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    Op amp integrator formula

    Hi,

    Can someone give me a simple example of this integrator formula, and confirm or correct what "s" is in this formula:

    Vout = - Vin (1/RCs)

    I understand that:
    s = jω
    ω = 2πf
    j = √-1

    Is j = - 1?

    The formula is from Op Amps for Everyone, page 429, circuit named A-27, Inverting Integrator.

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    I've struggled with: Vout = - (1/RC) * ∫ t 0 * Vin (t) * d t and cannot fathom what the ∫ t 0 means, what t in Vin is, or what d t refers to. I spent a whole day with both formulas getting nowhere but wrong answers and more and more frustrated. Help appreciated with simple examples of e.g. 1Vin, 10k, 10nF definitions and solutions to both formulas, please. I really have tried to figure both formulas out. Thanks.

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    Re: Op amp integrator formula

    No idea where you got the integral expression. I guess " ∫ t 0" is a garbled form of "integral from 0 to t", which would be the correct description of inverting integrator in time domain in contrast to the frequency- respectively s-domain description.

    Your comments about the frequency domain transfer functions are correct, except for j=-1. Correct is j²=-1. I presume that you are not familiar with imaginary and complex numbers. I don't attempt to explain it in few words, see e.g. https://en.m.wikipedia.org/wiki/Complex_number

    In the present context, the imaginary factor can be simply understood as 90° phase shift of a sine input signal.


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    Re: Op amp integrator formula

    Hi,

    Thanks. Good point about j² = - 1.

    Okay, other formula. For a 1kHz, 100mV (100mV from ground/0V) input square wave:

    I understand the notion of ∫ from 0 time to x time, but not what goes in there, e.g. Is it length of positive or negative signal such as a 1kHz square wave and number would be 0.0005s?

    Vin (t): What is t here, is it 0.1V for a signal with 100mV ampitude? Is it Vin at 0.0005s?

    d t I interpret as delta time, so would that be 0. 001s?



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    Re: Op amp integrator formula

    Integrator output voltage :Vo(t)
    Input :Vin(t)
    and

    Vo(t)=-1/(RC)∫Vin(t)dt

    where:
    RC are the components values and
    (t) is a time varying value of the variable
    with 'dt' --integrate with respect to time 't'.


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    Re: Op amp integrator formula

    Quote Originally Posted by d123 View Post
    Hi,

    Thanks. Good point about j² = - 1.

    Okay, other formula. For a 1kHz, 100mV (100mV from ground/0V) input square wave:

    I understand the notion of ∫ from 0 time to x time, but not what goes in there, e.g. Is it length of positive or negative signal such as a 1kHz square wave and number would be 0.0005s?

    Vin (t): What is t here, is it 0.1V for a signal with 100mV ampitude? Is it Vin at 0.0005s?

    d t I interpret as delta time, so would that be 0. 001s?
    I think you’re getting your brain tied in a knot. The upper bound for your integration is whatever you want it to be, it has NOTHING to do with the signal, v(t). If your signal is a square wave, that presents a special case, since a square wave is not a continuous function. For the square wave you’ll have to break it up into chunks. For instance, if you’ve got a 1KHz square wave, you integrate the first half-cycle amplitude from 0 to .0005, then the second half-cycle amplitude from .0005 to .001, and so on.


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