I am looking for a way to detect if the roughly 150mA I am running through salt water drops to zero, or a few mA. My enclosure presently has only passive components and gets power from a 12V battery. I use a 1 ohm current sensing resistor to measure the current.

I know the connections to my anode and cathode connection are prone to failure due to being in a long-term setup.

Its really a simple series circuit where I use a pot to drop the 12V down to about 1.5V to obtain the correct current for electrolysis.

I'm trying out constant current buck converters I bought on eBay right now.

 

I forgot to mention that I have no PCB. That's why I'm evaluating Buck Converter Modules like this. I'd like to get the voltage down so I can use smaller pots.

 

Sorry if the question seems lame, but this is for a really important project. It is for a box I am building for a friend who's Masters project involves marine preservation with warming ocean temperatures and is funded by the National Science Foundation. He is leaving for Hawaii in 6 weeks and adding this feature will allow an alert to be sent (probably just lighting an LED for now) when it's time to replace the cathode. Really trying not to add a PCB. I can probably mount a few components between the terminals inside the enclosure or use a small perf board. There are only 2 boxes for now. The project holds great promise!

 

Sorry, I'm trying really hard not to be flippant, but I can't resist:

...warming ocean temperatures...funded by the National Science Foundation.

<SARCASM>
My tax dollars at work...must be really important...
</SARCASM>

 

I did find this option, which I will try out:

Blown Fuse Indicator

This blown fuse indicator will work with a wide range of DC supply voltages from 5V to 50V. It illuminates LED1 when the fuse blows. With the fuse intact, Q1 is held off and there is no bias current available for the base of Q2. So the LED is off. When the fuse blows, a small current flows via the base-emitter junctions of Darlington transistor Q1, through its base resistor R1 and then via the load. Typically this current will be around 20μA and this turns on Q1 which provides base current to Q2 which then turns on to illuminate the LED.



Blown Fuse Indicator Circuit Diagram

The emitter current of Q2 is limited by Q3 which turns when the current reaches about 10mA, to shunt base current away from Q2. The three resistor values not given in the circuit are dependent on the supply voltage and can be calculated from the following simple equations:

• R1(kΩ) = V(DC)/0.02 = 560kΩ for 12V DC
• R2(kΩ) = V(DC)/2 = 5.6kΩ for 12V DC
• R3(Ω) = V(DC)/0.02 = 560Ω for 12V DC

R3 should be included for voltages above about 20V otherwise the heat dissipation in Q2 will be too great. At lower voltages it can be omitted. Any general purpose NPN transistors can be used for Q2 and Q3, provided they will handle the DC supply voltage. The PNP Darlington, Q1, could be an MPSA65, available from Dick Smith Electronics (Cat Z-2088).

 

Sorry, I'm trying really hard not to be flippant, but I can't resist:



<SARCASM>
My tax dollars at work...must be really important...
</SARCASM>

You have no idea. No place for that kind of crap here a-hole. Keep your immature thoughts to yourself. I thought this was a place for professionals. I am not a global warming believer, but right now the temps are higher and its killing coral reefs. This is a restoration and preservation project and could even allow rebuilding the oyster beds that were over harvested in New Orleans and allowed the floods caused by Hurricane Katrina. I went and volunteered in the recovery effort and know the meaning of the word important. Really important.

 

I like this idea better:

 

Hmmm, I'm a little out of my depth here, but have an idea that might work without too many components, although while I was sketching it up I see you found a few simpler ideas, so maybe this is overkill.

Anyway, this circuit uses an INA169 current sensor (drawn up using an Adafruit breakout,) which would replace your shunt resistor, and a MCP6542 comparator IC to provide an LED indication and 5VDC push-pull output which could be used as an alarm output if needed. Resistors R7 and R8 can be adjusted to change trip point, and R1 can be adjusted to control hysteresis. As drawn, this circuit is meant to trip around a 75mA threshold with roughly 11.3mA hysteresis. I've not breadboarded or simulated this circuit, so it bears careful double-checking... but I think it's all good.

Also, some or all of the caps may be redundant - parts of this circuit were copied from one I made previously which was running from a noisy 5VDC supply line. Running on batteries with not much switching or electrical noise around, you may not need them.

 

I did find this option, which I will try out:

Blown Fuse Indicator

This blown fuse indicator will work with a wide range of DC supply voltages from 5V to 50V. It illuminates LED1 when the fuse blows. With the fuse intact, Q1 is held off and there is no bias current available for the base of Q2. So the LED is off. When the fuse blows, a small current flows via the base-emitter junctions of Darlington transistor Q1, through its base resistor R1 and then via the load. Typically this current will be around 20μA and this turns on Q1 which provides base current to Q2 which then turns on to illuminate the LED.


View attachment 78028
Blown Fuse Indicator Circuit Diagram

The emitter current of Q2 is limited by Q3 which turns when the current reaches about 10mA, to shunt base current away from Q2. The three resistor values not given in the circuit are dependent on the supply voltage and can be calculated from the following simple equations:

• R1(kΩ) = V(DC)/0.02 = 560kΩ for 12V DC
• R2(kΩ) = V(DC)/2 = 5.6kΩ for 12V DC
• R3(Ω) = V(DC)/0.02 = 560Ω for 12V DC

R3 should be included for voltages above about 20V otherwise the heat dissipation in Q2 will be too great. At lower voltages it can be omitted. Any general purpose NPN transistors can be used for Q2 and Q3, provided they will handle the DC supply voltage. The PNP Darlington, Q1, could be an MPSA65, available from Dick Smith Electronics (Cat Z-2088).

1. If this a power then you should change the pin for B,C of Q1.
2. For the life and brightness of led, using 80% of rated current is better, the I= 20mA * 80% = 16mA.

 

A fairly simple solution is to use a reed relay and wind a coil around it so at the rated current it has a closed contact and drops out when current is zero. Reed relay can be used to drive a relay to turn on light, alarm etc.
I have used this for sensing brake light failure. When current is ok for two lights reed closes and lights led , when one fails led flashes on then is off. Coil of reed relay needs to be "tuned" for the application. No electronics involved.

 

First, when you or your friend works in an area that is controversial (that is, not 100% of the population believes in it) you need to accept some criticism from the population. After all, those tax payers are paying for the research and in many cases have the right to be cynical.

Next, your friend is supposedly a great researcher who is working on a project of great promise and leaving for Hawaii in 6 weeks. I would argue that he may be in trouble if he is relying on a friend who is relying on an Internet forum to solve a simple current sensing circuit.

Finally, if you are in need of help, don't insult the members of our community and call them names - you are the new member here. Take some time, get to know us if you wish. If you do, you will realize just how wrong you were in your post...

You have no idea. No place for that kind of crap here a-hole. Keep your immature thoughts to yourself. I thought this was a place for professionals.

This is a place full of free thinking, creative people - some professional and some not and plenty of "that kind of crap". Joeyd999 was even kind enough to put his comment in sarcasm brackets. So, if you care to stick around, chill out and stop acting like an a-hole. Note that I did not call you an a-hole. I just said that you were acting like an a-hole. If you stick around long enough, we can get to know each other and then I can decide whether or not you are one. Finally just to accelerate this relationship, my wife knows me pretty well and she calls me an a-hole all the time so I probably am one.

 

Johnsal,

I think you need to rethink your entire electrodes in a salt solution premise. Putting metallic electrodes in an aqueous salt solution, and driving them with DC to a potential (look at an electromotive series table) that causes electrolysis is going to dissolve them in a few hours. If the goal is to measure conductivity of the solution, you need to use AC (Zero net DC component), and then at a peak-to-peak amplitude of less than 1V.

Driving electrodes with a DC constant-current as you are talking about above is exactly the wrong thing to do...

Maybe this is an example of "Climate Science"???


Oh, and to answer your original flawed question: Why not simply put one of these in series with the electrodes?

 

You have no idea. No place for that kind of crap here a-hole. Keep your immature thoughts to yourself. I thought this was a place for professionals. I am not a global warming believer, but right now the temps are higher and its killing coral reefs. This is a restoration and preservation project and could even allow rebuilding the oyster beds that were over harvested in New Orleans and allowed the floods caused by Hurricane Katrina. I went and volunteered in the recovery effort and know the meaning of the word important. Really important.

We can't all be really important; some of us are just ordinary people, and have no idea how to help you with this global dilemma.

 

http://www.mod-tronic.com/Beede_meter_relays.html

should do exactly what you need,

 

Those meter relays really do look perfect for this. Should've known the product would already exist.