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Message started by raja.cedt on Sep 5^th, 2008, 2:59am

Title: ring oscillator
Post by raja.cedt on Sep 5^th, 2008, 2:59am

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

Title: Re: ring oscillator
Post by nano_RF on Sep 5^th, 2008, 10:19am

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

Hi,

This is pretty easy to visualize. Just take an Example of two inverter configuration. Start out with logic 0 at INV1 input --> this gives a logic 1 at INV2 input --> this gives a logic 0 at INV2 output and feded back to INV1 input which was already sitting at logic 0. So nothing fancy about it. It will latch to DC level due to complete 360 / 0 phase shift.

Regards

Title: Re: ring oscillator
Post by buddypoor on Sep 5^th, 2008, 12:19pm

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

..........
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

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

Title: Re: ring oscillator
Post by Frank Wiedmann on Sep 7^th, 2008, 11:23pm

I think that for the ring oscillator, the solution at some frequency might well be stable. However, very special initial conditions are required to reach this state, so in practice the ring oscillator will always latch to the dc solution.

Title: Re: ring oscillator
Post by raja.cedt on Sep 8^th, 2008, 11:08pm

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.

Title: Re: ring oscillator
Post by Frank Wiedmann on Sep 8^th, 2008, 11:37pm

I don't think that noise will be of much help in starting a ring oscillator with an even number of stages. The latched state at dc is very stable and the circuit will converge to this state from almost all initial conditions, so a small amount of noise is unlikely to change this.

Just try for yourself if you find any way to start up such an oscillator in simulation (for example by turning on the power in a certain way or by applying external pulses). It will probably be extremely difficult and it won't be any easier with a physical circuit.

To reliably start up an oscillator, you have to make the dc state unstable. A possible way to do this is shown in http://www.uni-stuttgart.de/int/institut/mitarbeiter/MA_Publikationen/groezing/2003_ESSCIRC_Estoril.pdf.

Title: Re: ring oscillator
Post by raja.cedt on Sep 9^th, 2008, 1:37am

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?

Title: Re: ring oscillator
Post by Frank Wiedmann on Sep 9^th, 2008, 7:18am

I don't know how to prove analytically my statement that most initial conditions will converge towards the dc state, but if you just try for yourself as I suggested, you will quickly see that it is true.

As mentioned in the article, one of the advantages of a ring oscillator with an even number of stages is that you can get quadrature signals from it.

Title: Re: ring oscillator
Post by Tlaloc on Sep 10^th, 2008, 8:54pm

Quote:

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?

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)?

Title: Re: ring oscillator
Post by loose-electron on Sep 11^th, 2008, 11:16am

Some items here:

1. Configuration: A ring structure with an even number of inverters in the ring is NOT an oscillator. Instead, it is a two state latched structure. Thats true for 2,4,6,8...... inverters.

2. Bias point vs. oscillation: For a ring structure with 3,5,7,9.... inverters, in a simulation it will generally find (and stay at) a VDD/2 steady state point and not oscillate with everything ideal. A quick fix for this is to put an IC (IC! not nodeset!) on one node in the ring at 0V or VDD, and that will get things started. In Ken K's book (desingers guide to spice and spectre) whats going on here gets covered in more detail.

3. A single inverter, in a ring configuration is not a ring oscillator, or a latch, it's two diode connected MOS devices connected between power and ground. Look at the circuit down at the transistor level and it will become evident there.

Title: Re: ring oscillator
Post by thechopper on Sep 11^th, 2008, 7:40pm

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