RC filter after voltage divider

Question

I'm trying to measure the resistance of water. My first attempt was to just use a voltage divider and measure the voltage between the water and a known resistance using the ADC on a microcontroller. The resistance of the water changed while measuring, probably because the electrodes are displacing the ions.

I think using a PWM signal with a virtual ground will solve this, though I have to filter out the PWM signal before measuring with the ADC. The problem is that the filter circuit affects the voltage divider, and always stabilizes at 2.5V no matter the resistance of the water.

The circuit I'm testing:

Is there another way to filter out the frequency of the PWM after the voltage divider, so I can measure the DC level on the ADC?

Answer 1

The ‘resistance of water’ isn’t exactly a thing, in the sense that current isn’t always proportional to voltage. You’ll need to decide under what conditions you want to make the measurement - AC vs DC, what distance between the electrodes, what kind of electrodes do you want to use, etc. You may well be measuring more about the electrodes than you are about the water. What are you trying to achieve? If you want to measure the level of ionic solutes then you may want to use relatively inert electrodes (carbon perhaps) and AC with no DC offset to minimise the deposition of ions on the electrodes.

Answer 2

There are many flaws in your design. A PWM shall not be used, rather a +/- bipolar square pulse voltage. The RC filter as you depicted of course affects the measurement, you need to buffer the input signal with Opamp, first.

You can use a dual voltage PSU, then use and ADC with chopper and external chopper driver, for example AD7730. It drives the external AC excitation bridge and does the filtering and conversion.

You can use analog switches to generate bipolar square pulse or H-bridge. Then you do reverse switch with analog switches the input signal to the opamp buffer/amplifier and do the math in the MCU.

If you don't need the AC excitation, then you can stay with single supply PSU, but without square pulse generator, only DC.

Answer 3

The resistance of the water changed while measuring, probably because the electrodes are displacing the ions.

If you keep the voltage applied I'm sure you'll notice the current change over time.

I think I'd just try something like this:

                        R
GPIO1-o ~~~water~~~ o-/\/\/-o-GPIO2
                    |
                   ADC

1. Apply 5V to GPIO1 and 0V to GPIO2, measure the voltage at the point ADC.
2. Apply 0V to GPIO1 for a while.
3. Apply 5V to GPIO2, measure the voltage at the point ADC.
4. Apply 5V to GPIO1 for a while.
5. Repeat 1 - 4 for as many cycles as you wish.

The ions will be pulled towards the electrodes during the readings. Steps 2 and 4 zero out the potential between the probes allowing the ions to reset back to equilibrium.

Reversing the potential on the probes has the benefit of reducing corrosion.

From the voltage readings you can work out current and then resistance.