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### Network # Phase shift oscillator help

#### george2525

Jan 30, 2015
170
would you not want R2 = 8R if possible?

if it was 9R for example, would the circuit not get unstable because of loop gain building up?

#### LvW

Apr 12, 2014
604
would you not want R2 = 8R if possible?
if it was 9R for example, would the circuit not get unstable because of loop gain building up?

No - for two reasons:
1.) The gain must be somewhat larger than the theoretical value (8). Otherwise, the oscillator would not start safely.
2.) It is not possible to have R2=8R because of tolerances. For R2<8R the circuit will not oscillate and for R2>8R it will safely start - however, with rising amplitudes. Therefore, hard limiting (clipping) or soft-limiting (diodes or other non-linear mechanisms).

#### LvW

Apr 12, 2014
604
If you are in the mood LvW, and have a fast method to derive this one that would be nice ;-)
Regarding your circuit without buffering:
Yes - everything correct.
The numerator of the loop gain function is imaginary.
Hence, also the denominator must be imaginary - in this case the ratio of both is real (requirement for oscillation).
Therefore, we set R(denominator)=0 and solve for w.

#### george2525

Jan 30, 2015
170
I was wondering if there was a fast way to derive T(s)?

Like with the buffered one it became easy because once we consider one stage/cube it and then put the gain in there we dont need to use circuit analysis

do you know of a trick for non-buffered RC networks in sequence?

#### LvW

Apr 12, 2014
604
I was wondering if there was a fast way to derive T(s)?
Like with the buffered one it became easy because once we consider one stage/cube it and then put the gain in there we dont need to use circuit analysis
do you know of a trick for non-buffered RC networks in sequence?

No - I dont know any "trick". I am afraid, you have to apply one of the classical methods (equations for node voltages or sequence of loaded voltage dividers,..) . It is a bit involved, indeed...

#### Ratch

Mar 10, 2013
1,098
Hi Ratch - yes, here are the results of my "fresh start:":

With reference to your post#14 (calculation in blue):
Third line from the bottom:
* Left side of the equation is imaginary and right side is complex (imag. plus real).
This requires to set the REAL part of the right side equal to zero.
This would immediately lead to w=1/[SQRT(3)*RC].

Not so, see below.

* Alternative: Continuing your approach:
Shifting the left imag. part to the right side of the equation, we get: 0=Im + Re.
In the second line from the bottom you have set all the imaginary parts equal to zero (Im=0).
The result is given in the last line of your calculation.

Yes, I did. See below.

* However, this is only the "first half of the story“.
Because the left side is zero you also must set Re=0.
Hence, we get as a second equation again: w=1/[SQRT(3)*RC].

Not so, see below.

* Finally, both equations must be fulfilled at the same time (because of 0=Re+Im)).
Equalizing both w expressions gives the condition: R2=8R. After inserting this requirement into
your w expression, we again get: w=1/[SQRT(3)*RC].

Yes, my last equation in post #14 is correct, See below.

* Comment: To me, it was clear from the beginning that w cannot depend on R2 because only the
CR sections determine the phase properties of the loop gain. So - how can we interpret your
result (involving R2) - even if it is only 50% of the calculations?
If you - from the beginning - replace the ratio R2/R by the gain A you will see that A will be not
part of the imaginary part and, hence, R2 would not appear as part of your result.
Why this surprising fact?
Because the resistor R at the inv. input node of the opamp plays a double role: It is part of the
CR network and, at the same time. part of the gain determining resistor ratio. Therefore, it is
good to use the term A instead of R2/R (and not to combine this ratio with the other R terms).

* Final comment: The oscillation frequency is determined by the time constant RC only.
That means: The resistor R2 determines the gain only with the requirement: R2>8R.
Of course, the oscillation amplitude will rise until it is limited by the supply voltage or by any

OK, here is my take on this. I set up an equation based on the assumption that the closed loop gain of the circuit had to be at least 1 in order to oscillate. I don't think you have any objections to that. Then I solved for omega based on the assumption that the phase was 0 at the frequency of oscillation. I don't think you have any objections to that either. Now, you assume that the real part of that equation must also be zero. There is no way that a positive real omega number is going to satisfy both Im=0 and Re=0 at the same time. Perhaps a complex number will, but that is not what we are looking for. For a unity feedback circuit like this, there is no reason to explore Re=0. The equation was already derived on the basis that the closed loop gain is 1. If you assume that a gain of 2 instead of 1 in the equation derivation, you still get the same w=1/[SQRT(3)*RC] for real terms, but a different omega value from the imaginary terms. So , I aver that my last equation for omega, assuming Im=0, in post #14 is the correct one for omega in this particular phase-shift circuit. And, R2 affects both omega and the gain. Furthermore, requirements like im+Re=0, or Re=0 are bogus.

Ratch

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#### Ratch

Mar 10, 2013
1,098
If you are in the mood LvW, and have a fast method to derive this one that would be nice ;-)

I can do it but it takes ages ;-(

View attachment 26607

View attachment 26608

You got the gain and frequency. What else do you need? What is T(s)? There is a lot of textual material on how those equations are derived. Why do it again?

Ratch

#### george2525

Jan 30, 2015
170
i agree its a waste of time. I just wish my circuits examiner felt that way. when they expect me to derive stuff like that in 10mins or so its hard not to make a mistake. Its not something I do for fun

#### george2525

Jan 30, 2015
170
yeah thats my take on it too. the numerator will always be 90 degrees no matter what R2 is so I cant see how it changes the w required for 90 down below

#### Ratch

Mar 10, 2013
1,098
Doesn't the factorization:

View attachment 26631

show that r2 is not a frequency determining component? It's only involved in the gain constant.

No so. In my post #14, I show that the term you mentioned is equated to 1, so R2 cannot be factored out. It is included in the equation for omega.

Ratch

#### Ratch

Mar 10, 2013
1,098
yeah thats my take on it too. the numerator will always be 90 degrees no matter what R2 is so I cant see how it changes the w required for 90 down below

Didn't I answer that concern in post #28 ? Didn't you understand my answer?

Ratch

#### The Electrician

Jul 6, 2012
117
This is from your post #14, showing r2 being factored out of the expression for t; r2 is involved in the 180 degree phase change, but its effect doesn't change with frequency, so the frequency of oscillation of the circuit won't depend on r2. Whether it oscillates at all will depend on r2, but not the actual frequency of oscillation.

Let's choose some values for components and plot the magnitude and angle of t. Let r=1000 ohms, c=.01 uF, and r2 = 8000 ohms for the blue plot, r2 = 16000 ohms for the red plot.

The magnitude changes when r2 changes from 8000 to 16000 ohms, but the phase angle does not change; the red and blue plots of phase are on top of each other: #### Ratch

Mar 10, 2013
1,098
This is from your post #14, showing r2 being factored out of the expression for t;

View attachment 26635

r2 is involved in the 180 degree phase change, but its effect doesn't change with frequency, so the frequency of oscillation of the circuit won't depend on r2. Whether it oscillates at all will depend on r2, but not the actual frequency of oscillation.

Let's choose some values for components and plot the magnitude and angle of t. Let r=1000 ohms, c=.01 uF, and r2 = 8000 ohms for the blue plot, r2 = 16000 ohms for the red plot.

The magnitude changes when r2 changes from 8000 to 16000 ohms, but the phase angle does not change; the red and blue plots of phase are on top of each other:

View attachment 26636
Look at the 3rd line from the bottom of the attachment which says t = 1

Ratch

#### The Electrician

Jul 6, 2012
117
What about the plots of phase vs. value of r2 shown in post #53, showing that the value of r2 has no effect on the phase vs. frequency of t?

#### Ratch

Mar 10, 2013
1,098
What about the plots of phase vs. value of r2 shown in post #53, showing that the value of r2 has no effect on the phase vs. frequency of t?
You are plotting a term called "t". You should be plotting the equation of " t = 1" .

Ratch

#### The Electrician

Jul 6, 2012
117
I don't know how to plot an equation. I only know how to plot a mathematical expression, such as the t of the left side of that equation.

The relevant thing to plot here is the loop gain--the gain AB, what you called
"the output of A1*A2*A3" which is also t. When the phase shift of a signal traversing the full path is 360 degrees, that's the frequency where oscillation will take place, if the magnitude of the gain is suitable.

The phase shift experienced by a signal traversing that path does not change with changes in r2, so r2 has no effect on the frequency of oscillation.

Perhaps you could try a simulation of the circuit in question to see if r2 affects the frequency of oscillation.

Here's further mathematical verification that the phase shift of t does not vary with r2: #### Ratch

Mar 10, 2013
1,098
I don't know how to plot an equation. I only know how to plot a mathematical expression, such as the t of the left side of that equation.

Ok, it is easy to do, but upon thinking about it, I don't think it proves anything to look at the absolute and argument values of all the terms gathered on the right hand side or left hand side.

The relevant thing to plot here is the loop gain--the gain AB, what you called
"the output of A1*A2*A3" which is also t. When the phase shift of a signal traversing the full path is 360 degrees, that's the frequency where oscillation will take place, if the magnitude of the gain is suitable.

Yes, and the freq of oscillation where the phase is 0 can change when the r2 changes.

The phase shift experienced by a signal traversing that path does not change with changes in r2, so r2 has no effect on the frequency of oscillation.

I beg to disagree. I calculated omega in terms of R2. You should prove my derivation wrong. Then I would be convinced.

Perhaps you could try a simulation of the circuit in question to see if r2 affects the frequency of oscillation.

Waste of time. Better to check my derivation.

Here's further mathematical verification that the phase shift of t does not vary with r2:

View attachment 26637[/QUOTE]

Tilt! You are banging on that single "t" term again. I aver that the proper way is to evaluate t = 1, and then find the omega where the imaginary terms disappear. Since R2 is a coefficient of one of the imaginary terms, it has a frequency determining function. Either describe where my method is wrong, or show that my calculation is wrong.

Ratch

#### LvW

Apr 12, 2014
604
Ratch - sorry, but you are not on the right track in this case.

General (preliminary) remark: If for any oscillator the frequency determining components are given (as in our case) there are still two unknowns to be determined: (1) The oscillation frequency wo and (2) the required gain A of the active deviice. And we have ONE equation to be fulfilled: The oscillation condition which is:
Unity Loop Gain at w=wo; T(jwo)=1.
Hence, we need TWO equations for finding both unknowns.
These two equations are derived from the real and the imaginary part of the loop gain:
Re(T)=1 and Im(T)=0.
Of course, the first equation is identical to R(T)-1=0.
As for each system with two unknowns, both equations (conditions) must be combined (fulfilled) at the same time.
That means: If we find two different expressions for wo from the two equations, both must be equalized to find the final and correct result for wo as well the gain A.

OK, here is my take on this. I set up an equation based on the assumption that the closed loop gain of the circuit had to be at least 1 in order to oscillate. I don't think you have any objections to that.

No - with one exception: It is the loop gain (not the closed-loop gain) which must be unity.

Then I solved for omega based on the assumption that the phase was 0 at the frequency of oscillation. I don't think you have any objections to that either.

No - of course, not - because this gives the first of the two necessary equations.

Now, you assume that the real part of that equation must also be zero. There is no way that a positive real omega number is going to satisfy both Im=0 and Re=0 at the same time.

No, I dont "assume" this. The equation Re(T)-1=0 is the second part of the oscillation condition (our second equation for solving the two-unknown system).

Perhaps a complex number will, but that is not what we are looking for. For a unity feedback circuit like this, there is no reason to explore Re=0. The equation was already derived on the basis that the closed loop gain is 1.

No, that is not true. Setting only the imag. part to zero contains not yet any condition for the real part.

If you assume that a gain of 2 instead of 1 in the equation derivation, you still get the same w=1/[SQRT(3)*RC] for real terms, but a different omega value from the imaginary terms.

Yes - and exactly this is the reason we cannot rely on only one of the different results obtained from the two equations. We have to combine (equalize) both results. Only then we find the two unknowns (wo and A).

So , I aver that my last equation for omega, assuming Im=0, in post #14 is the correct one for omega in this particular phase-shift circuit. And, R2 affects both omega and the gain. Furthermore, requirements like im+Re=0, or Re=0 are bogus.
Ratch

No - the result of both equations [Re(T)=1 and Im(T)=0] as well as all simulations clearly show that the oscillation frequency wo is solely determined by the frequency-determining RC-network (watch the commonly used name of this network!) and not by the amplifiers gain A.
By the way, this is the reason we are allowed to automatically reduce the gain using a non-linear gain determining element (diodes) for rising amplitudes. Otherwise, such an amplitude control would also influence the frequency and the method would be totally "obsolete".

EDIT: Ratch - here is another interesting point which prooves that solving only one of the two equations is not sufficient:

* You have set Im(T)=0 and solved for w. Now - would you expect that inserting your w value into the expression for T would result in a pure real function? This would be the case if your result (last line post#14) is correct.

* However, after inserting your result into T does NOT give a pure real function. The function still is complex with an imaginary numerator and a complex denominator (Re + Im). This clearly indicates that your formula for w in post#14 is not the final solution. Instead, it gives only one of two conditions which must be fulfilled for T(jw) real (Im=0). Hence, we must make use of the second equation which gives A=R2/R=8.

* Only after setting R2=A*R=8R in the denominator gives Re=0 in the denominator and all the "j" (or "i") cancel each other so that the function T(jw) is - as desired - a real function (with T=1 for A=8).
________________________
Sorry for the long answer.
(And I am off for one week now)

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