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I'm making a small circuit for detect water level of our home water tank .here is the diagram.

enter image description here

when water level is low i want to turn red led when water level is max i want to turn green one on and so on.but current circuit turn both red/yellow/green when water level is max because when water level high all terminals sink .

how can i turn only one led instead of all .for example when all terminals are sunk i want to turn only green led. how to turn only one led instead of all ?

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    \$\begingroup\$ That's a truly horrific circuit, by the way. Not only are no current limiting resistors as shown, as Wouter notes, but it passes DC through the electrodes so they'll be chemically etched away. It has the advantage of being much simpler than anything decent. \$\endgroup\$ – Spehro Pefhany Aug 29 '15 at 14:40
  • \$\begingroup\$ @SpehroPefhany can you clarify "they'll be chemically etched away" \$\endgroup\$ – Fast Snail Aug 29 '15 at 14:47
  • \$\begingroup\$ Seethis reference, for example. \$\endgroup\$ – Spehro Pefhany Aug 29 '15 at 14:56
  • \$\begingroup\$ @SpehroPefhany thanks but do all probes/metals etch.i thought some metals doesn't etch on the water when Electrolysis \$\endgroup\$ – Fast Snail Aug 29 '15 at 15:00
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    \$\begingroup\$ Not all electrodes will be etched away. Carbon is a good (and cheap) choice. \$\endgroup\$ – Wouter van Ooijen Aug 29 '15 at 21:45
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I hope you have series resistors in your circuit that you left out in your picture? Otherwise you are pushing an undetermined amount of current through the transistors and LEDs.

To 'block' a lower LED you could use this basic circuit (untested):

schematic

simulate this circuit – Schematic created using CircuitLab

As others have stated in the comments, if you use copper electrodes they will over time dissolve. How fast depends on the current, which you can minimize for instance by using darlington transistors and higher values for the base resistors.

But not all electrodes will be etched away, only those that can be oxidized easier than the abundantly available H+/H3O+. This includes copper, but not platinum or (cheaper and more readily available) carbon.

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  • \$\begingroup\$ Nice (and simple) idea. Would R3 be connected to the highest or lowest probe? Need to have think about this one - something in the back of my head is saying it won't give a single LED output. \$\endgroup\$ – JIm Dearden Aug 29 '15 at 14:45
  • \$\begingroup\$ @wouter van ooijen i didn't left resisters .that's my entire circuit i'm using 3v dc power.is there any problem in this circuit ? \$\endgroup\$ – Fast Snail Aug 29 '15 at 14:53
  • \$\begingroup\$ Had a think - and by my reckoning R3 goes to high probe, R4 to mid and the last one (not shown) goes to low. Then it will need another diode (D5) from the third stage (low) back to the collector of Q1 so that when the high stage is turned ON it will pull both the mid and low stages down. \$\endgroup\$ – JIm Dearden Aug 29 '15 at 15:00
  • \$\begingroup\$ You won't need that extra diode, when Q1 is active Q2 will be active too, and Q2 will disable the lowest LED. \$\endgroup\$ – Wouter van Ooijen Aug 29 '15 at 15:04

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