Reverse-engineering water detection circuit

Question

I am reverse engineering a PCB in an espresso machine and I am having trouble understanding the water detection circuit used for the steam boiler. The correct water level in the boiler is set such that the water just covers the tip of the probe.

The trace on C6 goes to a header pin that is connected to a stainless steel probe in the boiler. The probe is insulated from the boiler and the outside of the boiler is tied to the chassis, which is also connected to the AC earth. Presumably, this is to detect continuity through the probe with current passing through the water in the boiler to the chassis. There is no AC current source on the PCB.

There are also two traces going to GPIO pins on the microcontroller through the passive components pictured. You can see the traces connecting the passives on the PCB. From what I can tell, it looks like R7 is a current limiting resistor and R8 and R9 form a voltage divider and C7 is probably for signal smoothing, but I don't know enough to understand how the whole thing works together.

Any help understanding the circuit and how the microcontroller uses the two GPIO pins to sense the presence of water at the probe would be greatly appreciated!

Also, I know direct current passing from the probe to the boiler can cause corrosion. How might this circuit mitigate excessive corrosion on the probe?

An MCU can quite easily be an AC source. And the component going to the probe is the capacitor C6, so it must be a capacitive probe. Can you get an oscilloscope probe on it while it's operating? — Dave Tweed
– Dave Tweed, Commented May 30 at 16:04
Hi Dave, thanks for replying. Unfortunately, I don't have an oscilloscope at this time. The rest of the board is pretty simple, it's just this bit that is confusing me. — Philistino
– Philistino, Commented May 30 at 16:30
If you want to reverse engineer this you need to see the signal that goes to the probe. I would expect MCU generates some AC signal as it is common technique for water sensors. — Juraj Lonc
– Juraj Lonc, Commented May 30 at 18:09

SamGibson · Accepted Answer · 2025-05-30 21:02:35Z

Consider this circuit. I'm not saying this is exactly what they are doing, but maybe it's close. You would need an oscilloscope to be sure. You can measure the voltage at GPIO2 with a multimeter but even a 10MΩ input voltmeter will affect the voltage a bit.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The source at GPIO1 is a 0/5V 50kHz square wave. Vout goes to GPIO2.

With probe capacitance C3 = 10pF the Vout voltage is around 3.1V (a CMOS input 'high'). With probe capacitance C3 = 100pF the Vout voltage is about 1.5V (a CMOS input 'low').

With typical CMOS thresholds of Vdd/2 the circuit will detect the probe capacitance change if it is as I suggest.

There is no DC voltage across the probe due to the series capacitor (C2 in my circuit, C6 in yours).

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