OPAMP Circuit analysis

Discussion in 'Homework Help' started by xxxyyyba, Sep 23, 2014.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Hi!
    Here is my task:

    OPAMP circuit and output voltag eare shown in image below. Assuming that OPAMP is ideal, come up with a solution by hand.

    postavka.jpg

    Then change capacitor's value from 200nF to 33nF and output voltage will be:

    postavka2.jpg

    Explain why output voltages are different ( first has triangle shape and gradually decreases and second has cut off bottom and doesn't decrease).
    Any idea here? :)
     
    Last edited: Sep 23, 2014
  2. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    The circuit is an integrator with no way to discharge C1. Since one of the inputs is a +DC voltage, the integrator capacitor will slowly charge until the output goes to the negative rail of the op amp. If you extend the simulation time you will see that you eventually end with the same result with both capacitor values. It just takes 6 times longer with the larger capacitor.
     
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  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,963
    1,098
    Well the answer is very simple, smaller capacitor will charge faster. And notice that in this circuit the capacitor is charge/discharge via constant current.
    And this is why Vout voltage rump up or down depend on whether the capacitor is charged or discharged.
    C = Q/V = (I*t)/V --> V = I/C *t
    So 33nF will be charge/discharge in rate equal to 10μA/33nF = 303V per second and 200nF in rate 50V per second
     
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  4. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Here is what I have done:
    Since there are two sources, one DC and one time dependent, I used superposition method. When time dependent source is active, our circuit is:

    sema1.jpg

    v2(t)=0\rightarrow v1(t)=0

    i2(t)=0\rightarrow iin(t)=ic(t)

    iin(t)=\frac{vin(t)}{R1}

    v3(t)+\frac{1}{C}\int ic(t)dt+R1iin(t)-vin(t)=0

    For positive input voltage vin(t)=1V we have:

    v3(t)=-\frac{1}{RC1}\int vin(t)dt=-50\int 1dt=-50t,\,\, 0\leq t\leq 100*10^{-3}s

    For negative input voltage vin(t)=-1V we have:
    v3(t)=-\frac{1}{RC1}\int vin(t)dt=-50\int (-1)dt=50t,\,\,100*10^{-3}s\leq t\leq 200*10^{-3}s
    But I don't know what to do next :(
     
    Last edited: Sep 23, 2014
  5. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Solved it :)
    If someone is interested in solution, let me know :)
     
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