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Sensor conversion and scaling

revision 97382


Description

This article gives formulas for converting measured sensor signals to physical units, a function available in WB-MAI6 and WB-MAI2-mini/CC v.3. It also explains scaling lower and upper limits for nonstandard sensor signal ranges.

Terms and definitions

  • Minimum (Smin) and maximum (Smax) sensor signal values, for example 0–20 mA;
  • Measured (S) sensor signal;
  • Minimum (Pmin), maximum (Pmax) and calculated (P) physical quantity measured by the sensor. Minimum and maximum values define its measurement range and are specified in the sensor documentation.
    • P = Pmin + (S - Smin) * K.
  • Conversion factor (K), the ratio of the physical-quantity range to the signal range: physical units per signal unit.
    • K = (Pmax - Pmin) / (Smax - Smin).

Converting a sensor signal to physical units

Consider a 4–20 mA temperature sensor covering −40 to +80 °C. Calculate temperature at 12 mA.

Given:

  • Smin = 4 mA, Smax = 20 mA;
  • Pmin = -40 °C, Pmax = 80 °C;
  • S = 12 mA.

Solution:

  • Conversion factor: K = (Pmax - Pmin) / (Smax - Smin) = (80 − (−40)) / (20 − 4) = 120 / 16 = 7.5 °C/mA;
  • Physical quantity: P = Pmin + (S - Smin) * K = (-40) + (12 - 4) * 7.5 = 20 °C.

Range limits for nonstandard sensor signals

To convert standard analog signals to physical units, modules WB-MAI6 and WB-MAI2-mini/CC v.3 use the parameters Lower and Upper range limit, corresponding to the minimum (Pmin) and maximum (Pmax) physical quantity measured by the sensor.

For a nonstandard signal, such as 1.6–19.4 mA, selecting a 0–20 mA template range will produce incorrect conversion. Calculate the physical values corresponding to 0 mA and 20 mAand enter them in the respective parameters Lower and Upper range limit.

Sensor conversion and scaling illustration
Upper and lower range limits for a nonstandard 6–19 mA signal

Consider a temperature sensor with 6–19 mA output and −40 to +80 °C range. The template is set to 4–20 mA measurement. Calculate physical values corresponding to 4 mA and 20 mA.

Given:

  • Smin = 6 mA, Smax = 19 mA;
  • Pmin = -40 °C, Pmax = 80 °C;
  • S1 = 4 mA, S2 = 20 mA.

Solution:

  • Conversion factor: K = (Pmax - Pmin) / (Smax - Smin) = (80 − (−40)) / (19 − 6) = 120 / 13 = 9.23 °C/mA;
  • Physical quantity at the lower range limit: P1 = Pmin + (S1 - Smin) * K = (-40) + (4 - 6) * 9.23 = (-40) + (-18.46) = -58.46 °C.
  • Physical quantity at the upper range limit: P2 = Pmin + (S2 - Smin) * K = (-40) + (20 - 6) * 9.23 = (-40) + 129.22 = 89.22 °C.

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