What is the functionality of this Op-Amp circuit?

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Homework Help Overview

The discussion revolves around an operational amplifier (op-amp) circuit characterized by specific resistor and capacitor values. Participants are exploring the functionality of the circuit, particularly its behavior as an integrator and low pass filter, as well as its practical application and output characteristics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the output voltage in relation to the capacitor voltage and question the circuit's operation under various conditions, including the effects of input signals below ground. There are inquiries about the role of resistors in preventing saturation and the implications of using different power supply configurations for the op-amp.

Discussion Status

The conversation is ongoing, with participants sharing insights about the circuit's behavior and raising questions about its practical implementation. Some guidance has been offered regarding the effects of power supply types on output behavior, but no consensus has been reached on the overall functionality or output characteristics.

Contextual Notes

Participants are considering the implications of the initial capacitor voltage and the potential for output clipping based on the op-amp's power supply configuration. There is also mention of an image that is pending approval, which may provide additional context for the discussion.

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Homework Statement



For the attached circuit: R1=1Mohm, R2=10Mohm, C1=1uF. If the initial capacitor voltage is 0, find the output. Will this circuit work in practice? Explain. What does this circuit do??


What does this circuit do? This circuit integrates and is a low pass filter.
Will this circuit work in practice? I don't see why not.

The ouput voltage is equal to the voltage across the capacitor which in this case would be 0. I believe R2 is used to avoid saturation.

Am I correct in all this?
 

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Your image is still pending approval. If you need help immediately then upload on imageshack or something.
 
I think the image is ready now. Would the output voltage possibly be whatever the voltage is across the capacitor? In this case, 0??
 
The circuit is an integrator. Yea I believe the voltage out will be the voltage across the cap. Because the cap will charge to its max and become effectively an open circuit.

There is a formula for analyzing for an AC source for Vout.
 
Wouldn't the output be measured from the op-amp output to ground?

"Will this circuit work in practice?" What would happen if the input signal went below ground (negative)? Hint: How would this be effected by the op-amps' power supply?
 
Last edited:
You got me dlgoff. I don't know.
 
If the op-amp is a single supply type; say +15volts and ground, and the signal went below ground (negative) the output would get "clipped". Now if you use a duel supply type; say ±15volts, you can drive the input below ground and get a "clean" output.
 

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