Question about pole frequency in opamp

Your question is obfuscated in details.

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.

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.

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.

I think you need to understand op-amp design better, or get yourself a better cookbook.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

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.

If that pole is below the amp's unity-gain frequency, it adds (approaching) 90 degrees of feedback lag to the opamp's (nearly) 90 degrees of forward-gain lag, so the mess may go unstable.

John

Thanks for the infos.....

So i've used the open loop opamp gain in cut-off frequency equation (open-loop gain depends on opamp internal pole)

I've used this application note:

where i've found feedback resistor of 400Mohm max

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?