My other intergrater question for Daestrom

[Peter]

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!

 

[Peter]

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.

 

[John Savage]

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.

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 << C | |______|
gnd