Op-Amp Integrator

Try tackling the problem systematically...

Convert the circuit to the Laplacian equvilent (i.e. C becomes 1/(Cs) and derive the transfer function. The only special things you have to remember are:

a. For an ideal op-amp, the currents entering the inverting and non-inverting terminals is zero.
b. For an ideal op-amp, the voltage between the inverting and non-inverting terminals is zero.

Clue: In the s domain, a 1/s denotes an intergral.

Try this and see if it makes more sense to you.

 

Or that the voltage across a cap:

[tex]Vc = \frac{1}{C} \int i_c dt [/tex]

 

The [tex]R(f)=10M\Omega[/tex] resistor is there to make sure you do not have a small constant voltage on the inverted input (this would be integrated and you don’t want that). You can show that there is a frequency where the impedance of the capacitor equals the impedance of the R(f) resistor. Over this frequency the circuit acts as an integrator, so you can try to neglect that resistor in your calculations

 

The feedback resistor limis the gain at DC to 100. If it was not there the theoretical gain at DC would be infinit. Any small input offset would cause an integration that would eventually saturate the op amp. Then all the usual assumptions about op amps go to H@$^*.
Joe