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integrator circuit fault

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Derun93

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Hi, I share input and output signal below. I could not understand why the output signal is like that?
Please tell me my fault.

signals.pngintegral.png
 
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The capacitor is assumed to have no charge at time=0.

The capacitor charges from the output. The only way charging can take place, is if it can pass current in one direction or the other. The op amp is labelled 'ideal' therefore its input has infinite resistance.

Therefore charging current needs to be able to travel from or into Vin. However the schematic does not specifically indicate this is possible.

Therefore we must assume Vin is a voltage source whose unseen terminal is connected to 0V ground. And R1 is the total resistance in that path.
 

what CAD are you using?? Are you shure of the connections? IMHO there are some problems in the schematics and note that you are using uV as input signal, i think that they are too small!!! try at least some x10mV
 

Hi,

from the picture it seems the lower supply rail to br GND.
But from the diagram it seems the lower supply rail is at baout -30V.

****
280uV @ 10kOhms = 28nA.

28nA @ 100nF gives a constant rate of I/C = 28nA / 100nF = 0.28V/s

***
What is Ksec? kilo-seconds?

Klaus
 

from the picture it seems the lower supply rail to br GND.
But from the diagram it seems the lower supply rail is at baout -30V.

****
280uV @ 10kOhms = 28nA.

28nA @ 100nF gives a constant rate of I/C = 28nA / 100nF = 0.28V/s

Why from -25 to -28 V on the output graph has shape like a decreasing exponential ? It is not supposed to be a line with slope -1/CR until reaches lower supply rail ≈-28 V ?
 

Hi,

-1/CR? I don't think so.

Curve: Maybe caused by input currents, out of operating range....
Hard to say without knowing any specifications.

Klaus
 

What you posted is a subcircuit in ADS. In order to help you, the complete circuit, at simulation engine level, have to be shown.
 

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