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I found this design on the internet. It's a 180 degree phase shift oscillator. enter image description here

Actually, it is a 203 degree phase shift oscillator. Because the frequency is 6.5 kHz and the capacitive impedance is 24485.37 ohm. So, the phase shift of each stage will be 67 degrees not 60 degrees.

The total phase shift of the three stages of the high pass filter is 203 degree.

Is it OK to have more than 180 degree of phase shift?

If it is OK, what is the allowable range? How do I know the range at which the circuit oscillates?

Tom Carpenter
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Michael George
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    A lot of nice answers. If you want to build this you need some sort of AGC, else it will either not oscillate, or hit the power rails, and then you get a slightly different frequency. – George Herold Feb 09 '16 at 16:13

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Both of the other answers are correct, but you might find the numbers instructive.

Let's start at the output of the right-hand capacitor. It feeds a very high-impedance load, and the phase shift across the cap is, as you suggest, about 67 degrees (67.8, actually).

The next cap to the left is a bit different. Its output feeds the CR network just mentioned, and this affects the phase shift. Its phase shift will be 56.3 degrees.

The left-hand cap is like the middle one, but even more so, since it has 2 RC sections loading down the output, although the right-hand section doesn't have much added effect. The phase shift across this cap is 55.8 degrees.

The total phase shift across the 3 caps is then $$\phi =55.8+56.3+67.8 = 179.9\text{ degrees} $$ which is close enough, given rounding errors in the 3 component phase shifts.

So the formula given in the link works, and your concern was unfounded. Your impedance calculation was correct, but it did not take into account the complexity of a multi-stage network. The web site you linked did.

EDIT - A slight correction. "The web site you linked did." is only partly correct. The overall calculations of phase shift were performed correctly, but the illustration was sloppy. This probably results from the web page author never having actually calculated the phase shifts, but rather he depended on the formulas. He probably simply took it for granted that the individual phase shifts were equal, which is the starting point for the calculations. Sloppy work, but this sort of thing is fairly common for information presented for beginners. Conceivably the author knew that the phase shifts are different, but felt that tossing this in would complicate the principles too much. The technical term for this sort of teaching exposition is "Lies-to-children".

WhatRoughBeast
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    I think, the formulas as given in the linked document do NOT work for the shown circuit. The author has mixed two different circuit principles. – LvW Feb 09 '16 at 16:53
  • I prefer the term [Wittgenstain's ladder](https://en.wikipedia.org/wiki/Wittgenstein%27s_ladder) *chuckles* –  Feb 09 '16 at 23:23
  • I was able to correctly calculate the phase shift of 67.8 degrees for the first stage, but I'm having trouble calculating the phase shift for the next two stages. Could you share how you computed those values? – gsgx Oct 13 '21 at 01:46
  • I've asked about computing the shift for the other stages here: https://electronics.stackexchange.com/q/674222/5546 – gsgx Jul 16 '23 at 21:35
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No - it is not the classical phase-shift oscillator. The resistor from the inv. input to ground makes no sense. It must be replaced by a series resistor (in series with the last C) in order to allow a gain of app. -29 (or slighly larger).

(By the way: The referenced internet link contains more similar errors. Don`t blindly trust Internet sources).

However, the circuit can oscillate based on another principle: The opamp works as an inverting "Differentiator" (counterpart to an integrator) with a - more or less - constant phase shift of -90 deg. Both C-R chains must produce +90 deg at the desired oscillation frequency. But it must be emphasized that an oscillator with R-C lowpass blocks (instead of C-R high pass elements) - together with an inverting integrator - has much better properties (noise!).

EDIT: The oscillation frequency is w=1/(RC*SQRT(3)) and the oscillation criterion is fulfilled for Rfeedback=12*R.

EDIT2: Here is the loop gain:

T(s)=s³R²RfC²/(1+s*4RC+s²*3R²C²) ; Rf=feedback resistor.

Setting s=jw you find the oscillation frequency for T(s)=real (which means: Denominator must be imaginary, because the numerator is imag.).

EDIT3: In general, there are 4 different types of phase-shift oscillators:

  • Three C-R stages with a negative gain stage
  • Two C-R stages and an inverting differentiating stage (your case)
  • Three R-C stages and a negative gain stage
  • Two R-C stages and an inverting integrator.

It has been prooved that the last two alternatives are best.

LvW
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  • you mean Both C-R chains must produce +60deg.. do you? – Michael George Feb 09 '16 at 16:38
  • And the opamp works with phase shift of 180 deg? – Michael George Feb 09 '16 at 16:45
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    No - as I have mentioned, the inverting differentiator produces -(+90)=-90 deg and both C-R elements must produce together +90 deg for a total loop phase shift of 0 deg. – LvW Feb 09 '16 at 16:49
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    Michael - for your information: Theoretically, a differentiating circuit produces +90deg phase shift. here we have an inverting circuit with -180de+90 deg=-90 deg. Because we need a TOTAL phase shift of 0 deg, both C-R stages contribute TOGETHER +90deg (not each +45 deg. because they are not decoupled from each other). I can give you the loop gain formula (for the open loop) under EDIT2. – LvW Feb 09 '16 at 16:59
  • Thank you very much for the answer. It is really helpful. You were correct when you said: "The author has mixed two different circuit principles". I simulated the circuit, your frequency equation and oscillation criterion works. The author's equation does not work with his oscillation criterion. Thank you again for your answer, You are wonderful !! – Michael George Feb 09 '16 at 18:12
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    With or without the resistor between the inv. input and ground? It should work in both cases. The resistor is not necessary - but it can stabilize the opamp block if the phase margin is too small. – LvW Feb 09 '16 at 18:30
  • With the resistor between the inv. input and ground. It was like the circuit shown in the question. – Michael George Feb 09 '16 at 19:43
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    OK - no problem. But you can remove this resistor. Remember the principle of "virtual ground". A resistor between "virtual" ground and "real" ground has no effect on the oscillation. But it can improve the phase margin as mentioned. – LvW Feb 09 '16 at 20:21
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You are correct that the phase shift of a single RC stage using 1nF and 10kΩ is 67° at 6.5kHz.

If you look at the same page you linked to, you'll notice a formula that does give the correct result: $$f=\frac{1}{2\pi RC\sqrt{2N}}$$

I think your question is equivalent to asking why the above formula is not simply $$f\stackrel{?}{=}\frac{1}{2\pi RC \cdot N}$$

The reason is that when you cascade more than 1 RC stage together, the impedance of the following stages loads down the preceding stage. The simple phase delay formula only applies when driving a high impedance (e.g., an op-amp input). I won't work out the math but you could start from the formula in Wikipedia's phase-shift oscillator article.

Justin
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It can't be a 203° phase shift oscillator.
No matter what R and C values you have it always will stay a 180° phase shift oscillator.

If the frequency value doesn't result in 180° phase shift for the given values of Rs and Cs then it's not the phase shift value that is wrong but the frequency. The 180° determins the frequency. Not the frequency determines the phase shift.
I guess the 6.5kHz value is just an approximation.

According to the Barkhausen stability criterion if an amplifier with positive feedabck oscillates stable the phase shift is always 360° (or a multiple).
In this circuit 180° are already contributed by the inverting amplifier so that 360° - 180° = 180° necessarily have to be contributed by the triple RC-network.

Curd
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Yes it is properly designed.

The actual theoretical values are irrelevant, you can calculate a Zc of 24485.37 ohm but since a typical capacitors and resistors have 5% tolerance on their value, being so precise is meaningless.

The design toplogy of the circuit will make sure that it oscillates. For sure not exactly at the 6.5 kHz you calculated but at the frequency where the total phase shift (excluding opamp) is 180 degrees because that is what makes it oscillate.

Bimpelrekkie
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    I rather think, the total phase shift must be 0 deg (resp. 360 deg) for a circuit to oscillate. And the (inverting) differentiator can contribute only -90 deg. – LvW Feb 09 '16 at 16:55
  • You're forgetting that the opamp adds 180 degrees as we're feeding back on the inverting input. – Bimpelrekkie Feb 09 '16 at 20:00
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    No - I didn`t forget the inverting property of the opamp. The opamp works not as an amplifier but as an inverting differentiating circuit with a phase shift of -90deg. The rest of the necessary phase shift (for a total of zero) is provided by the first two C-R stages. – LvW Feb 09 '16 at 20:18