Thanks for the infos.....
>Real op-amps are not ideal, and your model is missing im****tant pieces.
>Most im****tantly, nearly all op-amps available today are internally
>compensated, which means that the dominant pole is determined by an RC
>pair thats inside the amp, and is not something that you can change
>short of changing the part number of the op amp you're using.
So i've used the open loop opamp gain in cut-off frequency equation
(open-loop gain depends on opamp internal pole)
>Second most im****tant, I don't know of any op-amps that are going to
>operate realistically with a 100Mohm feedback resistor. There may be
>new, ultra-low-power ones that could do this, but that seems awfully low.
I've used this application note:
http://focus.ti.com/analog/docs/techdocsabstract.tsp?familyId=1483&abstractName=sboa061
where i've found feedback resistor of 400Mohm max
>Finally (I don't know how much im****tance to attach to this), I don't
>think your A0/RC calculation means anything much in particular, so I
>don't think it's going to be much use to you.
>The pole caused by Rf and the capacitance at the inverting input, Ci,
>has its corner at a frequency of 1/(2*pi*Rf*Ci), irrespective of the
>opamp's gain. So rip A0 out of your numerator.
The pole is caused by Rf, but Rf value changes for miller effect, so
its value is
Rf '=Rf/(A0+1)=Rf/A0
the pole frequency becomes
fp=A0/(2*pi*Rf*Ci)
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