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opamp noise gain transfer function zero

Hello,

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 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?

The application note that I am looking at is on page 20 of:

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Cheers, Chris

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Apparatus
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Hi.

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.)

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.

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That looks like a usable app note.

Likewise.

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--Larry Brasfield
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Larry Brasfield

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