4-20 mA sensor with logarithmic scaling

Question

We have a liquid conductivity sensor/meter that can be configured to output its sensor value encoded onto a 4-20 mA current loop. The meter has options of outputting the 4-20mA loop as a linear scale or a logarithmic scale. We need the logarithmic scale so we can get finer resolution at the low end, but still have a wide full range.

When configuring the meter for logarithmic mode, it asked for the following inputs, and I show what we input:

Sensor Min : 0.02 uS/cm
Sensor Max : 2.00 mS/cm
Number of Decades: 4 (wasn't quite sure what to put here)

My question is: based off of these settings, what is the forumula to convert mA back into siemens/cm? I contacted the manufacturer for help, but none of their tech support seemed to have a concrete formula. One of their engineers was able to replicate our meter settings and "simulate" some readings, just to see what the current would measure. This was his table:

1) 4 mA = 0.02 µS/cm
2) 6 mA = 0.625 µS/cm
3) 8 mA = 2.00 µS/cm
4) 10 mA = 6.5 µS/cm
5) 12 mA = 20 µS/cm
6) 14 mA = 66 µS/cm
7) 16 mA = 200 µS/cm
8) 18 mA = 660 µS/cm
9) 20 mA = 2000 µS/cm

Based on these numbers, this is my best guess at a formula:

uS = 10^((mA-const)/4)
const = 20 - 4 *log10(uS_max)
uS_max = 2000

But this doesn't give me exactly the same values. The lower end of the scale is also a little bit wonky, because this equation relies on there being a sensor decade every 4 milliamps, but that breaks down at the lower end.

Can anyone help figure out what the conversion formula is?

The sensor is a Mettler Toledo InPro-7000 VP. The meter (the one in which generates the current loop) is the Mettler Toledo M300 (the older version, with buttons instead of the touchscreen).

Answer 1

Your table does not show a linear response to the log scale. Here is a general formula to convert a linear 4 - 20 mA to a logarithmic scale.

$$R=10^{\Bigl(\bigl(log(max) - log(min)\bigr)\cdot\frac{i-4}{16}+log(min)\Bigr)}$$

Answer 2

I suspect that there is an error in the table you have posted. The reading goes up by 4 mA per decade except the first one. That should be 4 mA = 0.2 µS/cm. Assuming that is the case then:

mA    µS/cm
 4       0.2
 8       2.0
12      20.0
16     200.0
20    2000.0

Therefore

$$ \mho = 0.2 \cdot10^{\frac {i-4}{4}} $$

where i is the current in mA.

---

Now that the make and model have been posted (without links to datasheets) it appears that the unit has a five decade range. See InPro 7000, page 6. The only mention of 4 - 20 mA is in section 8.3 of the M300 meter manual but there is no explanation of the scaling.

The Aout type can be Normal, Bi-Linear, Auto-Range or Logarithmic. The range can be 4–20mA or 0–20mA.

For 5 decades the scaling would have to be 3.2 mA/decade.

mA        µS/cm
 4.0       0.02
 7.2       0.2
10.4       2.0
13.6      20.0
16.8     200.0
20.0    2000.0

For this setup

$$ \mho = 0.2 \cdot10^{\frac {i-4}{3.2}} $$

where i is the current in mA.