Inverting Op-Amp Integrator and Low Pass Filter

[bengaltiger14]

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. I believe R2 is used to avoid saturation.

Am I correct in all this?

 

[berkeman]

This is homework, but I'll leave it here because it is similar to the recent "Bad Circuits -- Test Your Knowledge" thread.

To answer your question, you should look up the datasheet specifications on two typical jellybean opamps: LM741 and TL081. What is different about these two types of opamps? What issues can you identify with using such a large input resistor value R1 for this circuit? Is one of the two opamps listed preferable when using large input resistor values?

 

[unplebeian]

The function of R2 is to make the circuit behave close to an ideal integrator. This is apparent if you look at the bode plot of the transfer function of a low pass 1/(s+1/tau) or Z= R||C which is 1/(RCs+1).

The break off point is tau.

Berkeman is right. R1 seems unnecessairly high for a device that has high impedance inherintly.