*Help / Advice designing control circuit for 12V automated watering system

Hi, and thanks for reading, Ok, so I only have a little knowledge about this stuff, but I'm sure I can do this with the right help,

I've built garden boxes, where the water drains out to a reservoir at the bottom, what I want to do, is put a pump at the bottom which when activated, pumps the water back up into the soil again, the pump I plan on using is 12VDC, 1.05A 550l/h. The system would be powered by a 12V 100AH Solar recharged battery, (I plan on many more boxes if I can get this one right) I've found a tutorial on how to build a basic water detector, (http://www.youtube.com/watch?v=ofVwrIi1zII)
So, Basically, what I want to do is, put three of these sensors in the reservoir, one right at the bottom of the sump, to tell the pump to stop, when there's no water left, One at about halfway, To tell the pump when to start pumping, and one at about 80%, to tell 2 solenoids to close / open (one will be N/O and close, the other will be N/C and open) to redirect the pump flow from going to the sprinkler, into the water tank in the garden, (In cases of heavy rain where the reservoir will be overfilled), and the solenoids should return to normal again, after the water level has gone back under the half way sensor. So basically I guess I need some help in designing the circuitry for this? I could probably make the water sensors easy enough, but how to tell them how to contol the solenoids, and pump properly, I wouldn't have a clue! I'd also probably need switches to manually turn the pump on before the water reaches the halfway mark, and to manually open the solenoids to drain all the water from the reservoir to the garden tank. Is this too compicated to do?

Any help would be greatly appreciated!
thanx

 

personally I would just use 2 float sensors (1 for high level start and 1 for low lever stop), a couple 12V icecube relays, your pump, and a 12V power supply. Put it all into a waterproof enclosure and call it a day.

But basically if you want solenoids thats fine too..just use relays to control it all.

 

I had thought of some sort of float sensor initially, but found the detectors and thought making them might be more fun, and educating, as I'm just learning at this stuff,

Anyways, how would I wire it so that the float sensors trigger the pump to go on/off properly? (trigger pump on when water reaches halfway, trigger pump off when water drops below bottom sensor), The only way my little brain can see, would cut the power to the pump as soon as it falls below the halfway sensor, instead of when the bottom sensor is inactive,

Plus I would probably need another float sensor for the solenoids? and how would I wire that so it triggers the solenoids to activate when the water hits the top, and deactivate when the water drops back below the halfway sensor, or would I need to use yet another sensor for this?

Thanks for your input!

 

Im just reading information on Latching circuits with relays, which seem to be the answer!

 

So, I'd use the bottom float switch, as a latch on the relay contoling the pump, so that after the middle float switch trips the relay, starting the pump, the bottom float switch will keep the relay closed, until the water drops below it. Right? Now my concern is using as little as possible power for the relay signal, I dont want to waste my batteries unnescessarily, how can I drop the volts/amps from 12V down to the least usable voltage? or is this not possible/not an issue?

 

Im just reading information on Latching circuits with relays, which seem to be the answer!

You want to be careful. If the water is overfilled, what happens?

You need a full feedback loop and redundant sensors, or a drain float to empty the tank and stop the pump in the event of a relay failure, which is more likely than solid state failures.

What is the water level difference between empty, full, and overflow?

 

You want to be careful. If the water is overfilled, what happens?

You need a full feedback loop and redundant sensors, or a drain float to empty the tank and stop the pump in the event of a relay failure, which is more likely than solid state failures.

What is the water level difference between empty, full, and overflow?

The tank would have a manual drain on it if I needed it, How often would failure with relays be expected? how expensive would other options be? if its cheaper to use relays, I would probably rather just repair them if they break, providing its not too often!

The overflow system would be 2 solenoids that redirect the pump flow from the sprinklers to a water storage tank, tripped by a float valve at about 80%,

The reservoir holds about 100 litres, the pump would start watering at about 50% until empty, overflow would be triggered at 80%, and go back to normal at 50%, the pump moves 550l/h, so should be more than enough

What exactly is a full feedback loop & reduntant sensors?

Thanks,

 

I was going to ask the level of your expertise but you sort of answered it with your last question.

I set up a system almost identical to what you are speaking about years ago, back on the farm as a matter of fact. I was about 21,22 so it was along time ago and I was just teaching myself electronics.

That means it was all relay and switches. I used floats but just as sensors and it was for a stock tank filled from a spring. (Kept overflowing and left nothing but knee deep mud for fifty feet around the tank.)

Why don't you draw out a block digram and post it. Then we will have something to work from. It is not real difficult but as you stated, you want to learn, so start with a drawing, the best you can.

Also as thatoneguy stated, you need to be real careful. I am assuming you have some sort of sealed reservoir/tank, or maybe not re-reading the post. What would happen if it overfilled?

A manual drain is fine but what happens if you are not there?

I don't have anything on the system I set up, I'm 70 now and a lot of water under the bridge, as I stated, it is not too difficult but again, you need to do what you can. Start with the block digram.

Oh yes, the pump and the rest of the system require what they require in power. Changing voltage is not going to help. Just use the best, most efficient pump you can get or afford.

Luck
Roland

 

 

I tried to upload a block diagram but it didnt work, hopefully it will attatch to this message!

 

So the above is just the pump relay, I have to draw another one for the overflow system,

If it all fails, and goes without notice, it is not greatly detrimental, after the reservoir is overfilled, if the water doesnt get relieved, it will build up in the soil of the box, then overflow the top eventually, drowning watevers in the box. but it would never get to that stage at all,

Thanks

 

The voltage change I am concerned about is the relay, I dont want to power the relay with 12V and be burning more than I have to to keep it closed, Would I just use some sort of low-activating power relay? and still power it with 12V? or step the 12V down somehow so the relay system uses less power?

 

The block digram did come through. I'll have a look at it but think I am going to be missing for the afternoon. May be tonight before I can get back to you.

As to the relays, you can hold some relays at a lower voltage then the pull in voltage, but not much. Usually the current used to pull a relay in is a bit more then the holding current but I would be surprised if it helped much.

Most small relays like you are talking about use something like 30 or 40mA. That's 0.030A. If you are real concerned find the amp hour rating for the battery and divide it by .040, the higher current. That will tell you how many hours your battery will last. Of course you have to multiply that number, the 0.04A by however many relays you need to use.


Gota go now, later/

Roland

 

Hey StainlessSteel.

I didn't have to go to school after all, my granddaughter was on the bus so I just had a short run.

Okay, you did a neat job but unfortunately I'm not sure what you have there. I think I know what you are saying and want to do but not how that block digram is supposed to do it. Well, maybe but lets not go on maybe. Answer the questions and get back to me.

The way it looks to me is the tank must be somewhere between empty and half full. Is that what you wanted to show.

One other small item. Just like a car or truck, we normally use negative as ground. As drawn you have negative being the switched side of the circuit. Works, but certain conditions can cause it not to. Won't go into that right now, just turn your polarity around. Make the positive side of your circuit the switched side.

Are you trying to show the conditions as I said above, with the tank somewhere between half and empty?

Also while you are at it, do you have the pump, relay, and what are their specifications, amperage requirements? Voltage is marked as 12VDC. What is the solar panel output? Amp hour rating for the battery?

Get back to me and I will have had time to eat and think how best to help with the idea you have. OH, the open switch on top is not meant to be connected to the circuit, is it? I mean as drawn, the top contact is not supposed to be connected to the pump power, I hope?

Later, gota go feed my face, one of the disadvantages of living alone..
Roland

 

No that top switch is not meant to be on the pump circuit I just drew it wrong, basically the idea would be, the pump starts when the water trips float switch 2 (about halfway), and float switch 1 should hold the relay in until the water no longer trip it(water is empty), unfortunately I have to wait for the relays etc to come in the mail, I live in a false economy, and I would have to pay about $50 for what will only cost about 10 online!

 

Okay, not a problem on the wait for parts.

A situation I am quite familiar with trust me. Just return to your post and let me know when you get them. Do you have anything on the solar panel and battery or are they on order also?

Later, take care.
Roland

 

Hey Kingsparks, so I'm just a bit confused about Coil power, Coil resistance etc, How much amps would a relay with a coil voltage of 0.53W and a resistance oh 275 ohms draw? can you explain that to me?

 

Not real quick but in partial answer; The power in the coil, or used by the coil would be, in your example, .53watts. Two hours for a bit more than a watt used. The power is a result of the current times the voltage or in this case, since current is not given but resistance is we get the current by dividing the voltage, 12V, by the resistance, 275 ohms. That gives up the current of 0.0436 or a little over 40mA. That's milli Amps or 1/1000 of an amp. in this case 43.63/1000 of an amp.

Does that help.

Roland

 

Yes that helps a lot, so of course, current = V/R. Does this and any resistance reduce the current available further down the circuit?

 

Hi.

Depends on the type circuit. Of course the power source matters also but only in the limitations for suppling current.

The way the relay is set up is in parallel with the other systems, the pump as the only other power item at the moment. So in short it will not lower the voltage or limit the current as long as the supply is powerful enough to drive them.

That's why I ask for the amp hour rating on the battery and your solar panel output. No matter how you cut it, if there is not enough "juice" to run everything then your in trouble.

Hope that clears things up some. There is so much to something even this simple that it can be confusing to someone new.

Any questions or other doubts you have just ask. I won't say I can answer everything but there are a bunch of people on this forum a lot smarter then me and someone should know the answer.

Take care.
Roland