I have not been able to find a decent data sheet for a sensor like above.
From looking at it looks like it could be a differential amp. I am not sure if they make them in 3 prongs. I have only seen chips. Could someone give me some insight to the making of this sensor?
Mostly found google translated datasheets from china.
Ehh nm I am thinking it is just a transistor that limits voltage based on supplied current to the base of the transistor.
My guess is that the 3 pin package is an NPN transistor, type S8050 with "J3Y" marking, with the base connected to one set of traces. One or the other of the parts should be a high value resistor, but I can't make out the markings.
You could easily verify or disprove that with a multimeter. Only $0.29 at AliExpress!
Guess at the circuit, from the description on the "best data sheet" (water on the traces effectively shorts the base to ground, cutting off the transistor):
Shouldn't be a PNP?
5V is present on the base when the circuit is dry and a you add water you reduce voltage seen at the base in the laws of a voltage divider?
No, NPN. That is a voltage follower.
Tested it with my DMM and yeah a NPN.
I am having a little trouble wrapping my head around R1 and trace.
Trace is effectively an open until exposed to water. Why is the signal wire getting more voltage with more water?
Vout= (Vs*R2)/(R1+R2)
I am see Base=Signal wire voltage with my DMM.
Base is starting at a low voltage and going up with more water exposed.
Then you are making some egregious measurement error.
Signal wire=1.4V DC
Base =2.2 Emiter = 1.8 collector = 4.44 VS is 5VDC
Ok, but the behaviour of the voltage divider you describe is backwards. Did you make a mistake, or is the entire known body of physics wrong?
Base is starting at a low voltage and going up with more water exposed.
I am aware.
What are the actual resistor values on the board (make sure it is dry)? I was just guessing wildly.
With the sensor removed from the cruit here is some readings.
There are 2 sets of copper legs ofc.
Neither set of copper legs are conected directly to ground. One goes through a 10 ohm resistor to positive terminal of sensor.
Copper legs don't have connection to one another besides their corresponding group.
When reading to ground 6M ohms is read. This is most likely through the transistor.
So I assume based on this model the the less resistance between the copper legs the more voltage is allowed to go through the transistor.
I traced it out with my DMM
LM 1-6 represent the copper legs. I didn't have anything good to represent them.
Hmmm, I was looking at images of a different version of the sensor board. There is an LED on the one posted in reply #10, but not on the one I looked at most closely (similar to the "best data sheet" board linked in the OP).
I agree with the general circuit tracing for the image in reply #10, except for the missing LED and the stated values for the resistors.
The markings on the resistors can be read on the image posted in reply #10, and they are clearly 101, 101 and 102, which correspond to 100, 100 and 1000 Ohms, with the 1K Ohm resistor being a current limit resistor for the LED (board powered on).
If that is the board you have, then yes, as water allows current to flow to the base, the signal voltage should rise due to increased emitter current.
