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My other intergrater question for Daestrom

I didn't see a reply to my last message, so I'll repost it in case it got over looked. If someone replied, can you please repost it because I didn't see it.

Thanks

Something was really weird with MultiSim, maybe I had too many programs open the night I typed this question causing it to slow my PC down. After the op-amp went to the rail, there was a delay before the RC started rising. That really threw my understanding of an intergrater circuit off.

I tend to get baffled by the "outside" of the circuit and not think about the simpliest thing "the inputs are 'equal'", that explaination you gave made perfect sense.

You never got to the question about putting a cap from the resistor/capacitor junction (the inverting input) to ground. I saw a circuit with this capacitor and wasn't sure how to "add" that into the equation. Per MultiSim, it only changed things VERY little. Keeping in mind I use an ideal op-amp (no offset voltage, bias current, etc..) so when I calculate something and run it in MultiSim, my answers should be just about identical.

The caps aren't in parallel or series because I tried both and my numbers were out in left field.

Thanks again!

Reply to
Peter
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Well I have done that already and I believe it's there to 'roll' off any noise from the previous stage. As far as what it does to the ramp on the intergrater, it does change it a little bit. so I was wondering how that could be calculated into the circuit.

Reply to
Peter

The visual effect on the integrator's low freq output of a small capacitor from (-) input to ground will be minimal. The effect will be most visible on a Bode plot. Suppose you model the op-amp as an amplifier with a gain of -A (where A can be some number of your own choosing, try 200 to 4000). Then the effect of C1 (a small capacitor from the OA inverting input to ground) according to my back-of-an-envelope calculations is to lower the

-3dB corner frequency from its value without that capactor by a factor of x 1/(1 + C1/(A x C)) where C is the integrating capacitance. That is, the effect on Vout/Vin frequency response (for this simple OA model having an unrealistically perfect phase shift) is the same as that realised by increasing the feedback capacitor C by a factor of C1/(A x C), i.e., when C is replaced by C + C1/(A x C).

You should be able to confirm this by trying different values of A to see the effect on gain vs frequency at the high end of adding C1. With a step input, the effect on the slope of the ramp should be to lower it by the same predictable factor.

o---- Rin -------------------------- C ----------. Vin | | | C1 | ______ | | | | | | gnd `----(-)| | | | OPAMP|---------o .----(+)| | Vout C1

Reply to
John Savage

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