# 4-20 mA sensor with logarithmic scaling

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).

• That's 5 decades, by the way. – jonk May 15 '18 at 21:24
• I wouldn't trust a table handed to me. I'd want raw data. Questions I might have are: (1) Is there residual, unaccounted offset in the data? (2) Might there be several signal exponentials added into the data? (The sum of two exponentials cannot be modeled with one exponential, for example.) (3) How much signal-related gain variation is there? (4) What does the noise look like? Look at the distributions. (5) Are there standards? (Like 'freeze points' for temperature.) Or not? (6) What's the sensor physics? (*) etc. I'd want to document the development of a formula pretty well, if possible. – jonk May 16 '18 at 1:15

## 1 Answer

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.

• Your formula hits most of the data points from his table, except for the numbers with 6's (and obviously the 4mA number). There is some error. Also, I'm not quite sure if the original table had an error in it or not, because the sensor minimum IS 0.02 uS/cm and not 0.2/cm uS, so it would make sense to me that 4mA should indeed land on 0.02 uS/cm – user2913869 May 15 '18 at 21:05
• For 18 mA my formula gives 632 $\mho$ which seems pretty good. Has the sensor a name and a datasheet? – Transistor May 15 '18 at 21:10
• I just added sensor/meter information to original question. – user2913869 May 15 '18 at 21:14
• Some assumptions? (1) There is a residual, unaccounted offset. (2) There is uniform deviate noise which means that expectations about how much to weigh data points near the bottom is quite different than weights given to data points nearer the top, if linearizing first. (3) Therefore, the use of standard regression tools is likely to produce poorer results. A crafted design from the partial derivatives, and assuming the above, is unlikely to be readily found and needs to be custom crafted (my opinion.) Of course, I don't know the OP's domain. Just mine, where it seriously mattered. – jonk May 16 '18 at 1:05