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### Network # Phase shift oscillator help

#### george2525

Jan 30, 2015
170
Oh I like that. where did you get the 0.5 from?

could you explain that mathematically?

Thats led me to a very easy way to find w aswell

Each RC stage = (jwRC)/(jwRC +1)

numerator = 90 degrees

therefore we want denominator to = 30 degrees to make that stage = 60 degrees

phase of denominator = tan^-1(wRC/1) = set to 30 degrees

therefore wRC = tan(30) = 1/sqrt(3)

w = 1/(sqrt(3)RC)

#### george2525

Jan 30, 2015
170
ok i got it after using my one to get w first

just plug in w = 1/(sqrt(3)RC) to a single stage and cube it

thats saved me a lot of trouble, cheers

#### Ratch

Mar 10, 2013
1,098
Thats for Resistors

sorry I wont use a computer for maths so you'l have to put up with my writing

Your equation for omega at oscillation is wrong. R2 is not in your formula, but it should be because it affects the zero phase shift. What rational can you give for giving the real parts of the expression any value? You need to set the imaginary parts of the loop equation to zero so there is no phase shift. The zero phase shift frequency determines the frequency of oscillation. The gain determines whether it will oscillate. Did you look at my derivation and find anything wrong with it?

Ratch

#### george2525

Jan 30, 2015
170
yes you somehow got R2 in the denominator for T(s) from what i can tell

Remember that its only the real part of the denominator of T(s) that cant have any real parts

the R2 is in the numerator so it doesnt affect the maths for w

#### Ratch

Mar 10, 2013
1,098
My calculation is rather simple:
Each CR section contributes 60 deg phase shift at w=wo.
At this frequency the loss of this simple HP stage is -6 dB (factor 0.5).
We have three decoupled CR sections - hence the total phase shift is - as required - 180 deg and the total loss factor is 0.5^3=1/8. Hence the required gain must be A=-R2/R=-8 (or slightly larger for a safe start of oscillations)
Yes, I agree with your analysis. Using Steinmetz notation, the gain of the first two amps is (1/4)/_-120°. The gain of the last amp is -R2/(R+sqrt(3)*R*j) . Multiplying the gains together and setting them to 1 gives [(1/4)/_-120°][-R2/(R+sqrt(3)*R*j)] from which R2/R = 8

Ratch

#### Ratch

Mar 10, 2013
1,098
View attachment 26610

From this point onwards, ignore the numerator = -R2(SRC)^3

just deal with the denominator = R(1+SRC)^3 and then set real parts of that thing to 0

solve for w

Nope, you still have not looked at my derivation at post #14. I use the same term, but I set it to 1, which is the gain of the amp chain. That means that R2 cannot be ignored, and is a frequency determining element because it changes the phase.

Ratch

#### LvW

Apr 12, 2014
604
Oh I like that. where did you get the 0.5 from?
w = 1/(sqrt(3)RC)
Yes - correct. Good approach starting with 30deg for the denomionator only.
Inserting this frequency into the first-order highpass function gives the magnitude of 0.5.

I dont know what to do with your post #23. All the formulas at the bottom are wrong.
Any open qestions?

#### LvW

Apr 12, 2014
604
Your equation for omega at oscillation is wrong. R2 is not in your formula, but it should be because it affects the zero phase shift.
Ratch
Hi Ratch - this is a somewhat "critical" point. Let me explain my view:
* The Barkhausen condition (unity loop gain) gives the ideal/theoretical component values - for the frequency determining feedback network as well as the required gain. The corresponding oscillation frequency is calculated based on this condition. It is the frequency for steady-state sinusoidal output signals
* For a safe start of oscillations at t=0 we must make the loop gain slightly larger than unity (perhaps 1.1 or 1.2). In our case, the amplifier gain A=-R2/R should therefore be realized with a resistor ratio of app. R2/R=8.1 or 8.2.
* It is a known fact that for each oscillator the frequency during the start-up phase (rising amplitudes) is not identical to the ideal steady-state frequency (resulting from the Barkhausen condition).
* Now - in order to avoid hard-limiting (clipping) of the rising amplitudes and to allow a sinusoidal waveform (with some minor distortions) it is common practice to introduce some kind of non-linearity into the circuit resulting in "soft-limiting" (amplitude control). This method brings the circuit back to the ideal oscillation behaviour at nominal frequency (the poles which are in the RHP during the start-up phase are shifted back to the imag. axis of the s-plane). This can be accomplished, for example, using two anti-parallel diodes across R2 (or across a part of R2).

NOW THE QUESTION: Are we interested in the frequency during the start-up phase? In this case, it is true that this frequency depends on the amount of "over-fulfilling" the oscillation condition (R2>8R). Therefore, the expression which gives this frequency (valid for the short start period only) contains R2. But the steady-state frequency of the continuously oscillating circuit with (nearly) sinusoidal signals is NOT dependent on R2 - provided the whole oscillator circuit was designed properly.

#### Ratch

Mar 10, 2013
1,098
Hi Ratch - this is a somewhat "critical" point. Let me explain my view:
* The Barkhausen condition (unity loop gain) gives the ideal/theoretical component values - for the frequency determining feedback network as well as the required gain. The corresponding oscillation frequency is calculated based on this condition. It is the frequency for steady-state sinusoidal output signals
* For a safe start of oscillations at t=0 we must make the loop gain slightly larger than unity (perhaps 1.1 or 1.2). In our case, the amplifier gain A=-R2/R should therefore be realized with a resistor ratio of app. R2/R=8.1 or 8.2.
* It is a known fact that for each oscillator the frequency during the start-up phase (rising amplitudes) is not identical to the ideal steady-state frequency (resulting from the Barkhausen condition).
* Now - in order to avoid hard-limiting (clipping) of the rising amplitudes and to allow a sinusoidal waveform (with some minor distortions) it is common practice to introduce some kind of non-linearity into the circuit resulting in "soft-limiting" (amplitude control). This method brings the circuit back to the ideal oscillation behaviour at nominal frequency (the poles which are in the RHP during the start-up phase are shifted back to the imag. axis of the s-plane). This can be accomplished, for example, using two anti-parallel diodes across R2 (or across a part of R2).

I agree with what you say about introducing nonlinearity to stabilize the oscillator.

NOW THE QUESTION
: Are we interested in the frequency during the start-up phase? In this case, it is true that this frequency depends on the amount of "over-fulfilling" the oscillation condition (R2>8R). Therefore, the expression which gives this frequency (valid for the short start period only) contains R2. But the steady-state frequency of the continuously oscillating circuit with (nearly) sinusoidal signals is NOT dependent on R2 - provided the whole oscillator circuit was designed properly.

Yes, the amplifier chain in this circuit has to amplify a little greater than 1 to get started. Then whatever nonlinearity you have built into the circuit can take over and stabilize the oscillator. My calculations as shown on post #14 show that R2 is a frequency determining component in addition to being an amplitude determining component. If R2 is to be 8 times R, then the equation shown in my derivation will determine the frequency. Can you find anything wrong in my derivation?

Ratch

#### LvW

Apr 12, 2014
604
Your equation for omega at oscillation is wrong. R2 is not in your formula, but it should be because it affects the zero phase shift.

Ratch - the gain of the inverting opamp does not affect the phase shift of the loop gain. We cannot "tune" the oscillator by varying R2. The circuit can continuously oscillate at one frequency only - and that is the frequency where each CR section contributes 60 deg phase shift.

"Can you find anything wrong in my derivation?"
I will try to follow your calculation.

#### Ratch

Mar 10, 2013
1,098
Ratch - the gain of the inverting opamp does not affect the phase shift of the loop gain. We cannot "tune" the oscillator by varying R2. The circuit can continuously oscillate at one frequency only - and that is the frequency where each CR section contributes 60 deg phase shift.

"Can you find anything wrong in my derivation?"
I will try to follow your calculation.

My derivation in post #14 shows that R2 affects both the frequency and the gain. Show me where I went wrong.

Ratch

#### LvW

Apr 12, 2014
604
My derivation in post #14 shows that R2 affects both the frequency and the gain. Show me where I went wrong.
Ratch
As you wrote earlier- "please bear with me" because I needed some time to go through your calculation.
But now I can give some comments:
No - you were not "wrong", but you didnt finish the calculation.
Let me explain:
From the oscillation condition (unity loop gain ) you have derived an equation T(s)=1 which must be fulfilled at one single frequency w.
For solving this equation you have identified the imaginary parts and - as a first step - you have solved the equation IMG(T)=0. As a result you have found a formula for the frequency fulfilling the requirement IMG(T)=0.
But this is only the first part of the calculation because you have not yet used the condition that the remaining real part of T(s) must be unity.
Hence, you must insert the identified frequency expression into T(s) and set Re(T)=1.
This will result in the requirement R2/R=8.

#### Ratch

Mar 10, 2013
1,098
As you wrote earlier- "please bear with me" because I needed some time to go through your calculation.
But now I can give some comments:
No - you were not "wrong", but you didnt finish the calculation.
Let me explain:
From the oscillation condition (unity loop gain ) you have derived an equation T(s)=1 which must be fulfilled at one single frequency w.
For solving this equation you have identified the imaginary parts and - as a first step - you have solved the equation IMG(T)=0. As a result you have found a formula for the frequency fulfilling the requirement IMG(T)=0.
But this is only the first part of the calculation because you have not yet used the condition that the remaining real part of T(s) must be unity.
Hence, you must insert the identified frequency expression into T(s) and set Re(T)=1.
This will result in the requirement R2/R=8.

Perhaps so, but not necessary. We both derived, by other means, that R2/R needs to be minimum of 8 to barely sustain oscillation. Now, we can crank the R2 even high like 15*R. The frequency will change and the higher amplification will drive the amps into saturation. But, if we put a resistor in feedback path to lower the amplification, where there is now only a straight wire, then we will have a stable oscillator with a different frequency than at R2=8*R .

Ratch

#### george2525

Jan 30, 2015
170
LvW - im pretty sure that the un-buffered equations are correct. I have painstakingly derived them on paper using nodal analysis. I was wondering if there was a faster way

I guess a good question would be

is there an easy way to lump that RC network into a single component. The difficulty arises from the lack of buffers so you cant do that nice cubing of stages.

#### LvW

Apr 12, 2014
604
Ratch - without any calculation:
The loop gain is a product of a frequency-dependent block (3 CR sections) and a gain block, right?
Therefore, it is clear that the phase response is determined by the CR-sections only.
Hence, the gain A=-R2/R cannot influence the phase function at all.
That means: The frequency where the loop gain is real (imag. part of T(jw)=0) cannot depend on R2.
Therefore, something in your calculation (post#14) must be wrong.
Do you agree to this reasoning?
Perhaps I can find the error tomorrow. Now its time to stop "working and thinking" (11pm in Germany).

#### LvW

Apr 12, 2014
604
LvW - im pretty sure that the un-buffered equations are correct. I have painstakingly derived them on paper using nodal analysis. I was wondering if there was a faster way
.
Sorry - I didnt realize that you were talking about another circuit (unbuffered).
Perhaps we should not discuss two different circuits within one thread.

#### Ratch

Mar 10, 2013
1,098
Ratch - without any calculation:
The loop gain is a product of a frequency-dependent block (3 CR sections) and a gain block, right?

Correct, but they are not independent of each other.

Therefore, it is clear that the phase response is determined by the CR-sections only.

No, the components (R2) in the last amp also change the phase, That is because the last amp is amplifying a complex signal. The different amplification values caused by varying R2 can change the proportion of imaginary values to real values in the circuit, and therefore change the phase and frequency.

Hence, the gain A=-R2/R cannot influence the phase function at all.

My derivation says otherwise.

That means: The frequency where the loop gain is real (imag. part of T(jw)=0) cannot depend on R2.

Same answer as above.

Therefore, something in your calculation (post#14) must be wrong.
Do you agree to this reasoning?

No.

Perhaps I can find the error tomorrow. Now its time to stop "working and thinking" (11pm in Germany).

I look forward to it. You are 7 hours ahead of me. A fresh start is always good.

Ratch

#### LvW

Apr 12, 2014
604
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].

* 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.

* 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].

* 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].

* 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

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