I am planing to build a ROV for underwater use and at this point I am concidering to have some water leak detection system for the electronics enclosure that has to withstand underwater pressures up to 5bar / 70psi / 500kPa.

From my research I found a nice solution from Blue Robotics called the SOS Leak Sensor that goes together with some SOS Probes.

The design and the description of the probes let me think a bit about how they function. In a section of the product description it says:

Each probe can be re-used a number of times if dried out and re-compressed, but once the tip degrades you can also get replacement SOS Probe Tips.

So the probe tips are made of a sponge material that is probably enriched with some sort of salt (one could simply use NaCl for that I think), that is going to rise the conductivity of any water that is going to touch and get soaked into the tip.

If I am right I think to replicate those probes would not be much of a problem.

Moving on to the circuit board (SOS Leak Sensor):

Here is the circuit diagram from the documentation.

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So far I understand that if any probe is going to short out, the circuit is going to switch the signal - that is pulled down by R3 - to high. But what are R1 and R2 doing exactly?

I need to know this because I still didn't get the relation of the conductivity/resistance of the water to where the circuit detects the "short out" and at what threshold.

I got some samples of the water in my local area, where I am planning to bring the ROV in use, and have poorly measured the resistance of them and some tap and distilled water for reference. I've used a multimeter for this and I did not had any rig to hold the two probes at a constant distance to each other but I tried my best holding them with my hand, so my measurements are not so constant. But here are some significant results:

The sample water hat a resistance of something about 500kOhm (+-100kOhm) at a probe distance of about 5mm (tip to tip). I than added a pinch of salt into a little sample of the same water and measured about 40 to 30 kOhm.

So I am sure that I can get readings with homemade probes with at least 100kOhm resistance. But back to my question, how does these relate to the circuit shown above? do I need to make any adjustments to R1 and/or R2? And if I have to, how are the calculations for that?

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    \$\begingroup\$ Interesting problem. Since your ROV presumably has a wide variation in XYZ orientation, you will be needing multiple probes presumably in various corners. A simpler solution may be to monitor the pressure inside the chamber. \$\endgroup\$ – Trevor_G Aug 29 '17 at 15:11
  • \$\begingroup\$ Another alternative is to not worry about the pressure, vent the chamber and fill it with an appropriate oil. Though obviously all the individual components need to be able to withstand the pressures. \$\endgroup\$ – Trevor_G Aug 29 '17 at 15:16
  • \$\begingroup\$ I am planning to have a pressure sensor at the outside in addition to the water leak sensors on the inside. The point is that if I have a leak, my system should recover itself to the top of the water as soon as possible. If water gets into the electronics chamber, I am not worried about the electronics itself, since I am going to coat them with a suitable water protective coating ( silicone, wax, resin, ... ) as much as I can. but what I am worried about is the buoyancy of the ROV... \$\endgroup\$ – Ace Aug 29 '17 at 15:24
  • \$\begingroup\$ I've designed the floating elements and the electronics chamber (that is going to be filled with air) so that the whole system is slightly floating up. The reason behing this is that thatway I can maintain the depth by having thrusters blowing up and not down, so I dont push up any settlement. and dirsturb view that way... also if the system crashes, or if I have to trigger the emergency shutdown, the system will still float up to the water surface by itself to be recovered. So playing around with oil that 1. makes accessing the electronics more difficult and 2. adds weight, is not an option \$\endgroup\$ – Ace Aug 29 '17 at 15:28
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    \$\begingroup\$ In that case I'd pre-pressurize the chamber to greater than the target depth so the internal pressure is always higher. That way leaks will mostly be air leaking out, not water coming in, and it's easier to detect loss of pressure at any depth. Also gives you the added benefit of knowing seals are tight even before you hit the water. \$\endgroup\$ – Trevor_G Aug 29 '17 at 15:32

R1 (in addition to the resistance of the probe) limits the Q1 base current to a safe value, and probably is there to prevent killing Q1 with an accidental short to ground.

R2 pulls the base up to Vcc when the input is open or a high resistance. This assures that the transistor is turned off unless the base current exceeds a minimum value. In round numbers, this is 22 uA (0.6 V Vbe / 27,000).

If you know what Vcc is, you can calculate the equivalent probe resistance needed to cause the output to change. The transistor probably has a gain of around 100, but for decent saturation with a light load assume a value of 20-30.

The 240 ohms in series with the LED indicates to me that Vcc = 5V, so the base voltage is 4.4 V. The external resistance needed for 22 uA is (Ohm's Law) 4.4 / 22 uA = 200K. Minus the 1K already there = 199K. That is the point at which the transistor starts to conduct. But wait, there's more.

There's about 12.5 mA of LED current, plus 5 mA of R3 current, we'll guess another 1.5 mA of output current. That adds up to 19 mA of collector current. Round up to 20 mA, divide by a transistor gain of 20, and you need 1 mA of base current. Back to Ohm's Law, Rbase = 4.4 V / .001 A - 4.4K. Minus R1 = 3.4 K external sensor resistance for a firmly saturated transistor.

  • \$\begingroup\$ So if I did understand: If I use the same circuit as above, the sensor signal will go high when any of the probes measure a resistance between 0 and 3.4K ? I am not shure what "saturated" in terms of transistors means... but never the less, your answer is excellent! there are still some points I didn't get, but I think I'll figure them out. Thanks! \$\endgroup\$ – Ace Aug 29 '17 at 15:39

If you substitute the pnp transistor in your schematic with the equivalent model from here, then you will see that the resistors R2 and R1 are forming a volatage divider. Let's call the potential on the probe V_probe. Using the node methode you can find that the base NP diode will have Vcc on the anode and Vcc*R1/(R2+R1)+V_probe*R2/(R2+R1) on the cathode.

So my advice is to start with R1, which will determine the base current. Base on that you can calculate the R2, so that the base diode will turn on, once the probe reaches the required potential.


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