Re: How does LC phase shift oscillator work?

Hi,

I recall that for oscillation __all__ these condition must be fulfilled at some (sinusoidal) frequency:

a) Phase of loop gain is an integer multiple of 360º

b) Module of loop gain is greater than 1

c) Phase vs. frequency has negative slope.

Regards

Z

Re: How does LC phase shift oscillator work?

Quote:

Originally Posted by

**zorro**
Hi,

c) Phase vs. frequency has negative slope.

Hi Z

Why is that?

Re: How does LC phase shift oscillator work?

That necessary condition that makes that a linear system becomes a quasi-sinusoidal oscillator (i.e. **Phase vs. frequency must have negative slope**) is frequently overlooked.

Let's try an intuitive, non rigorous justification:

Imagine a linear circuit with an input and an output. Supose thet we connect a sinusoidal generator at the input and we see that at a particular frequency the ouput is identical (it has __exactly the same amplitude and phase__) to the input. If now we suddendly connect the output (instead of the external generator) to the input closing the loop, we can guess that the circuit continues in the state of self-sustained oscillation. The condition that output=input is met if the gain loop is 1, i.e., at that particular frequency the amplitude of the transfer function is 1 and phase is an integer multiple of 360°. This constitutes the well-known Barkhausen criterion for stability. (Footnote 1).

Now, suppose that for some reason there is a perturbation like an instantaneous small increase of the phase lag (equivalent to increasing delay) in the active device. Then, the frequency of that self-sustained oscillation has tendency to be lower (delay increases -> longer period). If at this new slightly lower frequency the lag increases even more, the frequency of the self-sustained oscillation decreases even more, going towards a state when self-sustained oscillation is not possible because the condition A(f)=1 is no longer met (Footote 2).

The situation described (phase lag is lower at a lower frequency) corresponds to a positive slope d(phi)/d(freq) in the characteristic of phase vs. frequency.

Something analog to this happens if the perturbation is a decrease in the phase lag and the slope is positive: the frequency of self-sustained oscillation tends to increase, moving away up to a condition where eventually it ceases (Footote 2).

Instead, if the slope d(phi)/d(freq) is negative, the effect of the perturbations is corrected by the network itself: the variation of phase vs. frequency has the slope needed for restore the original frequency of self-oscillation, where A(f)=1 is met.

In other words: if the slope d(phi)/d(freq) is negative, the oscillation has a stable equilibrium point. Otherwise, the equilibrium is unstable.

The analogy with the mechanical equilibrium of a sphere subject to gravity over a surface may be useful. If the surface is convex, there is a point of equilibrium but it is unstable. The surface must be concave in order to make stable the equilibrium.

I hope this justification sounds satisfactory. In my opinion, a more rigorous treatment including conditional stability requires the Nyquist criterion.

Regards

Z

Footote 1: if |A|>1, nonlinear effects limit the amplitude of the oscillation. The *descriptive function* is an useful tool for this. Thinking in self-sustained oscillation, we cuold say that for __some amplitude and frequency__ the output is *essentially* the same as the input.

Footote 2: or another stable equilibrium point at a different frequency is reached.

Re: How does LC phase shift oscillator work?

Quote:

Originally Posted by

**zorro**
Hi,

I recall that for oscillation __all__ these condition must be fulfilled at some (sinusoidal) frequency:

a) Phase of loop gain is an integer multiple of 360º

b) Module of loop gain is greater than 1

c) Phase vs. frequency has negative slope.

Regards

Z

In fact, condition c) as mentioned by Zorro is an absolute requirement that must be added to the well-known Barkhausen condition (see a) and b)).

It is surprising that this part of the (necessary) oscillation condition cannot be found in related books and papers - with one exception:

Randall W. Rhea: Discrete Oscillator Design (2010).

It is easy to show (by simulation) that circuits not meeting condition c) do not oscillate - although the loop gain is unity and a) and b) are fulfilled.

Re: How does LC phase shift oscillator work?

Hi all,

In my previous post I made the following statement:

Quote:

Originally Posted by

**zorro**
Now, suppose that for some reason there is a perturbation like an instantaneous **small increase of the phase lag** (equivalent to increasing delay) in the active device. **Then, the frequency of that self-sustained oscillation has tendency to be lower** (delay increases -> longer period).

It was observed that what I've highlighted needs a justification. I'll try to explain it now.

It can be useful to think in the pictorial view of a phasor. Let's take the possible self-sustained oscillation at angular frequency ω_{0} (where the two first conditions are met) as the reference for the phasor.

Let's consider first a perturbation producing a small __advance __in the phase. It can be a small increase of the phase lead, or added noise with a positive quadrature component. The phase advances some amount (say δφ) taking for it some time (say δt). This counterclockwise rotation of the phasor corresponds to an increase in angular frequency by an (average) amount Δω=δφ/δt .

If the transfer function of the loop has a positive phase vs. frequency slope, then at ω_{0}+Δω it has a lead greater than at ω_{0}, and the phasor advances its phase even more and more. Frequency increases moving away from the point of (possible) self-sustained oscillation.

If the perturbation produces (like in the example of the quote) a phase lag instead of an advance, then δφ and consequently Δω are negative, and the system goes away with a clockwise rotation of the phasor.

Instead, if the slope dφ/dω is negative, the phase characteristic makes that the phasor goes always towards its equilibrium point.

We can see the analogy with an inverted pendulum that losses its equilirium point for any small perurbation acting in any sense, and a regular pendulum that has a stable equilibrium.

I hope this is more clear now.

Regards

Z