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Feedback and frequency equation loop of ring oscillator

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DimaKilani

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Hello,

I got a question on what type of feedback loop does the ring oscillator (made of odd number of inverter) have? it is positive or negative feedback? and why?
I addition, the frequency of the ring oscillator= 1/2Nt. I know N is the number of inverters and t is the delay for each inverter. but where does number "2" come from?
Also is the relationship between input and output of inverter is 180 degree phase shift?

Regards,
Dima
 

Re: feedback and frequency equation loop of ring oscillator

In a ring oscillator the output is applied to the input after a given delay. The ouput signal is not summed to an input signal, then we cannot speak of positive or negative feedback.

If we have N inverters, the output will be put at the input pin after "t" seconds. Let's suppose this is the transition 0 --> 1. We need another "t" seconds to have the transistion 0 --> 1. This means the "1" state duration is "t" second and the "0" state duration is "t" seconds again, so the period T=2*N*t and f=1/T=1/2*N*t
 

Re: feedback and frequency equation loop of ring oscillator

An odd number of inverters has its low frequency output out-of-phase (180 degrees) with its input so I call it negative feedback.
But at high frequencies the inverters have phase shift and at the frequency the phase shift is 180 degrees then it oscillates so I call it positive feedback.
 

For as stable operating point we need negative feedback (180deg phase shift) for DC (therefore: an odd number of inverters).
The oscillation criterion reqires positive feedback - hence, oscillation occurs at a frequency which makes an additional phase shift of 180deg (due to added time delay effects).
 

The ring oscillator works with an odd number of inverters because if input is 1 then output is 0 and viceversa. In case of even number if input is 0 then also output will be 0, and if input is 1 also output will be 1. If you insert and additional 180 deg. phase shift the oscillation stops.
At T=0 (when the power supply is switched on) we start with 0 at the output, that is 0 at the input of the first inverter (this two points are connected togheher). After N*t seconds all inverters of the chain will have commutated according to the input, that means the output will be now 1. Then the input will be forced to 1, but the output will be switched to 0 only aftet N*t second will be elapsed.
So it's difficult to define this kind of feedback as positive or negative. The oscillator works based on time delay between input and output and the strong non linear behaviour of the inverters
 

The ring oscillator works with an odd number of inverters because if input is 1 then output is 0 and viceversa. In case of even number if input is 0 then also output will be 0, and if input is 1 also output will be 1. If you insert and additional 180 deg. phase shift the oscillation stops.

Sorry - but the opposite is true.
As we know - the frequency of oscillation is fo=1/(2*n*Td) with n=number of stages and Td=delay time per stage.

The ring oscillator does work because of the time delay properties only.
The total time delay determines the oscillation frequency and the given formula represents the case of an additional phase shift of 180 deg within the loop (in addition to the 180 deg due to an odd number of stages).
Thus, the given formula represents and fulfills Barkhausens oscillation condition:
Negative feedback for DC and positive feedback (360 deg phase shift) at one frequency fo only.
 

Possibly I didn't understand correctly what you said.

I'm going to try to explain better (I hope) what I want to say.

We have logical gates, so we will have only 0 deg phase or 180 deg. phase (360 deg = 0 deg.), then considering in input Vin=logical "0":

a Vo= "0" will have 0 deg. phase (with respect to the input)
a Vo= "1" will have 180 deg. phase

the opposite considering Vin= "1". Then each inverter has a phase shift of 180 deg. and a time delay "t".
This means:

let's suppose to have an inverter with the input set to "0", it will assert to its output "1". If now I put "1" at the input, for a time "t", also the output will be "1". After the time "t" has passes, the the output will switches to "0". If I connect directly this output to the input I'll have oscillation of period T=1/2t because I force the input to the output level, then I force the output to change logical state after a time "t"
If instead I connect an additional 180 deg., between the output and the input (that is in the feedback path), I'll have "0" (or "1" depending from the starting level) at both input and output so no oscillation cannot occour.

Barkhausens oscillation condition (considering unity feedback) are gain(w)>=1 and phase(w)=180 deg. They are fullfilled with no additional phase shift.
 

Barkhausens oscillation condition (considering unity feedback) are gain(w)>=1 and phase(w)=180 deg. They are fullfilled with no additional phase shift.

No - the Barkhausen condition requires a loop gain of unity (resp. somewhat larger) and a phase of zero (360) deg.
 

I agree with you that my last sentence about the Barkhausen condition is not correct.

However I'm not sure you can describe this circuit in terms of negative or positive feedback.
Even referring to the modified Barkhausen condition (that works properly f.i.. in wien bridge oscillator), I cannot see here a way to apply it.
The gates are always working in saturation. We have a hard non-linear function and a delayed feedback

Could you suggest me a reference in which this topic is treated in a rigorous (math) way ?
 

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