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Op amp integrator. design calculation not matching experiment.

  1. Sep 20, 2014 #1
    classic_circuit_1000.jpg
    For the above op amp circuit: Vin= squarewave with 10Vpp and 1kHz. C=0.01uF. Vcc and -Vcc is 15V and -15V. Using the above specs, I calculate R using these equations:
    C0328-Equation3.gif
    Requirement: Create a triangle wave with output voltage Vo to be 10Vpp.
    This is a gain of 1. Using the above, I find R=1/(2*pi*1000kHz*0.01uF)=15915 Ohm.
    I used resistors to get up to 15.8ohms to be the input resistance.
    Theoretically, this would give me close to the 10Vpp I want.
    Experiment:
    Using a 741 Op Amp and a 103 capacitor (0.01uF), I built the circuit and measured with O-scope. The output was 15Vpp. This is 50% off of the expected 10Vpp

    What is the explanation for my experimental values to be way off?
    Someone told me that the parts I'm using are probably not very accurate to their ratings, but it shouldn't cause this much of a difference.
    I've also tried different capacitors and still the same thing happens.
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. Sep 20, 2014 #2

    jim hardy

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    yet you used 15.8 ohms....

    also : You'll find out quickly how difficult it is to keep an AC integrator zero-centered.
    Place a few megohms around your capacitor so it'll have some DC feedback.
    Make sure RfeedbackCintegrator time constant is way longer than RinputCintegrator and you'll still have an approximately pure integration of your square wave input..
     
  4. Sep 20, 2014 #3
    I'm sorry I meant to write 15.8k. I did try a 2Mohm resistance across the capacitor. I'm still puzzled by the gain difference.
     
  5. Sep 20, 2014 #4
    Assuming a 50 % duty cycle, 10 Vpp, 1 kHz square wave input centered at 0 V, you apply ±5 V for 500 µs. That's an increment of:
    $$
    \Delta v_\mathrm{out} = \pm \frac{1}{15.8\cdot10^3 \cdot 0.01 \cdot 10^{-6}} \cdot 5 \cdot 500 \cdot 10^{-6} \, \mathrm{V} \approx \pm 15.8 \, \mathrm{V}
    $$
     
  6. Sep 20, 2014 #5

    jim hardy

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    Sanity check on the arithmetic:

    Opamp basics: Iin = Ifeedback

    Iin = 5volts/15.8K = 316 microamps
    so Ifeedback must be the same

    i=Cdv/dt,
    so dv/dt = i/C
    dv/dt = 316E-6/.01E-6 = 31600 volts/sec = 31.6 volts/ millisec,
    That's 15.8 volts every half millisec.

    About what you got ?
     
  7. Sep 21, 2014 #6

    NascentOxygen

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    It doesn't work that way for integrators.

    The gain of an integrator has units of volts per second per volt, or secs-1 for short.

    * anyone figured out how to get sub- and super-scripts in this new software?
     
  8. Sep 21, 2014 #7

    jim hardy

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    hmmm..... M. Y. and N. O. - i like your more scholarly approaches.


    if gain is the term before the integral, 1/RC,

    gain is 1/(15800 X 10^-8) = 1/.000158 = 6239

    which with five volts of input will integrate at 31645 volts/sec , 31.6 volts/millisec, 15.8 volts per half cycle of a 1khz square wave.

    Hmmm. same number i got by nuts&bolts method of evaluating currents.

    So that integrator, with an input of +/- 5 volt square wave centered on zero, should swing between 7.9 volts above and 7.9 volts below wherever its output is centered (remember initial condition with integral).

    OP reported 15V p-p, plenty close enough for 10% components and reading a 'scope screen by eye.

    I just love it when math and hardware agree.

    old jim
     
  9. Sep 21, 2014 #8
    I'd guess the OP sees ##\approx##15 Vpp since it's likely that the feedback cap starts out discharged. The opamp then just repeatedly swings from 0 V to one of its rails. IIRC one of the 741 packages can swing rail-to-rail if its output sees a high impedance (the oscilloscope probe should qualify).
     
  10. Sep 21, 2014 #9

    jim hardy

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    That's quite possible. Mother Nature loves to toy with us......

    Maybe he'll post a scope trace.
    With a 25K resistor , both with and without a couple megs of DC feedback..
     
  11. Sep 23, 2014 #10
    Thanks. I'm going to redo the circuit whenever I can. So from what I've seen here, my calculations and understanding of the gain is incorrect. Let me see if I can redo my math.
     
  12. Sep 23, 2014 #11

    jim hardy

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    oops deleted an accidentt...
     
  13. Sep 24, 2014 #12

    jim hardy

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    Gain for a sinewave will be different than for square. Integrating a sinewave doesn't change its shape like it does for square.
     
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