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Question about pole frequency in opamp

If i consider an opamp in corrent-to-voltage configuration with:
-ideal generetor current in inverting input
-non inverting input to ground
-feedback resistance equal to 100Mohm
-open loop gain egual to A=128dB
-capacitance common mode plus differential mode equal to 3pf
i'm looking for pole frequency that shoud be equal to A0/(2*pi*C*Rf)
120dB=20LogA0=>A0=2.512e6
fp=2e6/(2*pi*3e-12*100e6)
but this frequency is too high for this opamp (http://focus.ti.com/
docs/prod/folders/print/opa128.html) , so i think that i've made some
errors...what's wrong?
 
Thanks for the infos.....
Real op-amps are not ideal, and your model is missing important pieces.
Most importantly, 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 important, 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 importance 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)
 
In addition to my previous post, i've simulated an transimpedance
amplifier with:
1)Rf=10kohm
2)ideal current generator=0.2mA
3)OPA128 (with 1MHz as open loop bandwidth)
4)only parasitic capacitance(3pf)
IId this circuit cut off frequency is equal to
Vout=(-Rf*Iin)/(1+j(f/fc))
fc=A0/(2*pi*Rf*Ci)
with
A0=open loop gain at frequency pole
because:
Rf'=Rf/A0 for miller effect
Ci=parasitic capacitance (there is no miller effect for common mode
and differential mode parasitic capacitance)
So,i've obtained fc equal to 2,3Mhz (calculated and simulated), but i
not conviced: its strange that i have a bandwidh greater then open
loop gain...I know that open loop bandwith is obtained as voltage
amplifier (instead fc is obtained as transimpedance amplirier), but
i've some doubts on these results....what do you think?
 
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