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?

 

[Tim Wescott]

lionelgreenstreet@gmail.com wrote:

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?

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

 

[John Larkin]

On Sun, 29 Jun 2008 03:24:38 -0700 (PDT), lionelgreenstreet@gmail.com
wrote:

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?

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.....

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?