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Characterization of an Op-amp Function Generator Circuit

  1. Mar 24, 2015 #1
    1. The problem statement, all variables and given/known data
    I need to characterize the circuit below in order to compare measured data to theory. Can you determine any the theoretical relationships for the frequency, amplitude, gain of the two voltage outputs? I measured the output voltage of each op-amp while varying both of the variable resistors individually. I can't seem to find any theoretical relationships for this circuit online, and I definitely don't have the knowledge to do it myself.

    funcgen.gif
    Source: http://www.unirioja.es/cu/lzorzano/OPAMPcollection.pdf

    2. Relevant equations
    Don't know.

    3. The attempt at a solution
    Don't know what to do.
     
    Last edited: Mar 24, 2015
  2. jcsd
  3. Mar 24, 2015 #2

    FactChecker

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    A classical way to put the relationship into mathematical terms is to represent the electronic components with Laplace transformations in the S-plane. Then the frequency response can be analysed. The Laplace transforms are very powerful and can be applied to very complicated circuits, but it is a subject in itself. If that is not what you are looking for, then there may be a simpler way to analyse it using more basic EE fundamentals.
     
  4. Mar 24, 2015 #3
    Considering I don't understand anything of what you just said, using EE fundamentals would be very appreciated.
     
  5. Mar 24, 2015 #4

    FactChecker

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    Ok. Sounds like my answer was way off the mark. Sorry. I hope someone with EE background can help.
     
  6. Mar 24, 2015 #5
    Thanks for trying!
     
  7. Mar 24, 2015 #6

    berkeman

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    Yeah, Fourier Transforms will not be needed today... :smile:

    @lpuppy79 -- What have you learned about opamps so far? What are the characteristics of an ideal opamp? How do those characteristics help you to write the equations for the transfer function of opamp circuits like this one?
     
  8. Mar 24, 2015 #7

    berkeman

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    Also note -- that first opamp is being used as a comparator and not an opamp...
     
  9. Mar 24, 2015 #8
    Ideal Characteristics:
    High Input Impedance
    Low Output Impedance

    Yes, the first is used as a comparator.

    Is the second op amp essentially just an integrating circuit?
     
  10. Mar 24, 2015 #9

    berkeman

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    Yes, good!

    The first comparator is used to make a square wave output, and the 2nd stage opamp integrates that to make a triangle output. That triangle output is fed back to cause the 1st stage comparator to keep tripping to make the square wave.

    How would you go about figuring out the amplitude and frequency of the output triangle wave? Do you know what opamps you will be using in your lab measurements for comparison?
     
  11. Mar 24, 2015 #10
    vtri=-1/(RC)*integral(vsqudt) from 0 to T

    ftri=fsqu

    I already have taken the measurements. I used a LM358.
     
  12. Mar 24, 2015 #11

    berkeman

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    What did you use for your power supply rails? +/-12V? What was the square wave amplitude? When an opamp output pegs against the rails, the voltage depends on the opamp output stage. Some have "rail-to-rail outputs", and others like the LM358 are not... :smile:
     
  13. Mar 24, 2015 #12
    +/- 12 V as the power supply rails. The amplitude of both the square and the triangle wave varied based on the resistance of R2 so there was no constant square wave amplitude.
     
  14. Mar 25, 2015 #13

    NascentOxygen

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    Are you sure about this?
     
  15. Mar 25, 2015 #14
    While varying R3 from 100-1000 kiloohms with constant R2 (1 megohm) , the frequency's range was .311 kHz to 2.05 kHz, vsquare ranged 22.9-23.1 V , vtriangle ranged 200-272 mV.
    While varying R2 from from 0-1 megaohm with constant R3 (140 kilohms), the frequency's ranged from .02275-1.61 kHz, vsquare range 22.3-23.1 V, vtriangle ranged .253-18.3 V.
     
  16. Mar 25, 2015 #15

    rude man

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    At power-up, assume U1 (left-hand op amp) output V1 latches up to + 11Vdc, causing the integrator U2 output voltage V2 to crank downwards (at what rate? I mean, what is dV2/dt at the output of U2?)
    Next: at what negative V2 will U1 switch its output from +11 to -11V? Note that this depends on the relative resistances of (R1+R2) vs. R5.
    Then the process repeats in the other direction.
    With this info you should be able to compute the pk-pk voltages V1 and V2, and the frequency as determined by dV2/dt.
    Your data makes sense. The square wave pk-pk V1 should always be ~22V. The pk-pk of V2 depends on the setting of R2. Once R2 is set the pk-pk voltage V2 should not change as you vary R3. But at a high enough frequency the op amps no longer behave ideally so expect deviations from this statement at high frequencies.
    Example: set R2 = 500K, V2 pk-pk should be about 0.36V pk-pk.
     
  17. Mar 25, 2015 #16
    1. In all scenarios, dV2/dt would be -1/[RC]=-1/[(R3+R4)C], correct?
    2. V2=R5/(R1+R2+R5)*V1max?
    3. If those are correct, I don't understand how to get the frequency or V1pp?
     
    Last edited: Mar 25, 2015
  18. Mar 25, 2015 #17

    NascentOxygen

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    You could say that Vsquare was constant at about 23V when you tie this in with theory. Also, Vtriangle ideally would be constant.

    Again, ideal op-amps would see Vsquare here not changing, while Vtriangle changes over a wide range.

    Did you sketch V1 underneath V2 showing how they are related in time?
     
  19. Mar 25, 2015 #18

    NascentOxygen

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    You need to multiply this by a voltage, that voltage being V1's maximum

    Not quite. How did you calculate this?

     
  20. Mar 25, 2015 #19

    rude man

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    That's correct for {dV2/dt}/V1. You need to include V1 in dV2/dt.
    Right. Except what about the sign of V2? remember, V2 is + before the switch.
    You need to think more about how the circuit works: V2 ramps down until the + input to U1 crosses zero V. Then U1 output switches from +11 to -11 and V2 changes direction, now going up in voltage.
    I have to leave it at that. I practically told you what the pk-pk voltage V1 is already. Figure out how much time passes before V2 hits the voltage necessary to reverse the polarity of U1's + input every half-cycle. That times 2 gives you the period, and you do know how to calculate frequency given the period.
     
  21. Mar 26, 2015 #20

    NascentOxygen

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    Still wrong. (the sign change notwithstanding)
     
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