Does a ring oscillator work if the phase shift of the loop is not 180 degreee?

We know if the loop gain modulus of a negative feeback loop is larger than unity and its phase angle is 180, the circuit will oscillate, causing a very high input swing due to the ever-increasing signal loop. I wonder what if the modulus is larger than 1, but the phase angle is not 180. Will it still oscillate or end up in a static state?

Jasper Chow said:

We know if the loop gain modulus of a negative feeback loop is larger than unity and its phase angle is 180, the circuit will oscillate, causing a very high input swing due to the ever-increasing signal loop. I wonder what if the modulus is larger than 1, but the phase angle is not 180. Will it still oscillate or end up in a static state?

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Speaking about "loop gain" the phase angle necessary for oscillation is 360 deg (including the sign inversion for neg. feedback).
Regarding your question: Yes - the frequency of oscillation automatically adjusts so that the loop phase is exact 360 deg. (180 deg without the sign inversion) - provided the gain requirements are met.

LvW said:

Speaking about "loop gain" the phase angle necessary for oscillation is 360 deg (including the sign inversion for neg. feedback).
Regarding your question: Yes - the frequency of oscillation automatically adjusts so that the loop phase is exact 360 deg. (180 deg without the sign inversion) - provided the gain requirements are met.

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How does the frequency of oscillation adjust? we can find the frequency of unit gain in a Bode diagram, if at that frequency the phase shift is not 180 degree, the system is stable. But if they are not well mathced, no matter how you change the frequency, the sinosoid oscillation is impossible. My question's point is whether there can be an irregular oscillating signal?

I believe, the OP is not talking about "phase shift" rather than the number of signal inversions in a ring oscillator.

Besides fulfilling the oscillation condition as discussed by LvW, a ring oscillator must have a suitable DC operation point. In a standard ring oscillator, it's achieved by an odd number of signal inversions, resulting in an inverting overall DC gain. A non-inverting gain would result in a logic latch-up.

P.S.: As an additional remark, a ring oscillator is a non-linear circuit and can't be fully described by linear oscillator theory. To analyze the expectable output signal and find out if different transient solutions converge to a predictable, stable signal, non-linear analysis must be applied.

Jasper Chow said:

How does the frequency of oscillation adjust? we can find the frequency of unit gain in a Bode diagram, if at that frequency the phase shift is not 180 degree, the system is stable. But if they are not well mathced, no matter how you change the frequency, the sinosoid oscillation is impossible. My question's point is whether there can be an irregular oscillating signal?

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I know what you mean - however, read my remark "provided the gain requirements are met".
Let´s assume you are designing an oscillator for fo=1 kHz. For safe start of the oscillation you provide a loop gain of 1.2 at the desired frequency fo and a phase of -360 deg (0 deg).
Now let´s further assume that - due to some tolerances and other uncertainties (e.g. finite open-loop opamp gain) - the phase condition is met at 950 Hz.
If the loop gain at this frequency is still larger than unity the circuit will oscillate at 950 Hz. That´s what I call: automatic adjustment (because this is the only frequency that fulfills the Barkhausen condition).