# Analysis of 3 stage common source ring oscillator

I have been attempting to analyze the following common source ring oscillator: I am having some trouble here. Based on my understanding, each stage of the FETs provides 270 degrees of phase shift total. 180 from the inverting common source, and 90 from a pole due to the capacitor. In order to oscillate we need 360 degrees total. With only two stages of the CS FETs, the necessary phase shift for oscillation is there but would be stable at either rail. Why would it still be stable?

Now, at 3 stages, the circuit oscillates provided the gain of each stage is greater than or equal to 2. Can someone explain this as well? I do not understand why it is 2 instead of one, nor can I seem to understand the math behind this.

For a stable DC bias point we need at zero Hertz a phase shift of -180deg. This is ensured because we have three inverting stages in a closed loop (number n of stages is odd).

According to the Barkhausen criterion we need in total -360 deg (identical to zero deg) for a certain frequency. For this purpose, we have three 1st order lowpass stages - each with a cut-off frequency at app. $$\wc=1/R_DC_L\$$.

Hence, the oscillation criterion is fullfilled for another additional phase shift of -180deg at the desired frequency wo.

That means: Each stage must contribute -60 deg. - and the oscillation frequency is somewhat larger than the cut-off (which gives only -45 deg).

Comment: The mentioned -90deg per stage is achieved at infinite frequencies only.

• I see, thank you, I understand this now. Can you explain why the oscillation frequency is someone larger than the cutoff? How do I find this oscillation frequency? Additionally, according to the Barkhausen criterion, the loop gain should be somewhat greater than 1 to ensure oscillation. Why is it that this circuit requires the loop gain to be greater than or equal to 2? (This was stated in the section on 'Ring Oscillators' in the book I am reading, but gave no explanation as to why).
– rkj
Dec 2 '18 at 21:53
• At the 3dB cut-off frequency wc the phase shift is -45 deg (well-known fundamental fact). Then, it is clear that for a required phase shift (per stage) of -60deg the frequency is larger than wc. Regarding the other question: I cannot comment a sentence in a book without knowing the surrounding text.
– LvW
Dec 3 '18 at 8:14

Additionally, according to the Barkhausen criterion, the loop gain should be somewhat greater than 1 to ensure oscillation. Why is it that this circuit requires the loop gain to be greater than or equal to 2?

At the frequency of oscillation, each R-C network shifts the signal 60 degrees. Also at the frequency of oscillation, each C has an impedance that forms a voltage divider with its R. The three attenuations compound, and the circuit must provide gain to get the loop gain above 1.

A common configuration for a phase shift oscillator is one transistor or opamp with three R-C stages in a series between the collector and base (or output and input). In this circuit, the amplifying device must be configured for a voltage gain of 26 to compensate for the three R-C attenuators.