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opamp noise gain transfer function zero

Discussion in 'Electronic Design' started by Apparatus, May 22, 2005.

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  1. Apparatus

    Apparatus Guest

    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:
    http://www.analog.com/UploadedFiles...94843852500435775318349064202532445Fsect5.PDF

    Cheers,
    Chris
     
  2. 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.
    That looks like a usable app note.
    Likewise.
     
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