RC Integrator

Discussion in 'General Electronics Chat' started by jag1972, Oct 18, 2015.

  1. jag1972

    Thread Starter Active Member

    Feb 25, 2010
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    I am looking at the passive RC Integrator circuit, the Vout equation for this circuit is:  Vout= \frac {1}{RC} \int Vin \: dt . This only works for certain conditions e.g. when  RC > T . This is however misleading as most text books show a square wave which equals  5tau . This is totally confusing IMO as the vout equation does not match the output voltage at all. Is it just me that thinks this or is it misleading or have I made a mistake in my assumption.

    Jag.
     
  2. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    40
    The equation for Vout is valid when time, t, is less than the time constant, RC. That is t < RC = tau. This represents only the first portion of the response to the squarewave transitions. The circuits use as an integrator is limited in time. After time t it is no longer integrating.

    Another equation for the capacitor charging is an exponential voltage rise (or decay depending on initial conditions) to the applied voltage. The amount of time it takes to get over 99% of the way there is 5tau. The first one fifth of this equation looks very much like a straight line.
     
  3. jag1972

    Thread Starter Active Member

    Feb 25, 2010
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    Thanks for your reply
     
  4. jag1972

    Thread Starter Active Member

    Feb 25, 2010
    60
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    I have a series RC Circuit where  R= 1 k\Omega\: and C = 10 nF , the input to the circuit is a square wave with a period 1 ms.
    The input square wave is: \: x(t)=\:    5,\: 0\geq\:t\:> 0.5 ms\:  0,\: 0.5 ms\geq\:t\:> 1 ms .

    When I use the integrator formulae:  \frac{1}{RC} \int_{0}^{0.5 ms} \: x(t}\:dt

     tau = 0.1 ms , therefore after 5 tau, vC is 99.9% of 5 V. This can be determined from the following equation/model for vC:
     vC=5(1-\exp^{-5})
    However when I use the integrator formulae I do not get that value:  Vout = \frac{1}{RC} \int_{0}^{0.5 ms} 5\: dt
    The value using the integrator formulae is 25 V which is clearly wrong can some please tell me where I am going wrong. I am aware that the integrator only works as a true integrator when RC is large compared to T. But I would have thought that Vout should always be right.
     
  5. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    Let's start with recomputing tau.

    tau = R * C = 1000 ohms * (10 * 10^-9 F) = 1 * 10^-5 s = 0.01 ms = 10 us.

    In order that your circuit is integrating the time over which you are integrating must be less than tau. The time over which you are integrating is 50 * tau. Even at 5 * tau in your initial calculation, the integration is no longer valid. You can only do your integral from time = 0 -> 10 us. With times greater than 10 us, only the exponential equation is valid.

    Look at the exponential response of the integrator to the input square wave. At the transitions the output starts off rising linearly, that portion of the response is what is equivalent to the integral. Once the graph starts to change slope, the integration will begin to have errors. So, time < tau is stated as begin the only valid time for integration because the output is very close to linear, the expected result of the integration of a square wave.
     
  6. jag1972

    Thread Starter Active Member

    Feb 25, 2010
    60
    0
    Thanks for your reply, I thought that was the answer. BTW I made a mistake with the reistor value it was supposed to be  10 k\Omega .
     
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