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Water level relay?

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things

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Hi, we have a pool out the back, and we have the pump setup on a timer, due to the fact we are out of the house alot. The problem is, if the pool doesnt have enough water,, or the inlet somehow becomes clogged, the pump would run dry all day, which surely isnt good for it. So I was wondering if I could somehow connect a float to the inside of the inlet box, so if the water level was too low, it would flick a relay and turn off the pump. The only problem i see, is when someone is in the pool, the water level of course, varies ALOT, which would cause the pump to turn on and off alot. ANy idea's how I could get it so the pump only goes off if the float is low for say 5 seconds?

Thanks
 
One possibility would be to use a basic PIC chip .... You would have to program it or get it programmed .... but it could handle the necessary logic.
Basically, connect a sense pin on the PIC chip to interrogate the float switch, maybe every second or so, and program a condition such that a relay of some sort activates the water pump only after 5 successive tests of the sense pin were read to be high.... indicating that the pump intake was seeing a steady water level, and not a wave,....
Depending on the position of the pump intake filter screen, this scheme might also insure that the pump was not activated during a condition of no water in the intake basin ....
 
Hi,

I had to reverse engineer a circuit that works very well for me. The only thing I don’t like about it is that I needed to alter the operation of the relay such that when the circuit is on the relay is pulled into the normally open position. This means that any drop of power to the circuit will cause the relay to return to its normal position which in this case is closed and therefore the pump will automatically begin to work. Perhaps there is someone who can re-engineer the circuit.

Anyways here is how the original circuit it works.

No water in vessel.
The LED lights and the relay closes and the pump turns on.

When the water reaches the LOW sensor the pump continues to operate until it reaches the HIGH sensor.

Water Level Achieved.
When the water reaches the high sensor then the circuit switches off and the pump stops. When the water level drops to just below the low sensor then the pump turns on again and pumps until high level is reached.

To modify the circuit for my needs which appear the same as yours I dispensed with the LOW sensor and kept the HIGH sensor. Instead of having the relay close when the circuit is energized I connected it to the open position (moved the connection from Pin87 to Pin 87a) so that the pump is not operational. When the water reaches the high sensor then the relay is energized and the pump begins to operate.


Regards
 

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ANy idea's how I could get it so the pump only goes off if the float is low for say 5 seconds?

Thanks

Yes. It is a very common requirement.

The thing you are looking for is called "Timer". They have many variations. The most common one will keep its contacts closed for a certain time even the input signal (float sensor) opens. If the sensor kept open, then after the set time, its contacts opens.

If you use the timer contact to control a pump, then the pump will keep on running even if the float switch is open due to low water level or rapid water level changes. Say you have the timer setup for 10 seconds delay, then after the float is stable and open for 10 seconds or more, the timer contact will open and the pump stops.

Got it? These timers come in many different operating voltages and contact rating so you need to match the voltage to your pump circuit.
 
Where I worked they called those kinds of timers, on-delay relays, and they were quite reliable. The older ones were kind of large but the newer models were solid state and one used a DIP switch to set the time delay desired. They usually pugged into octal tube type sockets but I'm sure they come in other styles using just screw terminals.


Lefty
 
Where I worked they called those kinds of timers, on-delay relays, and they were quite reliable. The older ones were kind of large but the newer models were solid state and one used a DIP switch to set the time delay desired. They usually pugged into octal tube type sockets but I'm sure they come in other styles using just screw terminals.


Lefty

Yeah, I missed those pneumatic relays that would kept its contact closed *without* external power for 45 minutes and then open the contact. Very good for backup safety purposes.

The OP, however, would need to use a type called "off-delay timer relay".
 
Another solution...much less complicated ...Just use a long RC time constant circuit to turn the pump on ... with a solid state relay ...allowing maybe 5 or 10 sec. before the pump relay charges up to the turn on point.

The case where the switch is floating is such that the switch to ground is open. The capacitor charges to 5 V through the resistor, and the relay and pump turn on ....after the appropriate delay of 5 RC time constants or so.

For the case where the float switch is 'not' floating ... the switch to ground would be closed... the the capacitor is immediately discharged, and the relay and pump are turned off.

If the float is 'not' floating and is momentarily disturbed by a wave,
causing the switch to open for a second or two, then there will be a delay of several seconds before the pump would turn on .... due to the long RC time constant... allowing the wave action to die down.

The one condition to be careful about would be to insure that there is plenty of clearance between the floating switch position and the closed position .... You would not want the float to be floating, the high water condition, and have an inadvertent closure of the switch, which would immediately shut off the pump.
 

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User_88,

The input side of a SSR is basically an IR diode so it would operate say at 1.5V across its terminal. Therefore, one needs to use say 220 ohm series resistor to limit the current when connect it to 5V.

So let say the charging resistor is 220 ohm then. Can you estimate what value of capacitor to use in order that the capacitor voltage will reach 1.5V after 10 seconds?
 
User_88,

The input side of a SSR is basically an IR diode so it would operate say at 1.5V across its terminal. Therefore, one needs to use say 220 ohm series resistor to limit the current when connect it to 5V.

So let say the charging resistor is 220 ohm then. Can you estimate what value of capacitor to use in order that the capacitor voltage will reach 1.5V after 10 seconds?
Most SSRs have a built-in dropping resistor for the LED. Look at the specs. If the input is spec'ed as "3-32vdc", it has a resistor inside. If it specs the input as "1-30mA" it doesn't have one, and you need to add one.

Ken
 
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User_88,

The input side of a SSR is basically an IR diode so it would operate say at 1.5V across its terminal. Therefore, one needs to use say 220 ohm series resistor to limit the current when connect it to 5V.

So let say the charging resistor is 220 ohm then. Can you estimate what value of capacitor to use in order that the capacitor voltage will reach 1.5V after 10 seconds?

Not all solid state relays are the same .....
I am suggesting the use of a DC voltage controlled SSR...
something similar to the ones shown in the link here: **broken link removed**
It looks like the input control circuit has a little more to it than just an IR diode.

The relevant question here is at what voltage does the control voltage turn on the relay....
Assume the turn-on voltage is 3 VDC.

Assume also, that no significant current is required to turn on the SSR .... that it is essentially a voltage controlled device.

The first consideration would be to determine how much current to allow through the main resistor, R. ... A value of R=10k ohms would allow 0.5 ma to charge the Capacitor... using a 5 VDC supply. This relative magnitude prevents a large current from flowing during the 'float switch closed' condition.

From the standard formulation for charging a capacitor through a resistor:
a supply of 5VDC,
a desired period τ of 10 seconds to reach 3 VDC,
That is, an elapsed time interval of 1 time constant.
... The reasoning here is that 3/5=0.6, which is a duration of about 1 time constant on the exponential RC charging curve.
and utilizing the 10k resistor.
Therefore,
τ=RC, or:
10 sec= 10k ohms * C farads .... solving ... C=1000
µF

That is, at t=10 sec, using the specified values of R= 10k ohms and
C= 1000
µF, the capacitor voltage should be approximately 3 VDC,
and the solid state relay should turn on.
 
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The pump current is less when it's running without water. Sense the pump current using a sense resistor and comparator and use this to shut off the pump.
 
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