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Phase Shift Oscillator

  1. Oct 4, 2016 #1
    1. The problem
    FIGURE 4(a) shows the circuit of an oscillator and FIGURE 4(b) gives
    the gain and phase response of the op-amp used.
    When the circuit was simulated in PSpice using an ideal op-amp, the
    circuit oscillated at the designed frequency. However when the ideal opamp
    was replaced by the TL072 the frequency of oscillation was observed
    to be about 15% lower.
    Investigate this anomaly giving your answer in the form of a report of
    about 750 words. The report should attempt an analysis of the circuit
    and include the results of any PSpice simulations. You may wish to use
    the transfer function derived as equation (6) in the appendix to lesson
    3-1 to calculate the phase shift in the RC network.

    2. Relevant equations

    [tex]arctan\frac{w_c}{w}=60^o[/tex]

    3. The attempt at a solution

    so this is what i have written so far, but i don't actual now the reason why.
    I'm thinking it is to do with the input impedance of the TL072 causing load on the RC network?


    The figure show is a phase shift oscillator. the frequency determining network (FDN) RC network in the feedback loop is designed to produce 180° phase shift at the desired frequency. Each RC element of FDN will give 60° phase shift so that the three sections together give the required 180°. The 180° phase shift occurs between the input to a RC network and the output from the network. The inverting op amp with its 180°phase **** completes the unity loop gain at 360°. The phase shift occurs only for the selected frequency to which the FDN is tuned.

     
  2. jcsd
  3. Oct 4, 2016 #2

    NascentOxygen

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    Staff: Mentor

    List the ways that an actual OP-AMP can differ from the ideal OP-AMP.
     
  4. Oct 5, 2016 #3

    LvW

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    Where are the circuits?
    Don`t forget the (parasitic) phase shift contributed by the opamp.
     
    Last edited: Oct 5, 2016
  5. Oct 5, 2016 #4
    Thanks
    for your help I have included some screen grabs with more info.

    The ideal op-amp will have;
    • infinite gain
    • Infinite input impedance
    • no output impedance.
     

    Attached Files:

  6. Oct 5, 2016 #5

    LvW

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    The oscillation frequency of your circuit is designed for app. 60 kHz.
    From the frequency characteristics of the used opamp you can derive if (and how much) the parasitic phase shift of the closed-loop gain amplifier will influence the frequency of oscillation. That is the most important influence of the real opamp. In addition, you should analyse the slewing properties of the opamp used.
     
  7. Oct 5, 2016 #6
    I have never heard this in my study but after looking up
    so the inverting input will not give 180 deg?
    Because of capacitor between the output and the negative input.
     
  8. Oct 5, 2016 #7

    LvW

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    The circuit will oscillate at a frequency where the loop gain phase is -360deg.
    The inverting input will contribute -180deg and the non-ideal opamp will contribute another -X deg.
    Hence, the oscillator will have a frequency where the passive feedback network contributes -(180-X)deg.
    This will be a lower value than designed.
    It is your main task to find the value of X.
     
  9. Oct 5, 2016 #8

    NascentOxygen

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    Staff: Mentor

    The OP-AMP graphs show a voltage gain much lower than you might have hoped, and a phase shift markedly different from what you probably expected, in this circuit's region of operation.
     
  10. Oct 6, 2016 #9

    LvW

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    Additional hints: The opamps open-loop gain and phase function shows a remarkable phase shift at app. 60 kHz.
    From this you can derive an estimate for the closed-loop (unwanted, but unavoidable) additional phase shift X.
    This gives some information on the actual oscillation frequency.
     
  11. Oct 6, 2016 #10
    I calculate the frequency of operation to be 65 kHz.
    from the graph we can see the phase shirt is 90 deg.

    The question also says
    "You may wish to use the transfer function derived as equation (6)"

    the info given is
    (where ωo is the angular frequency at which the transfer function of the RC ladder is purely resistive) see attached.
     

    Attached Files:

  12. Oct 6, 2016 #11

    LvW

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    Any further question?
    Don`t forget: The 90deg phase shift is for the open-loop gain.
    But your opamp is eqipped with feedback leading to a much lower closed-loop gain (which, of course, effects the phase).
     
  13. Oct 6, 2016 #12

    rude man

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    You may wish to use the transfer function derived as equation (6) in the appendix to lesson
    3-1 to calculate the phase shift in the RC network.

    This won't be of any use if the transfer function is V4/V1 since this network feeds into a virtual ground.
    Model the op amp as k/s using its 60 KHz ol gain to give you k. Then write the equations of the total circuit which will comprise k, the r-c network and the feedback resistor. Use software as much as possible to solve the equations.

    If you don't get close to the empirical (test) results then add the op amp's input capacitance, but the impedance level of your r and c network is low enough that I would omit it (the input capacitance) at first.
    You could also add the finite output impedance of the op amp at 60 KHz.
    A messy set of computations which is why PSPICE is always used in the 'real world'!
     
  14. Oct 6, 2016 #13

    NascentOxygen

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    Staff: Mentor

    The equation is appropriate (providing we can regard the input as a virtual ground), because the phase-shifter's R has been repositioned so OP-AMP input current is proportional to V4.
     
  15. Oct 6, 2016 #14

    rude man

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    There is no virtual ground. The gain of the op amp at 60 KHz is far below infinity! The V4/V1 transfer function is useless, R "repositioning" or not.
     
  16. Oct 7, 2016 #15

    LvW

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    In any case, all theoretical considerations should/could be supplemented (confirmed) by SPICE simulations.
    And - as you know - it is the LOOP GAIN that matters only. Loop gain simulations involving opamps with feedback are exact enough if you open the loop between the opamps output and C3.
    Then, verification of the oscillation condition gives the actual oscillation frequency.
     
    Last edited: Oct 7, 2016
  17. Oct 8, 2016 #16
    So as far as i can figure i need an closed loop gain of 29 in order to maintain oscillation.

    I am a bit stuck What courses the phase shift in the real op-amp.
    why does the frequency setal 15% lower.
     
  18. Oct 8, 2016 #17

    LvW

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    Look at my answer in post#7.
    Do you know any amplifier that has no parasitic (unwanted) phase shift?
     
  19. Oct 8, 2016 #18

    rude man

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    Post 12 paragraph 1.

    Your op amp has a gain of only 100 at the frequency of oscillation, and the phase shift is close to 90 degrees lag which is characteristic of a unity-gain op amp in your frquency region of operation (never mind what that means if you don't already know).
    Additionally, finite input and output impedances could also add to the low frequency of oscillation but ignore that for your first analysis.

    Analyze the open-loop (per post #15) or closed-loop circuit per the usual KVL (or KCL) method using my suggested model for the op amp.
     
  20. Oct 8, 2016 #19

    LvW

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    Yes - I also expect that the primary influence on the frequency comes from the unwanted phase contribution of the amplifier.
    We remember the name: "Phase-shift"-oscillator!
     
  21. Oct 8, 2016 #20

    rude man

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    All op amps (except "undercompensated" ones) have around 90 deg phase shift from a very low frequency (typically < 100 Hz) all the way to the unity-gain frequency.
     
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