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For a particular current transducer, http://www.digikey.com/catalog/en/partgroup/ho-p-sp33-series/49533, it states an "Output rms voltage noise (spectral density)" of 6.13 uV / sqrt{Hz}.

If the frequency of interest is from 1 Hz to 100 Hz, what would be the resulting peak-to-peak voltage variation of the output?

Would it be:

6.13uV x (\sqrt{ 100 Hz - 1 Hz }) x 6.6 = 403uV,pp,

meaning the resulting output voltage of the transducer is likely gonna vary by ~403uV peak-to-peak from noise within the transducer?

(NOTE: The 6.6 presumably is some number related to statistical probably of the resulting peak to peak voltage)

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  • \$\begingroup\$ This document, www.analog.com/media/en/training-seminars/tutorials/MT-048.pdf, may have an answer. It looks like if the bandwidth is high enough and the low-noise corner frequency is low enough, the assumptions from my original post would be mostly correct. \$\endgroup\$ – plu Jan 18 '17 at 3:05
  • \$\begingroup\$ I don't think p-p value tells you much about band-limited white noise. It's a probability thing, so on rare occasions you may observe a outstanding peak value but has little to do with the performance of the system. You are much better off just do the RMS which is closely related to noise figure and bandwidth, as you calculated, the 403uV value. \$\endgroup\$ – user3528438 Jan 18 '17 at 3:07
  • \$\begingroup\$ Well, the focus actually isn't on the transducer's performance; my real application involves the sampling of this transducer's output with an ADC, and I figured that if I can work out the voltage fluctuations based on the transducer's inherent noise, then it'd be possible to choose an appropriate bit-resolution of the ADC. \$\endgroup\$ – plu Jan 18 '17 at 3:21
  • \$\begingroup\$ "Unless otherwise stated (e.g. “100 % tested”), the LEM definition for such intervals designated with “min” and “max” is that the probability for values of samples to lie in this interval is 99.73 %. For a normal (Gaussian) distribution, this corresponds to an interval between -3 sigma and +3 sigma. If “typical” values are not obviously mean or average values, those values are defined to delimit intervals with a probability of 68.27 %, corresponding to an interval between -sigma and +sigma for a normal distribution." \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 18 '17 at 5:44
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The 6.6 presumably is some number related to statistical probably of the resulting peak to peak voltage

When you have gaussian noise you have a picture like this: -

enter image description here

The RMS value of the noise occurs at 1\$\sigma\$ (one standard deviation) and that means the signal is constrained as a peak-to peak signal to this RMS number for 68% of the time. That's not great for estimating p-p amplitude so, some people go for 6 \$\sigma\$ i.e. multiply the RMS value by 6 to get p-p but that is only true 99.7% of the time. Some people use 6.6 \$\sigma\$ and that guarantees the p-p value is constrained in amplitude to 6.6 x RMS for 99.9% of the time. This is the general norm but, some folk might want to go much higher and maybe use 7 or 8: -

enter image description here

If the frequency of interest is from 1 Hz to 100 Hz, what would be the resulting peak-to-peak voltage variation of the output?

Your CT's data sheet appears to say that the spectral density is constant from DC i.e. there is no flicker-noise (low frequency) effects so, if you have a bandwidth of 1 Hz to 100 Hz then the output noise will be: -

6.13 uV x \$\sqrt{99}\$ = 61 uV RMS (402.6 uVp-p using 6.6 \$\sigma\$).

However, this assumes a perfect brick-wall filter above 100 Hz. If you have a 1st order, low pass filter shaping the upper spectrum, the voltage noise will be \$\sqrt{\pi/2}\$ times higher at 504.5 uVp-p or 76.4 uV RMS. The hike of \$\sqrt{\pi/2}\$ is due to the noise equivalent bandwidth (in case you want to look it up). Here is a table that relates the filter order to the factor you have to use for powers and, thanks to @carloc, correcting my earlier mistake, these values need to be square rooted when talking about voltage noise: -

enter image description here

Also, the above table is for a butterworth type filtering. If different filters are used, the values are somewhat altered.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Nick Alexeev Jan 18 '17 at 22:24
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    \$\begingroup\$ I belive you are still missing one point. \$\pi/2\$ factor applies to bandwidth and hence to noise power. So noise voltage, both RMS or PP will rise as \$\sqrt{\pi/2}\$ only, up to 76uV only. More, in my own experience an RMS-to-Peak-to-Peak ratio around 4 or 5 makes much more sense than the overkilling 6.6 you suggest. \$\endgroup\$ – carloc Jan 19 '17 at 8:17
  • \$\begingroup\$ @carloc I use 6.6 but I'm not making a case for using it or any number - it's down to the user (and the target system they are working) on to decide what is appropriate. I'll check on the square root thing shortly (hopefully). \$\endgroup\$ – Andy aka Jan 19 '17 at 8:25
  • \$\begingroup\$ @carloc I believe this page does show that the square root isn't needed for signals but I welcome any thoughts on this. \$\endgroup\$ – Andy aka Jan 19 '17 at 8:33
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    \$\begingroup\$ So, the document you linked states: "Equivalent noise bandwidth(ENBW) is defined as the bandwidth of a brickwall filter which produce same integrated noise power as that of an actual filter", in the above example the 99Hz bandwidth turns into approx 157Hz. From now on nothing changes...so \$6.13\,\mu\text{V}/\sqrt{\text{Hz}}\times\sqrt{157\,\text{Hz}}\approx \, 76 \, \mu \text{V}\$ \$\endgroup\$ – carloc Jan 20 '17 at 7:44
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To find the peak-to-peak voltage you must integrate the power over your bandwidth to get the rms and multiply your rms by 6 sigma to get ~Vpp. P=v^2/r. So you Vpp=sqrt(en^2*bandwidth)*6=370uV

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  • \$\begingroup\$ Yes, the 6 sigma will give you 97% and 6.6 sigma get you closer. \$\endgroup\$ – ping Jan 18 '17 at 5:01

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