Exact role of opamp in integration circuit..??

Discussion in 'General Electronics Chat' started by Himanshoo, Oct 11, 2015.

  1. Himanshoo

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    Apr 3, 2015
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    Hi guys please help me on this...

    Is integration equals amplification....?? if yes then why it happens like this..

    The text says “ the role of opamp in integrating circuit is used to give higher output voltage than a simple RC circuit used for integration..”

    question is that if opamp amplifies the output voltage then definitely the peak amplitude status of the output voltage changes…and as we know that in integration circuit the peak amplitude of output voltage depends(proportional) on Vin and time duration of input voltage …and if this(amplification) happens..then the output voltage is not exactly proportional to integration of input voltage (product of changing input voltage over time)…so according to this there shouldn’t be any amplification…


    If no amplification is desired then why opamp is used …what additional functionality it gives to the circuit …??

    please correct me...
     
  2. Jony130

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    Yes, this circuit is indeed a voltage amplifier Av = - Xc/R1. Op amp act just like a input voltage to current converter and this current is convert into voltage via capacitor. And capacitor voltage is proportional to the time integral of the current.
    http://www.electronics-tutorials.ws/opamp/opamp_6.html
     
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  3. crutschow

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    That text is inaccurate. The amplification of the voltage is not the main role of the op amp in the integrator circuit. Don't know were your read that, but it's obviously not a reliable source for technical information.

    The main purpose of the op amp in the circuit shown is to generate a true integration of the input, not the exponential integration that a passive RC circuit does (which is not true integration but sometimes called a "running average" integration).
    In a true integrator a DC input will give a linear voltage ramp output the continues indefinitely.
    In an passive RC integrator a DC input will give an exponential voltage output with a final value equal to the input voltage.
    The op amp performs this true integration function by keeping the voltage across the resistor constant ("-" op amp input at virtual ground) as the capacitor charges and the output voltage changes.
    This causes the current through the resistor to be a constant (for a DC input). This means an equal and opposite current goes into capacitor (due to the op amp virtual ground) generating the ramp at the output. This ramp will continue until the op amp output saturates, if not otherwise stopped.
    Note that the polarity of the ramp is opposite the polarity of the input, due to the inversion of the op amp.

    A interesting fact is that, with a DC input, the ramp output of the op amp will increase by a voltage (inverted) equal to the DC input in one RC time-constant.
    Also, for an AC input, the output voltage is equal to the input voltage at a frequency of [1 / (2pi*RC)], with a response slope of -6db/octave or -20dB/decade. The phase-shift of the output is a constant -90°, independent of frequency.

    Make sense?
     
    Last edited: Oct 11, 2015
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  4. Papabravo

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    Feb 24, 2006
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    Integration and amplification are separate concepts. The role of the opamp is to restore the losses that would occur in a purely passive circuit, and to prevent the integrator from becoming an excessive load on the signal source. It should be evident from the operational description of the integrator that one of the following must be true:
    1. The signal source must go both positive and negative. That is the capacitor must be charged and discharged continuously, OR
    2. The integrator must be reset periodically, to prevent saturation at the positive or negative rail.
    EDIT: and what @crutschow said above.
     
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  5. Himanshoo

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    Apr 3, 2015
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    ok if i assume that the output voltage is changing due to charging of capacitor…and for positive cycle of input voltage the output voltage starts negative..but what about the basics..since we all know about the laws of electrostatic induction…that a charge can’t exist alone…so if we think the right end of the capacitor to be negative charge then it means according to the laws that equal and opposite charge is induced on the other(left) plate of the capacitor …but then how can this be possible…because according to this scenario the left end of the capacitor is held at virtual ground….
    so first the point of query becomes that how exactly the capacitor charges despite its one plate is at virtual ground…?

    yes for dc input it may be right to some extent since dc voltages are not too much fluctuating ….but what will happen for other inputs such as sine, cos…etc…
     
  6. MikeML

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    The role of the opamp in an integrator is to make the current in the capacitor equal to the current the resistor.

    279.gif

    Left as an exercise for the student:

    1. Why is I(C1)visible and I(R1) is not?
    2. Why is V(out) larger in amplitude than V(in)?
    3. Why is V(out) delayed, and the opposite polarity with respect to V(in)?
     
    Last edited: Oct 12, 2015
  7. Jony130

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    Simply capacitor do not care where our " Virtual ground " is. Notice that current enters into the capacitor at "virtual ground" when we normally would have a positive plate. But thanks to op amp "action"... The output voltage must change to keep virtual ground at 0V and this is why the capacitor is charged by a negative voltage.

    AA.png
     
  8. crutschow

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    For a DC input it's not "may be right to some extent" it's completely right (within the op amp voltage limits).

    For an AC input it's also correct. It generates the integral of the AC input.
    For example a sine input gives a cosine output, a square-wave input gives a triangular wave output, etc.
     
    Last edited: Oct 12, 2015
  9. Himanshoo

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    you means its actually the opamp output voltage which charges the capacitor in order to maintain the virtual ground ...not the input current ..
     
  10. Jony130

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    Well, yes we can say that. Also please notice that, from the current point of view we have a series circuit. Only "one" current will flow in this circuit, from: +Vin --->R---->C--->opamp output---->-Vin
    As this diagram shows
    AA.png
     
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  11. crutschow

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    It's both.
    Yes, the output voltage does change to charge the capacitor and maintain the virtual ground.
    The flow of current if from the input resistor, through the capacitor, to the op amp output.
    As you know, to charge a capacitor the current into one terminal has to equal the current out of the other terminal.
     
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  12. Himanshoo

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    Ok the whole idea of involving a capacitor is to linearly charge the capacitor in order to correctly integrate the input waveform..for that we have to ensure that it do not charge exponentially ..as crutschow in previous post said...
    Now the reason i know that capacitor charge exponentially is that as the capacitor starts charging,the charges which are already built on capacitor opposes the further flow of charges..which makes the process slow and we get an exponential curve out of it...
    also somewhere I read that in order to charge a capacitor exponentially a constant voltage source is needed ..and to charge it linearly a constant current source is needed ..maybe thats why with the use of opamp more emphasis is made to make the input current constant..so that it could linearly charge the capacitor...
    so how does a constant voltage source and a constant current source charges a capacitor exponentially or linearly...??
    I am not sure of it...
     
  13. Jony130

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  14. crutschow

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    The voltage change induced on a capacitor is (I*t / C) where I is the current flowing into (or out of) the capacitor, t is time, and C is the capacitance value.

    Now it can be seen that if I is constant, as the op amp circuit provides with a DC input, then the voltage change with time is also a constant ( a linear ramp).

    In an RC circuit with a DC voltage input, the voltage across R varies as the capacitor voltage changes, so I (V / R) is also not constant.
    Thus the current into the capacitor varies with time. The variation is such that it gives a exponential change in voltage on the capacitor.
     
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  15. MikeML

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    Himanshoo, please explain the behavior of V(rc) and V(int) for time greater than 120ms.

    126.gif
     
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  16. Himanshoo

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    Nice and very lucid explanation thanks for it ....

    But i am think the other way...What i think is that ..for a voltage to be constant across a resistor there should be a potential difference across it and that potential difference should be constant...ok ...now lets assume a varying input voltage (sinusoidal) applied...so in the integrator circuit..the left end of the input resistor would be at varying voltage level..due to sinusoidal input ...and the right end of the resistor would be at zero volts(virtual ground)...so now technically speaking..in order to maintain constant current the potential at the two ends of resistor should vary in equal and opposite direction..but due to opamp the right end of the resistor is held firmly at 0 volt..so how a constant current is maintained if the left end potential of the resistor constantly fluctuates with the input voltage..??
     
  17. MikeML

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    With a sinosoidal (any time-varying) input, the current through R1 is not constant. Go back and study what I posted in post #6. Note that the current in R1 is equal to the current in C1. Go back and reread the first sentence in post #6.
     
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  18. Himanshoo

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    I know the point what you want to make.... but if we consider only current we know that none of the current flows in opamp and as Ibias is very low so most of the current will go into the capacitor... its a kind of a series circuit...but my point of query is the opamp contribution in making the current across input resistor constant....
     
  19. Himanshoo

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    Apr 3, 2015
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    If we talk about the capacitor voltage ..I want to confirm how the capacitor exactly charges....and how virtual ground is established...?


    lets assume that sinusoidal input voltage increase due to which an increasing current starts to flow in the resistor...now as the current is increasing with the input voltage the voltage drop at the resistor increases ...causing a low potential at the right end of the resistor...this low potential is more than 0 V at noninverting terminal..so opamp will try to lower it further and output a low voltage as well ....this low Vout appears at the right end of the capacitor...

    now i am confused what voltage will appear at the left end of the capacitor before reaching zero volts...
    please explain....
     
  20. MikeML

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    The opamp cannot make the current in R1 constant! All it can do is drive its output pin to a voltage. That causes a charging current into the capacitor. In order to satisfy the "virtual ground" at the right end of the resistor, the current through the capacitor has to match the current in the resistor...

    Unfortunately, the simplistic example of the DC input in post #7 has led you astray... All of the talk about "constant current in R1" is restricted to just that simplistic example; it is not the general case, which I showed in post #6.
     
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