EC measurement using Capacitor - from Github

Hi
I am currently using GitHub - wvmarle/Arduino_ECSensor: Library to operate EC sensor for Arduino. by @wvmarle

I got a 1413 calibration liquid for 25C and my current temperature
Is 30C... So I am expecting to get 1552.. The capacitor discharge
Circuit is reporting 2.43 with the recommended probe at 30C.
Which is 56% too high.

Could I simply correct the value by reducing it by 56% or is there
a new capacitor value or resistor value to change it to?
Now I am using 330 OHMS and 22nF CAp. as per the schematics shown.

image

How can I make this more accurate?

how accurate are those values?

The circuit reports 2.4
The Calibration Liquid is 1.5

Errors are consistent so that is good, I feel I need to increase the resistance using a variable multi turn pot so I can fine tune it. But I am not electronics trained so I am asking for advise

I meant did you measure the value of your resistors and cap ?
they might be 5% or 10% off and both resistors might have different real values.
that could help adjust your maths

author recommended to use metal film type resistors with low temperature coefficient and a 22nF film capacitor such as PP film or PET/polyester

Yes, resistors are around 5 ohms off of course. They are 1% resistors and the cap is polyestor. 1% is metal film and the blue color also means metal film it seems.

yes, if you bought them from reliable sources (have seen batches of components sold from the far East that were 20 or 30% off their supposed value. They were likely coming from rejected components that should have been trashed but got in the hands of not so respectable sellers)

After more research : microprocessor - Water conductivity measurements/ESP8266 - getting strange results - Electrical Engineering Stack Exchange

The article uses the same circuit and method and basically it states the circuit is not important as long as the discharge time can be counted.

Using a known calibration liquid say 1550 US at 30C the measured output from the circuit is offset by the percentage difference
Using 2 pin probe: Measured 2400 expected 1550 ,so the difference is 36%
2400 - 36% = 1550 us/cm

So my understanding is to simply reduce the value by 36%
each probe will have to be calibrated for the percentage difference.

Yes, you need to calibrate each and every one.

Not just for the resistor and cap, also for things like the actual threshold voltage of the input pin (it's somewhere around 0.5*Vcc) and the actual probe used.

Put five seemingly identical probes next to one another and you will see large differences in the resistance measured.