Hello,
I have an above ground pool. I have had the pool for 10 years now and for the third time, one of the hoses have either split or come off its connector leading to the pump to discharge the water outside the pool. After a few minutes of this, the level of the pool drops below the level of the skimmer and soon the pump runs dry. The mechanical seal inside the pump does not last long without water to cool and the pump/motor burn up.
I want to utilize the non-contact water/liquid level sensor to shut off the pump if it detects that there is no longer water in the pipe.
Can the sensor be used as a standalone or does the Arduino Uno board have to be used with it?
Does the Auduino Uno board come with a power supply?
The wire on the sensor looks to be at best a foot long. Can additional wire be added without interfering with the operation of the sensor?
Has anyone that has used or seen this sensor tell me if the sensor is sealed?
Thanks for your help.
Peter
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Hi,
Please find below answers to your questions and comments:
Sounds like a great plan to stop this (potentially expensive) issue from reoccurring. For such a project, you’ll most likely need the non-contact level sensor, some form of relay to turn on/off the pump, a power source for the sensor and logic of the relay and some kind of logic to operate all of it.
We assume you are referring to one of these sensors?
]Gravity Non-contact Liquid Level Sensor RB-Dfr-694]/:m]
]Non-Contact Liquid Level Switch RB-Dfr-396]/:m]
Since this sensor is a digital switch (see below for more details), you can most likely create a digital circuit that will take care of switching a relay (for the pump). That being said, you will also need to add hysteresis to prevent the relay from switching continuously when water level is at the sensor trigger point. All of this is certainly possible, but it may be simpler to just use a small microcontroller board instead as it will most likely make the prototyping process much easier.
The stand alone boards do not come with a power supply. That being said, most use cases are fine with simple USB power (not your case, most likely). You can find plenty of adequate power supplies here. Something like an inexpensive 9 V DC power supply should do the trick, such as the RB-Spa-103. Please note the official Arduino Uno recommended input voltage requirements are 7-12 V DC (limits are 6-20 V DC).
This sensor basically acts as a digital switch, sending out digital a voltage output (high or low) based on its state. The “high” level is VCC (typically 5 V DC for Arduino-based platforms. the sensor supports up-to 24 V DC) and the “low” level is 0 V DC. It has an output current of ~1-50 mA. Therefore, you can most certainly add some extra length to it. Of course, be mindful that adding long, thin wire will add resistance and make the voltage drop. Therefore, you may need amplification (a simple non-inverting buffer chip would do the trick) part of the way if it drops too low to be recognizable by your Arduino board.
The manufacturer reports the sensor has rated IP67 (dust tight, immersion up to 1 m depth). You can find more about IP ratings here.
We hope this helps.
Sincerely,
Hello,
Okay. I have the sensor, an Arduino board and a 9V power supply all on my list to buy. A couple of concerns. First is getting water proof boxes for both the sensor board that will be near my pool and the Arduino which will also house the relay. The relay I have is a 781-1C-12D by Automation Direct. It says it is rated at 1.4W. I was actually thinking of using a 12VDC 1A power supply so I can power both the relay and the Arduino board from the same power supply since the board can handle 12V.
I estimate I will need to run a wire between the sensor board and the Arduino of around 20 feet. You said I will need a buffer chip somewhere in the middle to boost the signal, helping with losses due to the wire length. Do you carry one that would work for this application?
Thanks,
Peter
Sounds great! 
Be mindful of sealed boxes since that can also cause overheating issues by prevent air flow and heat exchange. In cases where one of the components may produce heat in a confined space (such as the relay and microcontroller board) this can be a problem.
Typically the solution is to have some air flow where the water cannot get in or keep the box sealed but have a part made of metal that will be used as a heat sink/heat exchange.
According to this datasheet, this relay would require at least 12 V DC to operate on its input pin (to trigger the coil). As you may imagine, the Arduino cannot offer this kind of voltage (5 V DC only) and therefore would most likely not be able to trigger the coil directly.
The solution here is to either use a relay that is compatible with 5 V DC / microcontroller-level electronics or to add a driving circuit to your Arduino board so it can interact successfully (and safely) with your relay.
That should not be an issue since the power supply should provide enough power for both devices.
I’d also recommend adding a fuse since your pump will also be using the mains (maybe a 120 VAC/5A fuse or whatever is appropriate for your pump’s current rating).
I’d first recommend some cable of this length, try it out and see if it works well enough first. You may not need a driver chip for just 20 feet. As mentioned previously, this will depend on how much resistance your cables add. For such a short distance it probably will not matter for the signal. Typical 24 gauge (typical size for breadboard wires) has about 25.67 / 1000 ft (copper) or a total of about 0.5134 Ohm for your 20 feet of length.
You can see with this neat online calculator that your drop would be around 0.051 V DC (end voltage of 4.919 V DC) assuming a starting voltage of 5 V DC and a 24 gauge copper wire. If using aluminum wire instead (everything else the same), you’d get about 0.081 V DC or a final end voltage of ~4.919 V DC. Even with this drop twice (once from Arduino to VCC of sensor, second from sensor output back to Arduino input), you’d still be left with well over 4.8 V DC which is completely fine for reading a digital signal. It should also be noted this is assuming a worst case of the sensor using 50 mA. Of the two sensors mentioned previously, one uses 25 µA (far less than my “worst case” estimation) and the other uses 5 mA.
