CompactDAQ NI 9215 Accuracy

[KarenM]

I would like to verify that the absolute accuracy for the CompactDAQ NI 9215 calculated according to the datasheet specs is greater than 310 uV? I have crossed referenced my equation with the example listed in the datasheet for a similar USB model: USB-6210. I am wondering if this CompactDAQ model will surpass it in terms of accuracy in millivolt readings.

Here are my calculations. If someone could please verify this, that would be wonderful. The intended input is from +/-1V range.

Absolute Accuracy = Reading*Gain Error + Range*Offset Error + Noise Uncertainty

= 1V (0.0002) + 1 V (0.00014) + 7.7uV (assumed 256 uV output)

= 340mV + 7.7uV

Also, what is the best way to find the noise uncertainty? Thanks.

[David-A]

Hello KarenM,

The accuracy of your measurement depends on the signal you are measuring.  A signal mesured closer to 0V will be more accurate than a signal measured near 10V.  Additionally, a signal with a range of ±1V will be more accurate than a signal with a range of ±10V.  Thus, it is unclear to what you are referring to when you say absolute accuracy.

Regarding your calculations: the gain error looks alright if you are measuring a 1V signal; the range is confusing since you said your range is ±1V which would be a total of 2V, not 1; and the uncertainty is especially confusing since the noise is a function of your range, not the input.

It appears you also used the incorrect units in your final result; using your numbers the result would be 340 micro volts, not milli.

Would you be able to elaborate on what you mean by "find" the uncertainty.  Does that mean you want to use a more accurate device to verify the uncertainty of this device?  If so, you may want to investigate some of our PXI devices if so.  Does that mean you want to calculate the theoretical uncertainty of this device? If so it is (Vrange) x (Noise) / (Resolution)  which with a range of 2 Volts and using the Peak to Peak value on a 16 bit device is (2) x (2^7) / (2^16) which equals 3.9mV. 

Hope this helps!

David A.

[KarenM]

Hi David,

Thank you for your assistance. As you said, since the accuracy would depend on the working range of the module, I need to measure millivolt range, so the range I will be using and would like to focus on is +/-1 V. 

The equation comes from the datasheet for an NI USB-6210 data acquistion module that we are currently using, and the values are plugged in from the values found in the 9215 datasheet. Absolute accuracy is the term I see given on all the datasheets I've read to indicate accuracy of a signal within a working voltage range. I have used the example calculation given in the 6210 datasheet to arrive at the curious calculation of using 1V when 2V is the working range of the input signal.

I hope this clarifies my post from before.

In addition, I have found an "accuracy calculator" at (http://www.ni.com/advisor/accuracy/) to help with comparing calculations between modules, so that may be of help as well. Thank you.

[RCampbell]

Hi KarenM,

According to the 9215 specifications, the range is 10.4 volts.  So your calculation would be:

Reading * 0.0002 + 10.4 * 0.00014 + Noise Uncertainty

The RMS input noise for the 9215 is 1.2 LSB.  The 9215 has a 16-bit ADC, so the LSB is worth about 0.305mV.  Averaging over 100 points,

Noise Uncertainty = InputNoise * 3 / sqrt(100)

= (0.305mV * 3) / 10

= 0.0915mV

So if you expect a max reading of 1V, your max absolute uncertainty will be:

1V(0.0002) + 10.4V(0.00014) +0.0000915V = 1.75mV

I hope this clears thing up for you

Best Regards,