hi ,
    can any body please explain why ring oscillator with even inverters will latch at dc only.I found some techniques to avoid latching at dc and make it oscillate .But my question is it has two stable points one is at dc ,another is at some frequency .Then which one is the operating frequency.
THANK YOU

raja.cedt wrote on Sep 5^th, 2008, 2:59am:


No, it has NOT two stable points. Only from the pure mathematical/theoretical point of view you can define two signals (one with zero Hz and one with a period)  which fulfill the input-output conditions, however only one of both leads in reality to a stable condition - and that´s f=0. After power switch-on the circuit has no reason to try another operating condition than at zero Hz.
We have same situation in pure analog/linear amplifiers: When you have an opamp with positive feedback (for low frequencies) there always will be a frequency which enables a negative feedback situation (for higher frequencies), but this second situation leads never to a stable operating point.
Another example is a circuit consisting of an opamp with suitable gain and a notch-filter. Although there is a frequency which theoretically can fulfill the oscillation condition, the circuit will never oscillate but will go, instead, into saturation.
(I hope I was able to express myself clear enough).
Regards

hi
i agree with franks  answer.but my question is how can you say it will latch at dc.Means it has two operating frequencies .It can choose any thing out of two. while starting of oscillations noise contains all frequencies so is there any chance of operating  at that frequency rather than dc.

hi frank,
             i am happy with your answer but
1.how can u say that dc is very stable may be because of high loop gain towards  dc
2.lets say oscillations  start at very high frequency (because of noise )so it will try to reach dc or low frequency mean while does it latch to second operating frequency?

3.Is there any advantage of even inverter ring oscillator when compared to odd inverter ring oscillator?

Quote:

I find this very unlikely.  If you ever reach the f=0 state, what could drive you out it?  Remember that you have one transistor off with the other acting as a switch in every inverter.  The noise would have to be so strong that it would have to jar the these devices out of their bias state.  This could only be a large signal source such as capacitive coupling of other switching signals nearby, not any thermal noise.

The only way that I could visualize a stable oscillation (with only a single feedback loop) is to have two pulse trains moving through the inverter chain at the same time, only delayed by each other.  If you kick-started the first train by a one-shot, for example, that one-shot would have to release the starting point before the first train came to the end of the chain.  If this condition is met, then you would then have two pulse trains going, out-of-phase and delayed from each other.  If the first train came to the end before you released the one-shot, the chain would be DC stable again, and no oscillations would result.

If what I described is accurate, than the design of the one-shot could be quite tricky.  Why bother when an odd numbered chain starts so reliably with only one pulse train going through it (half the power)?

I agree with loose-electron...and in few words

even number --> bistable multivibrator circuit: its name indicates it will be stable in either of the two possible states that can be externally forced.

odd number --> astable mutivibrator circuit: the name indicates is not stable in any of the two possible states, so it constantly switches between those two --> oscillator.


Cheers
Tosei