# rlarios

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• Hi calebmcelhaney!1. I guess you can replace the relay with the SSR you mentioned. The 12V version draws like 20mA, well within the capacity of the transistor.2. You can omit the pump red LED circuit altogether. It is there just to show that there is not enough level at the pump reservoir. The green LED circuit is directly hooked up to +12V and Q3's collector.3. Well, think about it, the green LED forward voltage is about 2V. Subtracting these 2V from the supply voltage should give you the voltage drop across the resistor, 12V - 2V = 10V. Current through this resistor = 10V/470 Ohms = 21.2 mA. The power dissipated across this resistor is 0.0212^2 times 470 (i2r loss) = 0.212 Watts! A typical 1/4W resistor would run too hot when used close to its rating and may not last as long as the othe…

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Hi calebmcelhaney!1. I guess you can replace the relay with the SSR you mentioned. The 12V version draws like 20mA, well within the capacity of the transistor.2. You can omit the pump red LED circuit altogether. It is there just to show that there is not enough level at the pump reservoir. The green LED circuit is directly hooked up to +12V and Q3's collector.3. Well, think about it, the green LED forward voltage is about 2V. Subtracting these 2V from the supply voltage should give you the voltage drop across the resistor, 12V - 2V = 10V. Current through this resistor = 10V/470 Ohms = 21.2 mA. The power dissipated across this resistor is 0.0212^2 times 470 (i2r loss) = 0.212 Watts! A typical 1/4W resistor would run too hot when used close to its rating and may not last as long as the other components on the board. Therefore, 0.212W x 2 would be 424mW, and as there are no 424mW resistors, a 1/2W resistor should do the trick.Thanks for your comments and have a great day!rlarios

• Get yourself a float switch. That's the simplest I can come up with.Regards

• Transformer supplies 12Vac, which one rectified supplies 16.9Vdc minus 2 diode-voltage-drops, or about 15.5Vdc.

That depends on the mineral contents of water. You could use stainless steel, though.

Yeah! I'm aware of that broken link. I guess if you start building it now, you'll finish before I can come up with an alternate supplier.

The reason a transformer is used is to provide galvanic isolation. Regarding contamination, you can't imagine how much of it we are unconsciously getting into our system, if that is a concern.

• RTKatNO1,Sorry for not coming back to you sooner. I have said that I would write something about it to be part of the instructable. Unfortunately, I have gotten busy on work.If you are in a hurry, go read the comments, you'll find somewhere down the modification to the circuit to do just what you have requested.Thanks for your comments, sir.rlarios

Hi renzd5,Is this a request for Lado115 or myself?rlarios

Hi renzd5,rafael.larios@gmail.comrlarios

Sorry,I've been busy. Saw messages on my phone and it wasn't until today that I got to my laptop.rafael.larios@gmail.comWrite your own instructable sir, I'll post a link in mine pointing to yours.rlarios

• Lado115,Please review slide show on step 1 of this instructable.It explains all the basic theory behind CD4001 and how to wire it as an S-R flip flop.You're welcome, and keep up the good work.Regardsrlarios

• Lado115,This is the first prototype i did of this circuit before doing the PCB seen in the instructable.Look on the solder side. +12V line is at the top, while ground bar is at bottom. I used thicker wires for power rails. You can see the ac in connector on the right end of solder side picture (which is on the left end on component side picture).You can see a thick green wire that runs vertically between the rectifiers down to the ground rail. There is a thick bare short wire the runs from the rectifiers up to the ground lead of voltage regulator, and the (-) lead of capacitor seen in front of voltage regulator on the component side picture.Do you see the white wire at lower left corner of solder side picture? That's the connection between relay transistor emitter lead and the ground bar …

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Lado115,This is the first prototype i did of this circuit before doing the PCB seen in the instructable.Look on the solder side. +12V line is at the top, while ground bar is at bottom. I used thicker wires for power rails. You can see the ac in connector on the right end of solder side picture (which is on the left end on component side picture).You can see a thick green wire that runs vertically between the rectifiers down to the ground rail. There is a thick bare short wire the runs from the rectifiers up to the ground lead of voltage regulator, and the (-) lead of capacitor seen in front of voltage regulator on the component side picture.Do you see the white wire at lower left corner of solder side picture? That's the connection between relay transistor emitter lead and the ground bar at bottom of picture. By the way, the ground bar was made out of a paper clip.Hope this helps.rlarios

• Lado115,This is the first prototype i did of this circuit before doing the PCB seen in the instructable. Look on the solder side. +12V line is at the top, while ground bar is at bottom. I used thicker wires for power rails. You can see the ac in connector on the right end of solder side picture (which is on the left end on component side picture). You can see a thick green wire that runs vertically between the rectifiers down to the ground rail. There is a thick bare wire the runs from the rectifiers up to the ground lead of voltage regulator, and the (-) lead of capacitor seen in front of capacitor on the component side picture.Do you see the white wire at lower left corner of solder side picture? That's the connection between relay transistor emitter lead and the ground bar at bottom of…

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Lado115,This is the first prototype i did of this circuit before doing the PCB seen in the instructable. Look on the solder side. +12V line is at the top, while ground bar is at bottom. I used thicker wires for power rails. You can see the ac in connector on the right end of solder side picture (which is on the left end on component side picture). You can see a thick green wire that runs vertically between the rectifiers down to the ground rail. There is a thick bare wire the runs from the rectifiers up to the ground lead of voltage regulator, and the (-) lead of capacitor seen in front of capacitor on the component side picture.Do you see the white wire at lower left corner of solder side picture? That's the connection between relay transistor emitter lead and the ground bar at bottom of picture. By the way, the ground bar was made out of a paper clip.Hope this helps.rlarios

• Lado, If your logic is working properly, and the relay is not, try checking what voltage you have at the base of transistor. If you have 0V there and relay is still on then transistor possibly is shorted which makes relay to be on all the time. Replace transistor if this is the case. On the other hand, if you're working on a proto board, they are notorious for intermittent bad contacts when not new. I would suggest you get a perf board and solder everything in place. If your circuit is too messy, start on another board from scratch. This may save you a lot of time and frustration.Ground is the same reference for everything. IC1 pin 7, IC2 pin 7, relay transistor emitter lead, ground terminals going to the tanks, ground lead of voltage regulator, (-)minus leads of decoupling capaci…

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• Lado,All circuits use the same ground as reference. The relay needs more current than the other circuits, therefore, you need to make sure your power supply can deliver the power needed for the whole circuit.Wire the circuit first using an LED to simulate the operation of the relay. Once you verify that the logic works, I'll help you with the relay to turn on the pump.By the way, post a picture of your circuit. I may be able to see what's wrong and help you fix it.Have a good morning, afternoon or whatever. It's 2:32 am here.Regardsrlarios

• Lado115,The attached images are actual screenshots I got doing a quick simulation using National Instruments MultiSim. Images may not be in order, but the simulation proves that the circuit works per your requirements.Closed switches = electrodes are in contact with water.Open switches = electrodes are not in contact with water.Hope this helps you understand.rlarios

• Lado 115,Please review attached picture. Do some logic analysis for yourself so you understand the circuit. Make sure you have same common ground on both tanks and that your connections are solid. One of my readers once thought that both circuits were the same because he saw no difference.Remember, IC1B pin 5 is wired to both IC1A pin 3 and IC1D pin 12. IC1B pin 6 is wired to both IC2D pin 11 and IC2A pin 2.Resistors are 2.2M.Test the circuit on your workbench first before wiring the actual tanks. Make sure that inputs go to a high state when not touching water, and that they go to a low state when they touch water. Ground must be touching water.Relax, be patient and enjoy the process.rlarios

• Lado115,You're welcome! Go ahead, use 18 or 16AWG wire. This way you'll keep wiring resistance low. rlarios

• Lado115,This is the circuit I meant for you. Please review and let me know how it works. You'll need an additional CD4001.Best regardsrlarios

• Lado115,Sorry I was not specific regarding the modification.You would use the controller the way it is PLUS another IC wired the way I described in my previous message. This way, the upper tank controller would do the job or was intended to do. The second CD4001 would be wired the way just described where pump is turned on when upper level probe of the tank you get water from is touched by water level, and turned off once water is below lower probe.I'm at my bank right now, once I get back home tonight I'll send you a drawing showing how you should do the wiring.Have a good day.Rlarios

• Dear Lado115,In order for the controller to operate in such a way as to turn pump on when water reaches upper level sensor and turn it off when water is below lower level sensor do the following:Refer to step 3 of instructable. 1. Cut line that goes from IC1 pin 3 to IC1 pin 5, without cutting feedback line that goes from pin 3 to pin 12. This is very important, pin 3 and pin 12 must remain connected.2. Run a wire between pin 11 and pin 5. This should make circuit behave the way you want. Test it and let me know, Thanks for your comments. rlarios

• Hi Lado115,Based on your comments, would waves affect the high level sensor, or the low level sensor?I believe maybe the high level sensor could stop the pump sooner because of water filling up the tank. How high would the waves be? Would the differential of actual water level be too high with respect to sensor? I don't think it would be. However, if that's your request I believe I can come up with something.If it's the low level sensor the one that could be affected which would stop the pump sooner, please let me know.Rlarios

• Hi Lado115,Thanks for your comments.Where exactly would you like to introduce the delay and why?RegardsRlarios

• Hi there Bundolo,Two years ago, one reader had a kit installed to control a 1,600 gal tank located under a deck (6.5' height). The tank was used for irrigation somewhere in Oregon.I haven't tested what the maximum distance between high level electrode and ground could be. I guess that's something I'll test during this coming weekend.Thanks for your comments Bundolo.Rlarios

• Great work HamedI2!I will really appreciate if you could share a link about those titanium sensors.Thanks, sir.

Anand7878,Check step 3 of this instructable. Circuit schematic diagram is shown right there.Rlarios

Hi there Tycss,Yes, it can work in reverse. If you are in a hurry, check the instructions I gave to "Murray" about three years ago on how to modify the circuit to do what you just asked.I said back then I would write about these simple changes in the instructable, and I never did. Shame on me. I hope I don't forget this time.Cheers!

• Dear Anand7878,Circuit diagram is shown in the instructable.Or, do you mean something else? Regardsrlarios

• Hi Mark 62726,Thank you for your comments.You can have the CD4001 operate off 5V. You would need a 5V coil relay with contacts rated at least at 1A at 120Vac, and instead of the rectifier diodes and 7812 voltage regulator, you could use your 5V phone charger.rlarios

• First clean flux residue with a cotton swab soaked in alcohol. Make sure board is clean on component side as well.Let me know if it still doesn't work so we continue debugging your control board.rlarios

• DarshP5,Show me a picture of solder side of your PCB.rlarios

• it depends on what you mean by that. Circuit operates on dc and I've seen some varying low voltage can be used and then rectified in order to be detected. There are other circuits out there that operate off ac power using relays, in case you're interested look them up on YouTube. rlarios