Re: opamp noise gain transfer function zero

"Apparatus" <apparatus.home@xxxxxxxxx> wrote in message
news:1116724423.756440.16410@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Hello,
Hi.
> I am looking at an application note for a current to voltage converter
> utilizing an opamp. There is an input capacitance across the inputs to
> the opamp, which the application note says introduces a zero at f =
> 1/(2 Pi R_feedback C) in the noise gain transfer function. This doesn't
> make sense to me. Shouldn't a shunted cap introduce a pole by
> attenuating the input signal as the cap shorts?

The cap introduces a pole in the feedback transfer function.
This creates a zero in the closed-loop gain as defined from
the op-amp input terminal pair, which is where voltage noise
gain is normally considered to be input. (It also creates a
pole in that same transfer function, but its frequency is moved
up near the loop-gain unity crossover.)

> The only way I can
> resolve this is by thinking that what they mean is that at higher
> frequencies the capacitor shunts out the signal to be amplified, and
> therefore noise dominates, introducing an effective zero in the noise
> gain transfer function. Does this make sense?

I suppose it does. The feedback is attenuated by that
capacitor above the zero frequency, causing the op-amp
input noise to be amplified with less reduction via feedback.

> The application note that I am looking at is on page 20 of:
> http://www.analog.com/UploadedFiles/Associated_Docs/5618248504532094843852500435775318349064202532445Fsect5.PDF

That looks like a usable app note.

> Cheers,
Likewise.

--
--Larry Brasfield
email: donotspam_larry_brasfield@xxxxxxxxxxx
Above views may belong only to me.


.

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