I'm trying to find the optimal, or at least a good resistor to put into a voltage divider to measure a thermistor, and an electro-conductivity K10 probe. The range is 10 uS/cm to 1 S/cm, and an operating temperature of 0-70 Celsius.

So equations gave me a working resistance of ~80 ohms to 19M ohms. I'm thinking that given the range of resistance it might be difficult to have a single resistor to compensate.

Also I'm using a 1k thermistor, and was wondering if in that case a 1k would be the optimal resistor to pair with that as well?

____/\/\/\_____thermistor____ |||gnd
|     ?     |    probe
(5v dc)   mcu

I've played around; seems a 10k is more accurate in getting the value of the thermistor.

So if I have known solutions, and a known probe the resistance should be in theory known. A K 10 probe and a .15 S solution should come out w/ 1.5 ohm?, or would it be ohm/cm?

  • \$\begingroup\$ Can you please add a schematic to show which resistor you mean? \$\endgroup\$
    – auoa
    Jan 31, 2017 at 19:50
  • \$\begingroup\$ For that conductivity range, no single-scale metering is likely to prove satisfactory. You'll need to break it up into multiple ranges, or use a logarithmic converter. \$\endgroup\$
    – Whit3rd
    Jan 31, 2017 at 22:28

2 Answers 2


A silicon diode is a fine voltage = log(current) converter, with 58 milliVolts per decade of current (at room temperature).

Just connect the thermistor to a reasonably steady power supply, and connect other end to anode[arrow] of silicon diode, then ground the cathode[bar].


simulate this circuit – Schematic created using CircuitLab


If you use 1k thermistor, use it with just 1k pullup. Then make a lookup table in excel. The function is not very linear, but distinctive enough.

One of the reasons for such simple circuit is the cost. If you add many components, you will not gain accuracy (compared to lut), but you will waste money, which is nit very good for business.


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