I don't understand how to calculate then properly draw the output waveform for an Integrator.

Discussion in 'Homework Help' started by G14Richard, Dec 7, 2014.

  1. G14Richard

    Thread Starter New Member

    Dec 7, 2014
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    Greetings all, thank you in advanced for understanding my VERY newness to op amps.

    I've been tasked with drawing the input and output waveforms, indicating the voltage levels, of an integrator with a given frequency of 20kHz and an initial voltage of zero.

    For wave frequency, frequency = 1/Period, thus my period per wave (T) is 50 micro seconds. (Not 100% sure if I am applying this correctly to the equation).

    My Resistor is 10K OHMS, my Capacitor is 0.01 micro Farads.

    Again, initial voltage=0.

    Voltage in (square wave) is plus/minus 4 v.

    I've attached my attempt to answer the question, as well as my work.

    IntegratorQuestion.jpg

    A questions is, when I am calculating for Vout, I don't seem to reach the (expected) value of 2 v, either positive or negative. Am I supposed to reach a 2v output? My output is consistently either zero or negative 1. For this question, is this correct?

    Where am I going wrong - OR, is this the correct answer?

    Thank you.
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,142
    1,790
    You only reach a particular value on the output of the integrator if there is enough time. If there is too much time the output will go to the rail and stay there until the polarity of the input is reversed.
     
  3. G14Richard

    Thread Starter New Member

    Dec 7, 2014
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    Thank you for your reply.

    Based on your response, and given that my calculations are based on the time interval of 25 micro seconds for half the input wave (which I based on a full wave period of 50 micro seconds), this will afford me 1 volt maximum, always. If I were to extend the time intervals to, say, 50 micro seconds, I would reach my 2 volt (expected) Vout. However, given the frequency of 20kHz in the homework question, how does one justify extending the time periods for the calculations?
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    The justification is that you want the answer, if only to satisfy you own curiosity.
     
  5. G14Richard

    Thread Starter New Member

    Dec 7, 2014
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    0
    Thanks for your input, hopefully someone else will be able to answer the question or offer direction.
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    1,790
    Why do you think 2V is the expected value?
     
  7. G14Richard

    Thread Starter New Member

    Dec 7, 2014
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    The 4 examples I was given in class has had a Vout equal to 1/2 of the Vin. As my Vin is 4 V, I've been expecting a Vout of 2 v. Again, I am new to this and am basing my expectations on my (limited) experience - hence my question to this forum.

    If I performed the exercise correctly, which resulted in the correct output waveform, then I performed the exercise correctly and it would be good to know that. If my process and work is flawed, it would be good to know that too.
     
  8. Papabravo

    Expert

    Feb 24, 2006
    10,142
    1,790
    In response to a step function the integrator will produce a linear ramp. If you change the sign of the value of the step function, the ramp will reverse itself. The peak values of the ramp depend entirely on the length of time that the integral is taken over. If your square wave was a different frequency then you might expect to reach a different peak value.

    You should go back to the 4 examples given in class and ask why did all of these examples have a peak Vout of Vin/2. Was it the frequency of Vin or was it something else? Try doubling the frequency of Vin, try cutting it in half, what happens.

    I could give you a pat answer to your question or I could challenge you to answer your own question. You have all the information and tools that you need. I choose the latter; trust yourself!
     
  9. crutschow

    Expert

    Mar 14, 2008
    13,009
    3,233
    A quick rule-of-thumb for an integrator is that the output change becomes equal to the (negative of the) input DC voltage in one time-constant (R * C).
     
    Last edited: Dec 8, 2014
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