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op-amp integrator basic's

Discussion in 'Electronic Basics' started by [email protected], Feb 22, 2005.

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  1. Guest

    Hi, I am using a classic opamp integrator, my input is a 500us pulse at
    a freq. of 80hz. The signal can vary in amplitude from several mv to
    about 8volts.

    The integration time is 10ms, which is about 20x longer than the
    expected input. I noticed, if the input voltage exceeds ~4 volts, a dc
    offset appears
    at the output and increases as a function of input voltage. To mitigate
    the problem, I placed a 75k resistor across the .1uf, to discharge the
    cap before the next pulse. The integrator's C=.1uf and the r=100k.

    I have several questions:
    Would it be better to increase the time constant, rather than placing a

    resistor across the integrator cap?

    As far as integrating cap's is there a specific type used, is ceramic
    ok?

    I can provide a 5volt logic signal at the proper time for to control a
    switch
    which would discharge my cap, can a device be recommended?

    thanks boats_ranger
     
  2. There is no such thing as an "integration time" that can
    be considered a property of the integrator itself. If
    you are refering to the R*C product, that affects the
    gain of the integrator, which has units of (volts out)
    per (volt second product in).
    Perhaps your integrator is reaching the limit set by
    the supply. That would call for decreasing its gain.
    Yes, assuming that you mean decreasing the gain of
    the integrator and it is important to have an integrator
    rather than a low-pass filter.
    That depends on what kind of linearity, accuracy,
    stability, and insensitivity to acoustic noise your
    application requires. Not knowing any of that, I
    am in no position to gainsay your choice of cap.
    I like the 4066 for this sort of thing. But your
    supply levels may preclude its use and require
    something like a DG211 or its ilk. If the input
    has a known polarity, a single MOSFET of the
    right polarity can do the job, or, if offset at reset
    is not critical, a BJT can even be used.
    Anytime.
     
  3. Guest

    Hi, I have several more questions:

    In this situation, I'm integrating over a single pulse. I was under the
    assumption
    that a long integration time improved linearity (as a by product
    reduced
    the output voltage). Is this correct?

    If linearity then accuracy are my primary concern can a capacitor type
    be recommended? Is there a type to avoid?

    Which value should be used, 50hz or pulse width (500us) of the input
    signal to calculate the corner freq? ( I hope I stated this correctly)

    Can the 4066 conduct voltage in both directions?

    If I used a mosfet, would it be connected across the integrator cap?
    If so, the source would in my case be at ground.

    thanks, boats_ranger
     
  4. Yes, in general.
    Avoid the high-K dielectrics in that case. If high
    linearity is important, you may want to use a film
    capacitor, such as polystyrene or polypropylene.
    However, since you were willing to put a resistor
    across the integration capacitor, I suspect that
    your linearity requirement is modest and just
    about any capacitor would do.
    I do not know how to usefully assign a corner frequency
    to an integrator. Its dominant pole is so close to zero
    that the difference is not worth talking about here.

    I would start with the output that you wish to see for
    the largest input (expressed in volt-Seconds). This
    sets the gain your integrator will have. It is easy to
    convert volt-Seconds in, divided by the input R, to
    a charge. It is equally easy to convert that charge
    and the desired integrator output, to a C value. Or
    set 1/(R*C) == integratorOut / voltSecondsIn.
    Yes, and that is why it works out, often.
     
  5. Guest

    I would start with the output that you wish to see for
    for example: the input to the integrator is 1v
    r = 1000ohms
    c= .1uf

    output voltage =(1/(r*c)) * input voltage

    output voltage = (1/(1000 *.1uf) * 1volt
    output voltage = 10000v/sec
    My first impression is to use a N channel device,
    with the drain connected to the opamp " -" and the source connected
    to the opamp output pin. To turn the device on, the gate potential
    needs to be greater than source. To ensure proper operation, 5v may
    not be
    enough.

    If this is the case, it seem a 4066 maybe a better fit solution.?

    Your thoughts.

    thank's boats_ranger
     
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