Software calibration Dell will improve the accuracy of the calibration in a later version? (Associated U2713H)

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• Understand and improve the absolute accuracy

  Many sculptures absolute accuracy calculated the computation of absolute precision in the manuals of the user for the use of modules in series C the following entries:

  • TempChangeFromLastExternalCal = 70 * C
  
  
  • TempChangeFromLastInternalCal = 1 * C
  
  
  • Sigma/Std-dev = 3
  
  
  • # of samples = 100

  I have a few questions about these two values listed above TempChange

  1. Why such a low value of only 1 has the TempChangeFromLastInternalCal * C; modules are continually doing internal calibration automatically?
  
  
  2. Regarding TempChangeFromLastExternalCal, I read correctly that NEITHER is the calculation of accuracy, assuming that the material has seen a +/-70 * C delta temperature since the last time it was calibrated externally. that is calibrated to 25 * C, but runs at 95 * C?

  To improve the absolute accuracy, I'm curious to know if the following two statements would improve accuracy:

  1. The material exists in an environment temperature controlled with a + / delta 10 * C ambient temperature. Can I use the value of 10 for TempChangeFromLastExternalCal if it has been calibrated at room temperature outside?
  
  
  2. If I'm short so HAVE them + HAVE terminals and a channel not used by COM and the extent to which results can be subratcted of readings of all other channels, I essentially elimited the error of the formula of absolute precision shifting?

  For completeness, for absolute accuracy formulas are:

  Absolute accuracy = ± (VoltageReading * GainError + VoltageRange * OffsetError + NoiseUncertainity)
  GainError = ResidualAIGainError + GainTempco * TempChangeFromLastInternalCal + ReferenceTempco * TempChangeFromLastExternalCal
  OffsetError = ResidualAIOffsetError + OffsetTempco * TempChangeFromLastInternalCal + INL_Error
  NoiseUncertainity = (RandomNoise * sigma) /(# of samples)

  Hey Sean,

  Change in temperature refers to the temperature difference from the last time that you did these two calibrations and the current temperature unit. So if you did an internal and external calibration at 25 ° C, then you will find the difference between the current temperature and 25 ° C. This is valid for two years for external calibration (according to the data sheet) to answer your questions:

  (1) If you both internal as external calibrated at 25 ° c, and then assign TempChange 1 ° C if the environment is in a summer of 24 or 26 ° C.

  (2) if the room was 10 ° C when the measurement has been taken, both internal and external TempChanges is 15 ° C (because your calibration conducted at 25 ° C).

  (3) If you were doing an internal calibration at 10 ° C, then your internal TempChange would be 0 ° C and your external would be 15 ° C.

  Regarding tips, I do not have too. The best I can suggest is to use a lower range and a device that has a lower production, but I don't think that it would be viable for your application. The best try of course is calibrated as close as possible to the more uniform temperature.

  I found a VI that calculates the absolute accuracy of a device. Must be a lot of values, but it allows you to play with the values to see what you can afford. My calculations show you in your range, but be absolutely sure of these fluctuations.

  https://decibel.NI.com/content/docs/doc-9912

  Attached a photo of the values that I used to get this error. I'm sure that the devices are calibrated at 23 ° C. Here's the manual where I found the numbers that I used.

  http://www.NI.com/PDF/manuals/374231c.PDF

  Sorry for the confusion earlier, I hope that this post clears up a lot of information and if it's not please let me know.

  -KP