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26th July 2019, 14:30 #1
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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.

26th July 2019, 15:40 #2
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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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26th July 2019, 15:46 #3

26th July 2019, 18:04 #4
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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.

26th July 2019, 22:41 #5
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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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27th July 2019, 00:42 #6
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Re: input current of the integrator
how can the current "goes through" the feedback capacitor
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27th July 2019, 02:00 #7
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27th July 2019, 02:37 #8
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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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27th July 2019, 02:54 #9
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Re: input current of the integrator
VcmVout=Integral[I1]/C
Vout=VcmIntegral[I1]/C
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27th July 2019, 07:35 #10
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Re: input current of the integrator
Why not? Current aka charges are flowing to one of the plates of the capacitor, and the opamp 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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27th July 2019, 08:14 #11
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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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29th July 2019, 18:03 #12
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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?

29th July 2019, 18:16 #13
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Re: input current of the integrator
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?
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29th July 2019, 18:36 #14
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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).

29th July 2019, 20:02 #15
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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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29th July 2019, 20:06 #16
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Re: input current of the integrator
Yep that's exactly it.
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