# Phase shift oscillator help

#### george2525

Jan 30, 2015
170
Hello,

I am having a lot of trouble understanding the buffered phase shift oscillator below. I think if i can get this one it will help me a lot with others. The first figure shows the standard positive FB block diagram and its formula. I get this.

I can also accept that the shown value for B under the oscillator figure is correct. I understand that this is the portion of Vo fed back to the inverting amplifier input terminal (at V3)

but then it is implied that A = -R2/R which is causing me the trouble. Why is A (open loop gain) the closed loop gain of the ideal inverting amplifier? and not A3?

I realise that R2 is also providing feedback here but then shouldn't this also be incorporated into the value for B and not into the value for A?

Looking at the first figure (block diagram) it seems to me that Vo/Vi = A/(1-AB) ....ok so why is Vo/Vi = AB = "T(s)" in the oscillator figure!?

Can anyone clear this up for me?

1. what is the standard formula for B (feedback) here?

2. what is the gain labelled "Af" in the block diagram as applied to the oscillator?

3. What is A here and why is it -R2/R and not simply A3 (third op amp intrinsic gain)?

Thanks, any help would be great!

#### duke37

Jan 9, 2011
5,364
The circuitry of such oscillators is past my ability but -
A3 is a standard inverting amplifier with a virtual earth.
An ideal op amp works with the two inputs at the same voltage. The gain of the amplifier ensures that this is nearly true.
The ideal op amp takes no current into its input terminals, this is almost true for all op amps, particularly for those with fet input.
Since A3 has the + input at 0V, then the - input will also be at 0V.
The current going into R equals the current going on through R2 so the voltage gain is -R2/R.
The op amp output is there to supply this current.

#### george2525

Jan 30, 2015
170
Thanks, im ok with the last op amp. im also ok with all the others. Im confused about what is the actual gain/transfer function of the circuit. I understand how to make it oscillate in principle but the equations and how they relate to the circuit seem very contradictory.

#### LvW

Apr 12, 2014
604
I am having a lot of trouble understanding the buffered phase shift oscillator below. I think if i can get this one it will help me a lot with others. The first figure shows the standard positive FB block diagram and its formula. I get this.

I can also accept that the shown value for B under the oscillator figure is correct.

Where is B?

I understand that this is the portion of Vo fed back to the inverting amplifier input terminal (at V3)

but then it is implied that A = -R2/R which is causing me the trouble. Why is A (open loop gain) the closed loop gain of the ideal inverting amplifier? and not A3?

Why do you think that A is an open-loop gain? It is simply a gain stage with the gain A.
Note that it shows NOT the classical opamp symbol
And this gain can be realized (as one of several alternatives) with an opamp as shown in the figure. This opamp has the open-loop gain A3 and the closed-loop gain A=-R2/R.

I realise that R2 is also providing feedback here but then shouldn't this also be incorporated into the value for B and not into the value for A?

The circuit has two feedback loops: One resistive negative loop for realizing the required gain value - and one positive feedback loop for realizing the oscillation condition (loop gain T(s) )
Both feedback loops can be (and in this case: must be) considered separately.

Looking at the first figure (block diagram) it seems to me that Vo/Vi = A/(1-AB) ....ok so why is Vo/Vi = AB = "T(s)" in the oscillator figure!?

Sorry, this is nonsense. An oscillator has no external input.
It consists of a closed-loop only - and this closed-loop must be opened in oder to find the loop gain T(s).
Hence, it makes no sense to ask for a transfer function Vo/Vi.
The given expressions for beta and the loop gain T(s) are correct.

The oscillation condition is T(s)=1.
Hence, all you have to do is the following: Solve the loop gain expression for w (set s=jw) under the condition that T(s) is REAL and the magnitude is "1".
That means: Set the imag. part of T(jw)=0 and solve for w.

Any further question?

#### george2525

Jan 30, 2015
170

B is refered to as V3/Vi in the screenshot

ok so - is the feedback provided by R2 creating 180 degrees and the RC's prvide the other 180? am I correct there?

Ok so "A" is just the standard non-inverting gain -R2/R .Cool Ill take that

the Circuit diagram shows "Vo" and "Vi"

but Vi IS Vo right?

anyway if you look at the equations for V3/Vi and the other for Vo/V3 = -R2/R

multiply them together to get T(s) and that works out as Vo/Vi right?

Therefore looking at the labelled circuit T(s) = AB = Vo/Vi

I agree with you that it seems silly to ask for a transfer function A/(1-AB) but that model was included with the circuit hence the confusion. To me the entire circuit is simply T(s)

So from my understnding - B is whatever is being fed back into the inverting op amp config at V3

A is the standard inverting gain ratio -R2/R

and the entire circuit is AB

so where does the whole A/(1-AB) thing come in?

I mean - I understand the oscillation conditions in the context of A/(1-AB) via Barkhausen etc

but what is bothering me is "where" is the A/(1-AB) in this circuit (the bold letters here) coming from???

Thanks

#### LvW

Apr 12, 2014
604
When an amplifier A has a feedback factor B (first drawing) the closed-loop gain is A/(1-AB).
As you can see, when the loop gauin T=AB is unity, the denominator is zero - and the closed-loop function approaches infinity.
This is the oscillation condition - and because the circuit is oscillating by its own, we do not need any input voltage.
In contrary, any input signal would (a) disturb the oscillation condition or (b) superimposed to the oscillation signal. In any case, this would make no sense.

Note that for negative feedback AB is negative and the denominator is (1+AB). This is a stable amplifier.

#### george2525

Jan 30, 2015
170
ok, so is it possible to calculate the closed loop gain of this circuit? I realise it is infinity but can we represent the A/(1-AB) in terms of the actual components here?

based on the equations it would be

(-R2/R)/(1-(-R2/R)((SRC/(1+SRC))^3)

if that is true then how would you prove that?

#### LvW

Apr 12, 2014
604
Writing a gain function requires at first to DEFINE the gain (input and output).
In your case it would be: vo/vi.
Now - look at you circuit: Vi is directly connected to vo.
Therefore, i wrote: It makes no sense to write a closed-loop gain expression for an oscillator circuit.

#### george2525

Jan 30, 2015
170
Ok great. So an oscillator circuit only has a loop gain comprised of AB = T(s) unless you were to try and drive it with something and attach an output somewhere?

that makes sense to me

I think the inclusion of the complete A/(1-AB) formula has caused all the confusion.

#### LvW

Apr 12, 2014
604
Yes - the gain of a closed-loop system is A/(1-AB), however, this expression looses its meaning for the case AB=1.
Even the mathematics does not allow this case.

#### george2525

Jan 30, 2015
170
Thanks for the help

I have managed to solve a couple of circuits based on this!

#### Ratch

Mar 10, 2013
1,098
Hello,

I am having a lot of trouble understanding the buffered phase shift oscillator below. I think if i can get this one it will help me a lot with others. The first figure shows the standard positive FB block diagram and its formula. I get this.

No wonder. The last op-amp in not a buffered amp. It is part of the frequency determining components. After all, it changes the phase by 180°, doesn't it?

I can also accept that the shown value for B under the oscillator figure is correct. I understand that this is the portion of Vo fed back to the inverting amplifier input terminal (at V3)

The feedback to the input is a dead short. That tells me that beta is 1.

but then it is implied that A = -R2/R which is causing me the trouble. Why is A (open loop gain) the closed loop gain of the ideal inverting amplifier? and not A3?

The gain of those three op-amps is not -R2/R1

I realise that R2 is also providing feedback here but then shouldn't this also be incorporated into the value for B and not into the value for A?

R2 is providing feedback to the last inverting op-amp. And, it is also a frequency determining element.

Looking at the first figure (block diagram) it seems to me that Vo/Vi = A/(1-AB) ....ok so why is Vo/Vi = AB = "T(s)" in the oscillator figure!?

You are interpreting the block diagram wrong. The 3 amplifiers are changing the phase and the feedback is a dead short of 1, and does not alter the phase as the block diagram appears to suggest.

Can anyone clear this up for me?

I will try.

1. what is the standard formula for B (feedback) here?

It depends on what the circuit components are. In this case it is a shorting wire with a beta of 1.

2. what is the gain labelled "Af" in the block diagram as applied to the oscillator?

It is the closed loop gain with feedback applied as opposed to no feedback.

3. What is A here and why is it -R2/R and not simply A3 (third op amp intrinsic gain)?

It is not -R2/R. It is easy to calculate if you know complex arithmetic.

Thanks, im ok with the last op amp. im also ok with all the others. Im confused about what is the actual gain/transfer function of the circuit.

I can see that. You should ask me to show you how to calculate the gain and frequency.

I understand how to make it oscillate in principle but the equations and how they relate to the circuit seem very contradictory.

That is due to your misunderstanding

B is refered to as V3/Vi in the screenshot

But, is it?

ok so - is the feedback provided by R2 creating 180 degrees and the RC's prvide the other 180? am I correct there?

The components attached to the op-amps are changing the phase.

Ok so "A" is just the standard non-inverting gain -R2/R .Cool Ill take that

Don't be too hasty to accept that.

but Vi IS Vo right?

Yes

I agree with you that it seems silly to ask for a transfer function A/(1-AB) but that model was included with the circuit hence the confusion. To me the entire circuit is simply T(s)

You might be surprised.

So from my understnding - B is whatever is being fed back into the inverting op amp config at V3

No

but what is bothering me is "where" is the A/(1-AB) in this circuit (the bold letters here) coming from???

It is the closed loop feedback equation for a single feedback loop. See http://www.learnabout-electronics.org/Amplifiers/amplifiers31.php

Ratch

#### LvW

Apr 12, 2014
604
Hi ratch, I must admit your post partly confuses me - and hopefully not also the OP.
The first figure shows a block diagram with an amplifier A and a feedback block "beta". And the second figure shows one possible realization with A=-R2/R1. Here "beta" is composed of two buffers and three CR sections. Thats all. I think, the only misunderstanding (at the OPs side) was his assumption that A was the open-loop gain of an opamp. But it is simply a finite gain value to be realized using any amplifier type (BJT or opamp or OTA...).

The feedback to the input is a dead short. That tells me that beta is 1.

Which feedback are you referring to? I think, the OP is right when he assumes that the two opamp buffers together with the three CR blocks form the positive feedback factor B resp. beta. I think, he never spoke about the "dead shorts" for the buffers.

The gain of those three op-amps is not -R2/R1
Who has claimed this?

You are interpreting the block diagram wrong. The 3 amplifiers are changing the phase and the feedback is a dead short of 1, and does not alter the phase as the block diagram appears to suggest.
I dont understand this sentence.

It depends on what the circuit components are. In this case it is a shorting wire with a beta of 1.
I think, it is clear that beta (the feedback network of the whole circuit) is composed of two buffers and three CR elements. Who spoke about the local feedback factor for the buffers?

It is not -R2/R. It is easy to calculate if you know complex arithmetic.
Comparing both figures, I think it is clear that "A" of the first figure is realized (in the second figure) using an inverting amplifier with the gain -R2/R. Why do you think, this would be wrong?

But, is it?
Yes.

Don't be too hasty to accept that.
But it is correct!
______________
LvW

Last edited:

#### Ratch

Mar 10, 2013
1,098
Hi ratch, I must admit your post partly confuses me - and hopefully not also the OP.
The first figure shows a block diagram with an amplifier A and a feedback block "beta". And the second figure shows one possible realization with A=-R2/R1. Here "beta" is composed of two buffers and three CR sections. That`s all. I think, the only misunderstanding (at the OPs side) was his assumption that A was the open-loop gain of an opamp. But it is simply a finite gain value to be realized using any amplifier type (BJT or opamp or OTA...).

Please bear with me because I have never done this before. First, one has to decide whether one is building an amplifier or an oscillator. The circuit above shows three amplifiers in series, each of which change the phase of the signal. The rc in the first amplifier changes the phase -60°. Same for the second for a total of -120°, The rc between the 2nd and 3rd amp changes the phase another -60° for a total of -180°. Then the last amp changes the phase by -180° for a total of 360° or 0°. The total gain around the loop has to be at least 1 in order to support oscillation. Using that criteria, the frequency can be found as shown below.

Notice that the last amp is not a buffered amp, but a critical part of the loop gain path which determines frequency. Also r2 is a frequency determining component, which it would not be if the last amp was truly a buffer amp.

Ratch

#### george2525

Jan 30, 2015
170
actually R2 should dissapear because -R2/R all sits on that numerator. Im not sure how you got it in the denominator like that

w = 1/(sqrt(3)RC)

ive done it on paper if you want to see it

#### george2525

Jan 30, 2015
170
R2/R must = 8 to make the gain = 1

im guessing this is because there are drops at the capacitors which have to be made up for.

#### Ratch

Mar 10, 2013
1,098
actually R2 should dissapear because -R2/R all sits on that numerator. Im not sure how you got it in the denominator like that

w = 1/(sqrt(3)RC)

ive done it on paper if you want to see it

R2/R must = 8 to make the gain = 1

Yes, please show it. My derivation above shows how I did it. I would like to see the calculation for R2/R = 8,

Ratch

Jan 30, 2015
170
thats for w

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#### george2525

Jan 30, 2015
170
Thats for Resistors

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

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#### LvW

Apr 12, 2014
604
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)

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