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Phase shift oscillator help

Discussion in 'Electronics Homework Help' started by george2525, May 1, 2016.

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  1. george2525

    george2525

    165
    0
    Jan 30, 2015
    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?

    or answer the following please!

    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!




    loopgain.PNG



    phaseosc.PNG
     
  2. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    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.
     
  3. george2525

    george2525

    165
    0
    Jan 30, 2015
    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.
     
  4. LvW

    LvW

    604
    146
    Apr 12, 2014
     
  5. george2525

    george2525

    165
    0
    Jan 30, 2015
    Thanks for the reply

    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???

    I hope that made sense

    Thanks
     
  6. LvW

    LvW

    604
    146
    Apr 12, 2014
    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.
     
  7. george2525

    george2525

    165
    0
    Jan 30, 2015
    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?
     
  8. LvW

    LvW

    604
    146
    Apr 12, 2014
    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.
     
  9. george2525

    george2525

    165
    0
    Jan 30, 2015
    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.
     
  10. LvW

    LvW

    604
    146
    Apr 12, 2014
    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.
     
  11. george2525

    george2525

    165
    0
    Jan 30, 2015
    Thanks for the help

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

    Ratch

    1,093
    334
    Mar 10, 2013
    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?

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

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

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

    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 will try.

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

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

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

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

    That is due to your misunderstanding

    But, is it?

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

    Don't be too hasty to accept that.

    Yes

    You might be surprised.

    No

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

    Ratch
     
  13. LvW

    LvW

    604
    146
    Apr 12, 2014
    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...).


     
    Last edited: May 3, 2016
  14. Ratch

    Ratch

    1,093
    334
    Mar 10, 2013
    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.
    George2525.PNG

    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
     
  15. george2525

    george2525

    165
    0
    Jan 30, 2015
    actually R2 should dissapear because -R2/R all sits on that numerator. Im not sure how you got it in the denominator like that

    the final answer should be

    w = 1/(sqrt(3)RC)

    ive done it on paper if you want to see it
     
  16. george2525

    george2525

    165
    0
    Jan 30, 2015
    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.
     
  17. Ratch

    Ratch

    1,093
    334
    Mar 10, 2013

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

    Ratch
     
  18. george2525

    george2525

    165
    0
    Jan 30, 2015
    thats for w
     

    Attached Files:

  19. george2525

    george2525

    165
    0
    Jan 30, 2015
    Thats for Resistors

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

    Attached Files:

  20. LvW

    LvW

    604
    146
    Apr 12, 2014
    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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