \begin{array}{rcl} \omega_0 &=& \sqrt{\dfrac{R_1C_1 + R_2C_2 + R_3C_3}{R_1R_2R_3C_1C_2C_3}} \\ \text{f}_0 &=& \dfrac{1}{2\pi} \sqrt{\dfrac{R_1C_1 + R_2C_2 + R_3C_3}{R_1R_2R_3C_1C_2C_3}} \end{array}. $$

When \$\quad R_1=R_2=R_3=R\quad\$ and \$\quad C_1=C_2=C_3=C\quad\$:

$$ \text{f}_0 = \dfrac{\sqrt{3}}{2 \pi RC} $$

However, according to all online articles including Wikipedia, the formula for the oscillation frequency is

$$ \text{f}_0 = \dfrac{1}{2 \pi RC \sqrt{6}}. $$

I did an experiment with the exact circuit I attached above with \$R=1k\Omega\$, \$C=100nF\$, \$R_i=1k\Omega\$ and \$R_f=33k\Omega\$ by using TL084 opamp. I observed the oscillation period as 7.4ms.

According to the formula I derived above, it should have been

$$ \tau_0 = 2 \pi (1k\Omega) (100nF) / \sqrt{3} = 362.76 \text{ns}. $$

And according to the other's formula, it should have been

$$ \tau_0 = 2 \pi (1k\Omega) (100nF) \sqrt{6} = 1.539 \text{ms}. $$

Finally, my questions are:

  1. Why is the formula I found above is different than the other's formula? Where did I make the mistake? Did using a an extra opamp for buffer affect it?
  2. Why does the period of my oscillator differ so much from what the both formula say?
  • \$\begingroup\$ As explained below - the last formula is correct. Perhaps it is interesting for you to know that you can drop the buffer amplifier and the inverting amplifier. In this case, the last passive RC section is to be replaced by an inverting integrator stage (classic MILLER intergator). However, in this case the time constant (inverse of the oscillation frequency) is identical to the first formula as given by you (involving SQRT[3]). \$\endgroup\$
    – LvW
    Commented Apr 22, 2014 at 14:34

4 Answers 4


Your 2nd and 3rd equations are incorrect.

The 1st equation is correct but the 2nd equation should be

$$V_2 = V_1\frac{\frac{1}{sC_2}||(R_3 + \frac{1}{sC_3})}{R_2 + \frac{1}{sC_2}||(R_3 + \frac{1}{sC_3})}$$

In other words, you didn't take into account the loading of the succeeding stages.

According to my morning algebra exercise, for uniform resistor values \$R\$ and capacitor values \$C\$,

$$\frac{V_+}{V_i} = \frac{1}{1 + 6sRC + 5(sRC)^2 + (sRC)^3} = \frac{1}{[1 - 5(\omega RC)^2] + j[6\omega RC - (\omega RC)^3]}$$

The phase shift is \$180^{\circ}\$ when the imaginary part of the denominator vanishes thus,

$$6\omega_0 RC = (\omega_0 RC)^3 \rightarrow \omega_0 = \frac{\sqrt{6}}{RC}$$

For the chosen resistor and capacitor values, the frequency is

$$f_0 = \frac{\sqrt{6}}{2\pi \cdot 1k\Omega \cdot 100nF} = 3.898kHz$$

To verify this calculation, I simulated the phase shift network and plotted the transfer function:

enter image description here


Your transfer function is not correct. You computed the transfer function as if the three RC networks were completely independent, i.e. as if there were buffers with an infinitely high input impedance between them. I believe that if you compute the transfer function correctly, you should be able to arrive at the correct formula.

  • \$\begingroup\$ Evidently, we came to the same conclusion at the same time. \$\endgroup\$ Commented Apr 22, 2014 at 13:46
  • \$\begingroup\$ @AlfredCentauri Yes, so chances are that we are both correct :) \$\endgroup\$
    – Matt L.
    Commented Apr 22, 2014 at 14:08

By using the fast analytical circuits techniques or FACTs, it is possible to determine the transfer function of the above circuit just by calculating the circuit time constants \$\tau\$ in various conditions. First, set \$s=0\$ and observe the passive circuit (we consider all \$R\$ to be equal as all \$C\$) and determine the gain in dc (all caps are open). You find \$H_0=1\$. Then, reduce the excitation to 0 V (short \$R_1\$'s left terminal to GND in the original circuit) and "look" at the resistances offered by all the caps when temporarily removed from the circuit. Then, alternatively placing all caps in their high-frequency state (replace the cap. by a short circuit), determine the 2nd- and 3rd-order time constants as shown in the below sketch:

enter image description here

Once you have all these time constants (please realize that you obtain them by inspecting the circuit, no algebra!), you combine them in the following way:


by combining all these results, we obtain


This is a 3rd-order polynomial expression that we can factor in the following form considering a dominant low-frequency pole and two coincident poles (by comparing the various time constants):

\$H(s)=\frac{1}{(1+\frac{1}{\omega_p})(1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2)}\$ in which

\$\omega_p=\frac{1}{6RC}\$ \$Q=6\frac{\sqrt{6}}{29}\$ \$\omega_0=\frac{\sqrt{6}}{RC}\$

These expressions are gathered and tested in the below Mathcad sheets:

enter image description here

enter image description here

giving the following frequency responses

enter image description here

If you now consider the magnitude expression nicely derived by Alfred Centaury, then the attenuation at the resonant frequency \$\omega_0\$ is exactly 29.248 dB for \$0.1\;\mu F\$ and \$1\;k\Omega\$. Considering a 100-\$k\Omega\$ for \$R_i\$ then \$R_f=2.9 M\Omega\$, the below circuit can be simulated showing nicely-sustained oscillations:

enter image description here enter image description here

Please note the .IC statement for cranking up the circuit at the beginning of the transient analysis.

One thing that could be misleading is that the frequency depends on the passive elements arrangement. For cascaded RC integrators as we have have here, the resonant frequency is \$f_0=\frac{\sqrt{6}}{2\pi RC}\$. However, if you go for the more classical CR differentiator configuration as in https://en.wikipedia.org/wiki/Phase-shift_oscillator then the frequency is \$f_0=\frac{1}{2\pi RC\sqrt{6}}\$. Now if you insert buffers between the CR stages (no loading of the individual cells), the oscillation frequency becomes \$f_0=\frac{1}{2\pi RC\sqrt{3}}\$ as already pointed out.

You have seen how the FACTs could lead you to the result just by inspecting the passive network. If the final polynomial form diverges from the raw transfer function (\$H_{ref}\$ in the Mathcad sheet), it is easy to go back to the small sketches and correct the guilty one. This is the "divide and conquer" strategy promoted by Dr. Middlebrook when he formalized the FACTs and his extra-element theorem or EET (https://en.wikipedia.org/wiki/Extra_element_theorem). If you are interested by the technique, have a look at this tutorial (http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf) and exercise yourself solving problems documented in http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf. Acquiring the skill of FACTs certainly requires a bit of time and efforts, but once you understand its power, you won't go back to the classical approach!

  • \$\begingroup\$ Is there an intuitive method to see no zero in this case? \$\endgroup\$
    – emnha
    Commented Jul 24, 2020 at 10:26
  • \$\begingroup\$ Yes, place one of the energy-storing element in its high-frequency state (a short circuit for a capacitor) and check if a stimulus applied in this condition would create a response. If there is a response, there is a zero associated with the considered energy-storing capacitor. No response means no zero. In this example, any cap. shorted to ground will shunt the signal and won't let the stimulus propagate. There are no zeroes in this circuit. \$\endgroup\$ Commented Jul 24, 2020 at 12:34
  • \$\begingroup\$ Thanks for the answer. What would be the basis for doing that? Like why shortens capacitors not open? Also how the method you gave above calculate Ho with this RC phase lag oscillator network? electronics-tutorials.ws/oscillator/… \$\endgroup\$
    – emnha
    Commented Jul 24, 2020 at 13:10
  • \$\begingroup\$ I would recommend you take a look at my APEC 2016 seminar or in my book dedicated to the FACTs. \$\endgroup\$ Commented Jul 24, 2020 at 15:57

Possibly the mistake you have made is in not using the same circuit as the Wikipedia article. Ignoring the extra op-amp you inserted (which is trivial), the phase shifting network in the wiki article is series capacitance and parallel resistance - you have used series resistance and parallel capacitance.

Given that each RC stage roughly accounts for a phase shift of 60 degrees, reversing the R and C would naturally only produce a phase shift of 30 degrees.

Doesn't that make a difference or am I being stupid?

BTW I did a simulation and found it to be about half-way between the two theoretical values using sqrt(6) and your sqrt(3)!!!

  • \$\begingroup\$ No - it is not a "mistake". The only advantage of CR high pass stages is that you don´t need a buffer for decoupling of the inverting amplifier (the first series resistor of the inverter can act as a load for the last CR section). And - of course - both alternatives (RC lowpass or CR high pass sections) work at a frequency which gives -60 deg phase shift. \$\endgroup\$
    – LvW
    Commented Apr 22, 2014 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.