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I want to be able to capture a differential output from a shunt resistor. The shunt resistor differential voltage will range from 2.5mV to 50mV at 50Hz.

My initial idea was to send this signal through a unity gain differential amplifier and then send this output to a series of passive high-pass filters, active low-pass filters, and amplifier stages.

This will be done in order to apply a 40dB gain to the signal. Hence 50mV will be amplified to 5V. However, I want to have a resolution of around 5uV. I came across the AD8675 which has an input noise of 2.5nV/sqrt(Hz). Which looks like it will be good enough.

Can I just use the AD8675 for all stages of my design? (i.e. differential, active filters and amplification stages)

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  • \$\begingroup\$ What is the bandwidth of the signal? Is any protection required e.g. for surge or fault conditions? \$\endgroup\$ Commented May 7, 2023 at 12:23
  • \$\begingroup\$ @TimWilliams for 'now' this signal is a 50Hz sine wave. I need to confirm what other frequencies are coupled to this, however, this is my starting point. I do not expect the bandwidth to exceed 100kHz (0Hz to 100kHz) \$\endgroup\$
    – JoeyB
    Commented May 7, 2023 at 12:28
  • \$\begingroup\$ So your signal is not a sine wave! Is this a mains connected thing? Do you expect harmonics? Up to what multiple? Transients? Interference? What about surge or fault conditions? Mains can supply thousands of amperes short-circuit! \$\endgroup\$ Commented May 7, 2023 at 12:30
  • \$\begingroup\$ ...Reviewing your recent post history, I think you're already aware it can supply thousands of amperes; the question then is, is that your nominal or peak/worst-case range? And how does that affect the accuracy of your measurement -- do you need to measure peaks accurately, or can they be discarded while prioritizing nominal (maybe ~30A?) currents instead? \$\endgroup\$ Commented May 7, 2023 at 14:11

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My first thoughts are these; you ought to "draw" a signal that is typical in wave-shape but, towards the low end of the amplitude values you want to detect. Then, you should explain what it is about the signal that is important to maintain when amplified and filtered. I mention these things because I believe you may be misinterpreting what the input noise of the amplifier means.

It doesn't mean you can amplify a sinewave (low signal bandwidth) and get an improvement in noise compared to a signal of twice the bandwidth.

What it does mean is that if your amplifier channel has a noise-bandwidth of (say) 100 kHz, the equivalent input noise is 2.5 nV (RMS) × \$\sqrt{100000}\$ = 790 nV RMS and, the \$6\sigma\$ p-p value will be 4.7 μV p-p. That's the equivalent input noise.

I want to have a resolution of around 5uV

If that's an equivalent input resolution then you are close to the numbers I used above. You should make your own calculations based on the estimated noise-bandwidth of your amplifier.

Of course you may not understand noise-bandwidth so please ask if you need more detail.

Also, have you taken into account the low-frequency noise of the op-amp (100 nV p-p)? It's quite a good op-amp so maybe this won't be a showstopper.

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  • \$\begingroup\$ What is 6 Sigma? \$\endgroup\$
    – JoeyB
    Commented May 7, 2023 at 11:02
  • \$\begingroup\$ If you have RMS noise it will remain within a peak-to-peak value of 6x RMS amplitude for 99.7% of the time: read this answer. This might also be useful: analog.com/media/en/training-seminars/tutorials/MT-048.pdf \$\endgroup\$
    – Andy aka
    Commented May 7, 2023 at 11:11
  • \$\begingroup\$ I am still not understanding the pdf in the 'read this answer'. Is there a video that explains this better? \$\endgroup\$
    – JoeyB
    Commented May 7, 2023 at 12:19
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    \$\begingroup\$ I don't use videos on-line. Maybe break it down to what you do understand and ask a new question about RMS noise and it's equivalent statistical p-p value @JoeyB \$\endgroup\$
    – Andy aka
    Commented May 7, 2023 at 12:22
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    \$\begingroup\$ Sigma is a relative measure of statistical variation in a sampling. "Six sigma" simply means it's very rare (one in a million, say) for something to fall beyond this bound. If you still don't understand, I'm not sure how much help we can be here, but I would encourage taking a course on statistics. \$\endgroup\$ Commented May 7, 2023 at 12:29

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