Greetings,

I am trying to design a circuit that will turn on an LED every time a pump motor is running. The motor is 24V and the LED will need to operate at 24V as well (I can do the LED portion of the circuit). I just need a way to detect that the pump is running. It is always powered on, but reacts internally to low pressure in the fluid side of the pump, and then powers up the motor to pump more fluid and re-pressurize the fluid side.

It is for my sailboat, and will help produce some warning if a water line below decks lets go while the engine is running and we can't hear the pump running to endlessly replenish pressure in an open line (which happened in my trip to Panama).

Many thanks,

Paul

 

all you need to do is put an LED in series with a resistor, both in parallel with your pump. here is an LED calculator http://led.linear1.org/1led.wiz
I assumed a normal red LED (2V drop, 20mA) and the result is that you should use a 1.2Komh resistor in series with the LED

 

Ugh, I don't like that calculator strantor!

It's OK for just one LED where the voltage supply is at least a volt above the Vf of the LED, but it gives bad recommendations if calculating an array and the supply voltage is nearly evenly divisible by the LED Vf.

Also, you misread the OP's problem statement - power is ALWAYS applied to the pump. The pump is switched on and off via a pressure switch (low-pressure switch). They want to be alerted when the pump turns on. Simply connecting an LED across the pump won't do that.

 

Here are the ways I can think of to detect the motor is running: a pressure detector in the fluid side of the pump or a noise detector that "hears" the pump motor. Even if you get inside the pump and reach the contacts that control the power to the motor, you still don't know the motor is running; you only know it's being given power. A pressure detector would be the best because it would tell you what you really want to know, i.e., there is fluid pressure.

 

Also, you misread the OP's problem statement - power is ALWAYS applied to the pump. The pump is switched on and off via a pressure switch (low-pressure switch). They want to be alerted when the pump turns on. Simply connecting an LED across the pump won't do that.

Yeah I didn't get that part. by "power is always applied to the pump" I figured OP meant "power is always applied to the pump circuit" because if power was literally always applied to the pump, then it would always be running. Hadn't considered that the circuit may be literally inside the pump. if the pump were opened, you could probably find where the motor leads connect to the circuit (circuit is likely just the switch) and connect the LED there.

 

I think that the current drawn by the pump will increase by a large amount when it is running, compared to when it is switched on but idle. The first step might be to look into any information on that from the pump manufacturer, or failing that, measure it. Please ensure that any ammeter you use for this is capable of handling the current. The running current may be big, and there is likely to be an even bigger surge on startup.

You could not simply put a resistor in series with the pump to get enough light to run an LED directly, as in a 24V circuit the power lost in this way would be significant. An arrangement with a low resistor and a comparator may be possible, but would require care to make it reliable in the possibly aggressive situation of a boat's electrical system.

Possibly a reed relay might work, using a low-voltage, high-current coil. Otherwise, these days a Hall effect sensor might be a better bet. LEM is one of the well-known makers, but there are others. http://www.lem.com/hq/en/content/view/13/86/


Commercial current sensor units, rather than basic sensors are certainly made by some industrial control manufacturers for AC mains use. I'm not sure about 24V DC, but as this voltage is used quite a lot in industry you might get lucky.

 

I am assuming that CityGuy does not want to open the pump--may be far more difficult than an electronic solution.
Another way to do this is to put a properly sized rectifier in series with the 24V line (low side) and detect the voltage drop when the pump is running. The voltage drop is detected via an op amp (LM358) that turns on the LED. Not difficult, but the downside is that the pump will operate at 4% lower voltage.

WARNING--adding anything in series with the wiring adds another point of failure--I would not want his boat to sink...

 

Hello,

Perhaps monitoring the current drawn by the pump with a hall sensor could work.
There are several sensor applications on this page from the EDUCYPEDIA:
http://educypedia.karadimov.info/electronics/sensorshall.htm

Bertus

 

If the pump draws a significant current, it might be possible to detect it simply by voltage drop in the wire that feeds it. For instance, 14 gauge wire has a resistance of 3 milliohms per foot. So if you connect 2 sensing wires on the feed to the pump and you can get them separated by 10 feet of run, you'd have .03 ohms between them, which would give you 30 millivolts of difference per amp of current flow, and that's easily sensed. Note that there's no voltage drop in the wires that feed the sensor, as there the current is very small.

 

Suggestion:

Wrap a few turns of one leg of the pump wire around a soft iron slug. Glue a reed switch to one end of the slug.

Pump turns on, electromagnet, reed switch activates.

Number of turns is going to be dependent upon pump current draw. Use 50% to 100% more turns than required to activate the switch.

 

Further clarification:

Is it a JABSCO SensorMax pump 32755-0000 running at 24v. I sent off an email to their tech support group for a schematic of their pressure sensing circuit/switch.

Many thanks,

Paul

 

Further clarification:

Is it a JABSCO SensorMax pump 32755-0000 running at 24v. I sent off an email to their tech support group for a schematic of their pressure sensing circuit/switch.

Many thanks,

Paul

Funny, I replied to this post and it didn't show up! Anyhow, if I read the spec correctly, the pump should draw close to 8A when operating. This is plenty of current such that you should be able of make the above mentioned electromagnet with only a few turns.

 

I will give the Electromagnet/reed switch a try and see if it works. It should be pretty simple to implement in the engine compartment.

 

Suggestion:

Wrap a few turns of one leg of the pump wire around a soft iron slug. Glue a reed switch to one end of the slug.

Pump turns on, electromagnet, reed switch activates.

Number of turns is going to be dependent upon pump current draw. Use 50% to 100% more turns than required to activate the switch.

It might actually be more sensitive if you place the reed switch inside the coil without the slug. The reed switch is made from magnetic material which will concentrate the magnetic field in the coil. It's the magnetic field going from one end of the switch to the other and across the contact gap that closes the contact. A magnetic shunt (such as an iron nail) on the outside of the coil close to the two ends of the switch will also increase the magnetic field through the switch.

 

It might actually be more sensitive if you place the reed switch inside the coil without the slug. The reed switch is made from magnetic material which will concentrate the magnetic field in the coil. It's the magnetic field going from one end of the switch to the other and across the contact gap that closes the contact. A magnetic shunt (such as an iron nail) on the outside of the coil close to the two ends of the switch will also increase the magnetic field through the switch.

Didn't consider that. Good point.

Edit: *But*, its a really small switch, and a really fat wire!

 

@CityGuySailing: How'd it work out? Any luck?