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Phase Shift Oscillator Op-Amp Neg Feedback

Discussion in 'Electronics Homework Help' started by toptip123, Oct 16, 2016.

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  1. toptip123

    toptip123

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    0
    Oct 13, 2016
    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.

    So I am struggling with this question
    I Have calculated the RC networks output to be 65kHz.
    I can see from the bode plot that the OP has a phase shift of 90 deg at 65kHz
    How can i work out the real frequency output?
    I am aware that in order for the system to oscillate the loop feedback must be 360 deg at unity gain.

    Any help on this would be appreciated I have also uploaded some images.
    Thanks again.
     

    Attached Files:

  2. LvW

    LvW

    604
    146
    Apr 12, 2014
    Some days ago you´ve got - in another forum - 30 replies to the same question.
    In one of my last answers I did ask you:

    Do you know how to handle a BODE plot - in particular, how to derive a phase plot from the magnitude plot?
    Otherwise, you cannot solve your problem (without circuit simulation).

    Unfortunately, you did not reply to this question. So - I am not sure if you will have more luck in THIS forum. We can help you only in form of a question-answer game.
     
  3. toptip123

    toptip123

    9
    0
    Oct 13, 2016
    I felt overwhelmed by the amount of people in that forum putting in bit and pieces of information that I found hard to understand.

    I am not confident I this subject as you can tell.

    I have been researching about BODE but I still don't understand how use them to work out the real frequency.
     
  4. LvW

    LvW

    604
    146
    Apr 12, 2014
    In this case, I think it is important for you to become familiar with the BODE diagram. In the mentioned alternative forum a corresponding link was given to you.
     
  5. toptip123

    toptip123

    9
    0
    Oct 13, 2016
    I am sorry I can't find this.

    If the loop gain needs to be -360 deg dose that mean the circuit will oscillate when the RC network provides(-360+180+90)=-90 deg??

    I know the BODE plot shows the gain and phase shift of the op-amp over a frequency sweep.
     
  6. LvW

    LvW

    604
    146
    Apr 12, 2014
    Please notice that your oscillator contains an opamp WITH FEEDBACK.
    Any negative feedback will drastically change the frequency-dependent behaviour of a "naked" opamp.
    This is the main purpose of negative feedback:
    Changing (reducing) the gain, increasing the bandwidth and "flattening" the phase.
    Therefore, for an opamp with feedback the unwanted (parasitic) phase shift will certainly be far below the 90deg - but it will not be negligible.
    At first - become familiar wth the BODE plot and with the three gains it can contain:
    (a) Open-loop gain Ao, (b) Closed-loop gain Acl and (c) Loop gain.

    Most important for you: Closed-loop gain (magnitude and phase).
     
  7. toptip123

    toptip123

    9
    0
    Oct 13, 2016
    Yes I have seen this when researched BODE plot with neg feedback.

    This is the gain of the op amp alone

    This is the feed back gain RC Network
    I beleve this is 33 = Rf/R = 33/1 Is the phase shift -180deg

    this is the overall circuit gain.

    When I simulate in spice how do I introduce noise to start oscillation?
     
  8. LvW

    LvW

    604
    146
    Apr 12, 2014
    LvW said:
    (b) Closed-loop gain Acl
    This is the feed back gain RC Network
    I beleve this is 33 = Rf/R = 33/1 Is the phase shift -180deg

    No - that is not correct. As the name indicates: "Closed-loop gain" is the gain of the amplifier with the closed feedback loop.

    LvW said:
    (c) Loop gain.
    this is the overall circuit gain.

    No - also not correct. "Loop gain" is the turn-around gain of the complete loop (product open-loop gain and feedback factor).

    When I simulate in spice how do I introduce noise to start oscillation?

    Who told you that you need noise?

    Toptip - with all respect, I`ve got the impression that you need to stuy basic of opamp circuits first. In a forum like this it is not possible to teach you fundamentals of feedback sysytems. We are here to answer specific questions. Are you really not able to look for some knowledge sources which can tell you something about terms like "closed-loop gain" or "loop gain"?
     
  9. toptip123

    toptip123

    9
    0
    Oct 13, 2016
    When I simulate in spice how do I introduce noise to start oscillation?

    Who told you that you need noise?

    It is in my reading material that, in real life the circuit would pick up noise from the surroundings which will initiate the feedback and oscillation until it settles at a loop gain of 1 in phase -360deg.

    The closed loop gain is the loop close (with feedback). The CLG will flatten, reduce the gain and increase the bandwidth where the CLG intersects with the open loop gain it will roll off the same. At this point is it unity gain?

    The open loop gain is the op-amps gain

    loop gain is the product of all the gains
     
  10. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    I have seen simulations of oscillators which need to be started with a some sort of signal. Once started, the loop gain will keep it going.

    In a practical circuit, no noise is required because it is there with non ideal components. It is necessary that the amplifiers are biased into a conducting condition, the bias may change in operation to reduce the loop gain to one.
     
  11. LvW

    LvW

    604
    146
    Apr 12, 2014
    Noise would play a certain role only if there would be no other "kick-off" sources. However, each oscillator will be powered (switched-on) at a particular time. And it is this switch-on transient which causes the start of the oscillation process. For simulation purposes you can (a) give a capacitor an "initial condition" (causing a kind of imbalance) or (b) introduce a short voltage pulse into any of the grounded nodes or (c) use a power supply which is switched on at t=0 (as in real life) instead of a fixed supply.
     
  12. toptip123

    toptip123

    9
    0
    Oct 13, 2016
    Thank

    My simulation works, went with changing the initial value of one of the caps. Although the ideal op-amp stops oscillating at a point and the real will carry on and on?

    So in order to oscillate the loop gain must be 1 V/V
    The closed loop gain is 33 V/V
    How can I use the BODE plot to estimate the frequency.

    Thanks
    TIPTOP
     
  13. toptip123

    toptip123

    9
    0
    Oct 13, 2016
    Thank

    My simulation works, went with changing the initial value of one of the caps. Although the ideal op-amp stops oscillating at a point and the real will carry on and on?

    So in order to oscillate the loop gain must be 1 V/V
    The closed loop gain is 33 V/V
    How can I use the BODE plot to estimate the frequency.

    Thanks
    TIPTOP
     
  14. LvW

    LvW

    604
    146
    Apr 12, 2014
    Three questions:
    * What do you mean with "closed-loop gain"? Which input?
    * For which function do you intend to create a BODE plot?
    * What is the general oscillation condition?
     
  15. toptip123

    toptip123

    9
    0
    Oct 13, 2016
    With the real op amp the oscillation start to build after 10ms but stops dead at 14.2ms
    the frequency is 59.3 Hz

    With the tl072 the oscillations start to build at t=0 and continuing to oscillate the frequency is 56.5 Hz

    at the negative input ie R4/R3 giving 33 gain
    the RC network has a gain of 0.29 so R4/R3 is required for loop fain of 1.

    does the plot need redrawing to include the RC network?
     
  16. LvW

    LvW

    604
    146
    Apr 12, 2014
    In short:
    * A closed-loop gain cannot be defined because the circuit IS indeed a closed loop withoput any input.
    * The oscillation condition (to be fulfilled at one single frequency only) is Loop gain=1.
    * This condition must be controlled using the BODE plot. Hence, you must create a BODE plot for the loop gain.
    * Do you know how to get the loop gain? (This is a DEFINITION which can be found in each relevant textbook!).
     
  17. toptip123

    toptip123

    9
    0
    Oct 13, 2016


    The loop gain is the product of the gains of the amplifier and the feedback network.

    I am not sure how to do this.
     
  18. LvW

    LvW

    604
    146
    Apr 12, 2014
    "not sure" means you have some rough ideas, don`t you?
    Tell us about your first approach.
     
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