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    input current of the integrator

    An integrator is built with an opamp and a feedback capacitor acossing the output node and negative input node of the opamp. The input current charges the negative input node, and the negative input node will manitain virtual ground, and the output voltage has a ramp signal.

    Where is the input current going? The input current can not across the feedback capacitor, and the input gate of the transistor is a high impedance. Thanks.

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    Re: input current of the integrator

    The input current is required to have zero average. Please notice that an integrator must be either equipped with a reset switch or operated with external negative feedback, otherwise it runs into saturation.

    Review real world integrator applications for clarification.


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    Re: input current of the integrator

    Quote Originally Posted by shanmei View Post
    Where is the input current going?
    The input current can not across the feedback capacitor
    No.

    Input current goes to feedback capacitor(integral capacitor).


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    Re: input current of the integrator

    Thank you for the feedback. Sure, a reset switch is needed to prevent saturation.

    But during the integration process, how can the current "goes through" the feedback capacitor? For example, if we apply a dc current.



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    Re: input current of the integrator

    The question was answered in post #3. It's elementary electronic law, I = C*dU/dt

    If you apply a DC current, the integrator output ramps respectively until it reaches the saturation limit.


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    Re: input current of the integrator

    how can the current "goes through" the feedback capacitor
    It's true that the capacitor needs to charge through a sensible value of impedance. If it sees near-infinite ohms in either direction then it isn't workable. Possibly the incoming signal comes through a very high resistance (or unknown resistance). In that case a proper op amp schematic needs to include a component (example, resistor) which creates a workable RC network with the capacitor.


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    Re: input current of the integrator

    Sorry I still do not understand.

    Assume the input current is dc current, it can not go through the feedback capacitor as I1, and it can not go into the negative input node of the opamp ( the gate of transistor), then where it goes? Thank you.


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    Re: input current of the integrator

    Understand that AC current is DC current until it switches direction. 50hz AC is DC for 10ms periods.

    So a cap is perfectly happy passing DC current...until its voltage as described by the equation in post #5 causes circuit saturation or cap failure. Hence in the long run caps can only supply AC.


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    Re: input current of the integrator

    Vcm-Vout=Integral[I1]/C
    Vout=Vcm-Integral[I1]/C


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    Re: input current of the integrator

    Quote Originally Posted by shanmei View Post
    .

    The input current can not across the feedback capacitor
    Why not? Current aka charges are flowing to one of the plates of the capacitor, and the op-amp drains charges (of the same polarity) from the other plate, resulting in a rising potential difference between the two plates (which is equal to the integral of current/capacitor)


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    Re: input current of the integrator

    when you charge or discharge a cap - current flows into and out of it - building or reducing charge on the plates ...


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    Re: input current of the integrator

    Thank you for all your feedback.


    Why node Vx ( negative input node of the integator) does not change when the current charges Vx node, because it is is a low impedance node?

    For an integrator, the input impedance is 1/(SC)/A. For example, if A=10000, then the input impedance of the integrator is still infinite for dc frequency?



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    Re: input current of the integrator

    Quote Originally Posted by shanmei View Post
    Thank you for all your feedback.


    Why node Vx ( negative input node of the integator) does not change when the current charges Vx node, because it is is a low impedance node?
    It does not change only if your gain is infinite. If your gain is finite, it does change.


    For an integrator, the input impedance is 1/(SC)/A. For example, if A=10000, then the input impedance of the integrator is still infinite for dc frequency?
    Doesn't it look like a capacitance of value A*C? (Miller effect!) And the impedance of a capacitor at DC is infinite anyway.


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    Re: input current of the integrator

    Because keeping the + and - inputs equal (by sourcing or sinking output current/voltage) is the textbook behavior of an opamp.

    Though it has limits in its ability to do that (output current, bandwidth, voltage).



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    Re: input current of the integrator

    Sure, the feedback loop of force negative and positive node of the opamp to be equal.

    Input current charges the negative node of the opamp, making it has a voltage rise, then the feedback loop sinks the output current leading to the voltage on the other plate of the capacitor ( the output of opamp ) to drop. This voltage drop has a bootstrap effective that making the negative input node to drop until it is equal to positive node voltage. Does it operate like this way? Thank you.



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    Re: input current of the integrator

    Yep that's exactly it.


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