# Phase Shift Oscillator

#### LvW

Apr 12, 2014
604
Don't be silly. You will never get the formula in the book that way!

It was not my primary goal to find a circuit which corresponds to the "book formula". (Although I gave him a suitable circuit using an integrator stage, see my post#8).
The questioner has shown us a circuit and he gave a formula which does NOT correspond to the circuit.
Hence, in post#2 and post#8 a correct formula was given.
Regarding "silly": What did you do? You have calculated for a single unloaded RC section the frequency for a phase shift of 60 deg (by the way: rather involved; it is much simpler using the Re and Im of the denominator from 1/[1+jwRC]). In your graph you have multiplied this phase function with a factor of "3" - assuming again three decoupled RC sections (ignoring the comments in post#12 and #14).
In order not to confuse the questioner I repeat: The given formula in post#1 does NOT correspond to the shown circuit.

#### BobK

Jan 5, 2010
7,682
I have done some simulations with LTSPICE.

The result: The sqrt(6) formula is correct. It always overestimates the frequency because a real op amp will add an additional phase shift.

sqrt(6) = 2.45.

With the given values my simulation comes out to 7558 Hz or 2.37 / 2*pi*R*C. If lower the frequency by a factor of 10 by upping the capacitor to 100n I get 773 Hz or 2.43 / 2*pi*R*C.

Also, I used a JFET input opamp and higher (1 + 29.1M) as the input / feedback to make the loading of filters negligible.

Bob

#### LvW

Apr 12, 2014
604
Yes - the loading of the last RC sections caused by the inverting opamp is one of the disadvantages of this circuit.
Of course, one could add a buffer amplifier for decoupling - nevertheless, we still have the problem that frequency tuning of the circuit is rather complicated.
These aspects have to be evaluated knowing that there are other oscillator circuits with one or two opamps which have much better properties.

#### BobK

Jan 5, 2010
7,682
As you can see from my different results for 7.5KHz and 750Hz, the phase shift introduced by the opamp is also a consideration, especially as the frequency gets higher.

Bob

#### LvW

Apr 12, 2014
604
As you can see from my different results for 7.5KHz and 750Hz, the phase shift introduced by the opamp is also a consideration, especially as the frequency gets higher.
Bob

Yes - that´s right. Parasitic phase shift of the opamp(s) is another aspect to be considered. Also this effect speaks against the oscillator topology under discussion because we need a gain of "29".
Other oscillator structures work with opamp gains of "3" or even lower - with reduced phase errors.

#### Arouse1973

Dec 18, 2013
5,178
Did you read posts #12 and #14 before making that statement? There are certain conditions that have to be met before you can just add on sections like that.

Ratch

Nope! I just like Laplace's comments

#### Ratch

Mar 10, 2013
1,098
Laplace,

I am suspicious of the blue curve above because a single section low pass filter with R=5 k and C=10 nf gives a break frequency of -45° of around 3 kHz. The above curve shows the break frequency of -45° at around 900 Hz. Here is my plot of it.

Notice that the the -60° occurs at 30000 radians or 5000 kHz.

OK, so we have something written in a text book that gives the frequency of a 3-section phase oscillator as sqrt(3)/(2*pi*R*C). We know beyond a doubt that the correct frequency of 3-section phase oscillator with zero source resistance and infinite load is sqrt(6)/(2*pi*R*C). They are both just as easy or just as hard to calculate, so why not go with the correct one?

Ratch

Last edited:

#### LvW

Apr 12, 2014
604
Laplace,
I am suspicious of the blue curve above because a single section low pass filter with R=5 k and C=10 nf gives a break frequency of -45° of around 3 kHz. The above curve shows the break frequency of -45° at around 900 Hz.

Ratch - I think, I can clarify the point. Laplace did multiply his phase (in blue) response of a single RC sections with a factor of "3" - assuming that all three RC sections would be decoupled.

#### Ratch

Mar 10, 2013
1,098
LvW,
No matter how you cut it, The sqrt(6) formula is going to give you a different frequency than the sqrt(3) formula.

Ratch

#### Arouse1973

Dec 18, 2013
5,178
I wonder what does making each stage have an impeadance of 10 times more than the last section do? This quite common on cascaded filters without having buffers.

#### Ratch

Mar 10, 2013
1,098
I wonder what does making each stage have an impeadance of 10 times more than the last section do? This quite common on cascaded filters without having buffers.

What I cannot understand is how the text book even suggested that formula. The calculated frequency using the sqrt(3) formula is always going to be wrong and different from the sqrt(6) formula no matter how the sections are loaded and coupled or not. And wrong by a ratio of 1/sqrt(2).

Ratch

#### The Electrician

Jul 6, 2012
117
According to the Barkhausen stability criterion, the phase shift of the 3-stage RC network must be 180° for oscillation to occur so the phase shift of each RC stage must be 60°. Now it happens that tan(60°)=1.732 so the ratio of the imaginary to the real portion of each RC voltage divider output will equal 1.732 at the oscillation frequency. It takes some complex algebra with imaginary numbers to calculate the RC phase shift. Can you do that?

The phase shift of each RC stage is not 60°. See this page:

http://wps.prenhall.com/chet_paynter_introduct_6/0,5779,426330-,00.html

second paragraph under Figure 18-3

#### The Electrician

Jul 6, 2012
117
What I cannot understand is how the text book even suggested that formula. The calculated frequency using the sqrt(3) formula is always going to be wrong and different from the sqrt(6) formula no matter how the sections are loaded and coupled or not. And wrong by a ratio of 1/sqrt(2).

Ratch

When all the stages are buffered as in Figure 15.16 of this pdf:

http://www.ti.com/lit/ml/sloa087/sloa087.pdf

Then the SQRT(3) formula applies (the RG resistor loads the last RC stage, so the formula isn't exactly right even with the buffering).

If there are no buffers between stages, then the SQRT(6) formula applies.

http://www.play-hookey.com/oscillators/audio/phase_shift_oscillator.html

#### The Electrician

Jul 6, 2012
117
I wonder what does making each stage have an impeadance of 10 times more than the last section do? This quite common on cascaded filters without having buffers.

The frequency of oscillation becomes SQRT(321)/5 * 1/(2 Pi R C)

The required gain becomes 9.272 rather than 29.

Apr 12, 2014
604

#### Excalibur

Jul 7, 2013
9
I have also checked in some books and indeed the formula i did give is wrong. it made the assumption that each RC network produced a phase shift of 60 degrees. But thanks to Laplace for showing me where the formula comes from.
Also, the correct formula for calculating the frequency is the one with sqrt (6) . Thanks to all

#### LvW

Apr 12, 2014
604
But thanks to Laplace for showing me where the formula comes from.
Do you want a simpler derivation for the frequency with 60deg phase lag (that means: Phase shift of -60deg)? Here it comes:

* For a RC lowpass stage we have H(jw)=1/(1+jwRC)
* Total phase shift PHI=PHI(numerator)-PHI(denominator) = 0 - PHI(d)=-PHI(d)
PHI=-PHI(d)
* tan(PHI)=tan[-PHI(d)]=-tan[PHI(d)] = -Im(d)/Re(d)=-wRC/1=-wRC
* For PHI=-60deg we have tan(-60)=-1.732
* Hence: -wRC=-1.732 and
w=1.732/RC.

#### Laplace

Apr 4, 2010
1,252
* Total phase shift PHI=PHI(numerator)-PHI(denominator)
Learned something new today. This website is just wonderful!

However, I went back to my linear circuits textbook to review the Rules of Complex (Phasor) Algebra and found it was not so much learned but rather being reminded of something I had long ago forgotten.

#### Excalibur

Jul 7, 2013
9
Do you want a simpler derivation for the frequency with 60deg phase lag (that means: Phase shift of -60deg)? Here it comes:

* For a RC lowpass stage we have H(jw)=1/(1+jwRC)
* Total phase shift PHI=PHI(numerator)-PHI(denominator) = 0 - PHI(d)=-PHI(d)
PHI=-PHI(d)
* tan(PHI)=tan[-PHI(d)]=-tan[PHI(d)] = -Im(d)/Re(d)=-wRC/1=-wRC
* For PHI=-60deg we have tan(-60)=-1.732
* Hence: -wRC=-1.732 and
w=1.732/RC.
Interesting way to do it indeed. Thanks.

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