Interpretation of Linear Variable Differential Transformer (LVDT) voltage output

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ca2n
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Good day to all Physics Forums members,

I am currently undertaking a project which involves the use of a Linear Variable Differential Transformer (LVDT) to measure the distance an object makes between two points. My setup is as follows:

  1. LVDT Type: Solartron DC50 with sensitivity of 6.158 mV/V/mm at 10 VDC
  2. Power supply to LVDT: 12.12 VDC (const. and cannot be varied)
  3. Voltage output from LVDT fed to dataTaker data acquisition system

In an attempt at verifying the calibrated sensitivity of the LVDT, I performed a number of height measurements using gauge blocks. I took 6 equally-spaced height intervals between the maximums of the LVDT range and plotted the LVDT voltage output vs. height. Here is what I encountered:

  1. The curve plotted was NOT linear as I anticipated it to be. As a matter of fact, the curve fit a polynomial equation of degree five with R2 = 1
  2. When comparing the LVDT output voltage with the LVDT calibrated sensitivity values, discrepancies occurred, e.g. a 20 mm difference between two gauge block heights returned a 20.5 mm difference calculated from the voltage output of the LVDT. The discrepancies grew as the ends of the LVDT were approached.

My questions are:

  1. Could the fitted curve be beneficial for my use seeing as I've determined the end-to-end output values of the LVDT, OR,
  2. Should I, instead, just get the LVDT calibrated and use the obtained sensitivity value.

To be frank, I would highly prefer Option 1, if it is in fact a justifiable solution.

Any help is much appreciated. Thanks. :smile:
 
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From your quoted numbers, the accuracy is 2.5% of reading. Considering that your supply voltage is over 20% high, that isn't bad at all. If 2.5% is within your needed accuracy, use it as is.

I see no reason to "...get the LVDT calibrated...", you just did when you measured the gauge blocks. Use those numbers if you need the accuracy.

Cheers,
Tom