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Op amp integrator

  1. Dec 17, 2011 #1
    http://pokit.org/get/8119a4b8a37171c67971c3fb13b33063.jpg [Broken]

    Consider picture above.

    I understand how an integrator behaves when the capacitor is full. It(capacitor) acts as infinite resistance ergo open loop so we get positive saturation at the output.

    What I don't understand how come we are starting at 0? Doesn't capacitor act like a short circuit in the very first moments of its charging? Effectively making this a voltage follower? (at very first moments).

    I understand what voltage follower does in situation when the Vin is connected to + lead and we have short circuit between output and - lead. Question risen what would happen if I grounded a + lead and connected Vin to - through a resistor like so:

    http://pokit.org/get/4b7a8c8dca68aacc548f3a3b2d748715.jpg [Broken]

    Have I just short circuited the whole op-amp? As it isn't there?

    Getting weird results from simulations.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 17, 2011 #2
    T the start, with the switch open, the voltages at the - input and + input of the op amp are zero and there is no charge on the capacitor. The output voltage is also zero.
    When the switch is closed there is a voltage across across the input resistor and current flows from the output, into the capacitor and through the input resistor.
    This current will be constant and is equal to 1volt/ input resistance. This current charges up the capacitor until Vout is saturated at the +supply voltage. (+15V)
  4. Dec 17, 2011 #3
    This doesn't answer my question which is, at the very first moment after the switch is closed, which you say current flows from output through capacitor, is output voltage different from 0?
  5. Dec 17, 2011 #4
    If the capacitor was uncharged at the start then Vout would be zero.
    One way to see what is going on is to imagine the switch is closed for a short time so that the capacitor gains a small amount of charge and Vout has increased but has not reached +15.
    If the Switch is now opened the charging will stop and the capacitor will remain charged to thi level which means that Vout will have a steady value.
    When the switch is closed again charging will continue from this point.
  6. Dec 17, 2011 #5
    Oh I understand thank you. I misplaced few parts.
  7. Dec 17, 2011 #6

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    Isn't the op-amp saturated in its negative voltage supply?
    Apparently that is 0 [V].
  8. Dec 17, 2011 #7
    I would say that Bass's diagram is correct.
    the input voltage is -1 therefore current flows from Vout through the capacitor through Rin to the -V which means that Vout is increasing +vely
  9. Dec 17, 2011 #8
    Yes, since it acts a comparator after the capacitor is charged, and -1 is on the inverting lead, it will go to positive saturation.
  10. Dec 17, 2011 #9
    I agree with you Bass, if Vin was +1V then Vout would be -15V
  11. Dec 17, 2011 #10

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    Am I misunderstanding?
    After the switch is closed, -1 [V] is on the inverting lead.
    Before I believe +15 [V] is on the inverting lead.
  12. Dec 17, 2011 #11
    hmm why would it be +15 on the inverting lead, before closing of the switch?


    here is the complete animation.
  13. Dec 17, 2011 #12
    The graph shows V = 0V before the switch is closed. No reference is made to any voltage across the capacitor at t=0 so I take it to be zero (V out = 0 on the graph)
  14. Dec 17, 2011 #13

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    Ah, sorry, my mistake.
    I interpreted the circuit to have +15V applied on Vout, but in retrospect that actually makes no sense.

    Now I see that initially no voltage at all is applied, so Vout will initially be zero (if you wait long enough).
  15. Dec 17, 2011 #14
    At t = 0 Vout will be whatever voltage there is across the capacitor at t =0.
    it makes sense to take this to be zero but it may not be the case....depends what happened for the capacitor before t = 0 !!!!!!!
  16. Dec 17, 2011 #15
    You suggested that the capacitor 'behaves like a short circuit' and you produced your second circuit diagram with 'Key = A' as the capacitor.
    You have made a good attempt to produce another model.... but ....a capacitor is not exactly a switch.... it charges up !!!!
    And your circuit diagram is incorrect.... the -1V is not connected to earth (ground)
    I have produced a diagram which ,I think, shows what you were getting at!!!
    Can you see that your model certainly gives the current that would be flowing (1V/100Ω)
    but it does not correctly model the capacitor charging up which would give a Vout increasing .
    Try this on your software programme.... I imagine it will give you I = 0.01A but not much else.
    Change KeyA for a capacitor and see what happens.

    Attached Files:

    • int.jpg
      File size:
      23.9 KB
  17. Dec 17, 2011 #16

    Yes, I did that. I used -1 in original simulation, don't know why I set it back to 1. I am trying to have a deeper understanding of the op-amp so I am asking questions like this.

    I do now understand that the acting of the capacitor as a short circuit, is not something to be taken literally.

    I think I have a clear picture how this circuit behaves when the switch (1st picture) is closed.

    It acts like a short circuit, but not like the classic short circuit. I think we understand each other.

    Thank you for your assistance, you have been more than helpful.
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