@MisterBill2 There should be no need to connect the NO. When the charger is operating the coil remains energized and the LED's are isolated from the battery. When the charger shuts down the relay will drop out and the NC will close, thus powering the LED's.

I'm in agreement with Bill - you don't want a 12V regulator, but rather a simple resistor to limit the current to the LED's.

You mention that you understand LED's are "Current" driven, not "Voltage". LED's have a characteristic known as Vf, or "Forward Voltage". We don't know what the Vf is on your LED Lamps. If they're designed to operate on 12V then you don't even need a resistor. There's already some sort of current limiting circuitry built into the lamps. Besides, a 12V regulator generally needs a "Head" voltage (some volts higher than the regulated output). Haven't messed with regulators in quite a while, so off hand I'll take a wild guess and say that for a 12V regulator to work the input voltage needs to be 2 to 3 volts higher than the regulated output (Head voltage 2 to 3V). That's probably wrong, but for the sake of argument, a 12V regulator powered from a 12V battery doesn't work. At some point the battery voltage is going to drop, and you won't have 12 regulated volts supplying the LED's.

SW1 ? ? ? Why the switch? Is that so you can test the lamps? So you can turn them on if needed? Personally I see it as a wasted use of excess components. The variable resistor - I get that from earlier comments that you want to be able to trigger the lighting during "Brown-Outs" (BO). If you have a lot of BO in your shop then I can see its usefulness. However, by switching the DC side with a relay as opposed to the proposed AC relay - the DC relay in general is not designed for long term operation. By "Long Term" I mean being active for days, weeks, months, etc. An AC relay is designed for that sort of operation.

As Bill pointed out - a "Wink-Out", where the lights flash off for half a second or so - with the charger controlling the relay, there's a likelihood that during such WO's the LED's will not flash on. The charger will probably have a capacitance on its output that will tend to carry the output for a few seconds. MY 13.8V PS can power an LED (single 5mm type) for a good 30 to 40 seconds after I've switched it off. YOUR LED's at 12V will probably continue to run for 2 or 3 seconds after switching the PS off. (my PS).

My opinion (and opinions are like arseholes, everybody has one and many stink) is that running an AC relay on the mains is the way to go. When power BO's, lights may or may not come on. But when power fails outright the lights will definitely come on. As for using a DC relay and a diode to block reverse current from holding the relay active - that's a workable solution. But I don't know, and don't think it would bother the charger. However, the presence of the DC relay on the charger might cause issues. Batteries have ESR (Equivalent Series Resistance). The smart charger may depend on reading that ESR. The presence of the relay will definitely confuse that aspect of the charger. Which is another reason why I would opt for a mains controlled relay.

 

Tonyr1084 -
SW1 is for allowing the voltage across the coil to reach mains voltage for it to trigger, using a momentary switch.
This shorts the resistor, putting full mains AC across the coil.
Once the relay is latched and the switch released, the coil's voltage / current is dropped by R1, dropping voltage and heat at the coil. According to Ian0 (I think) the holding voltage of the coil is much lower than the trigger voltage, and I'm keen to reduce both operating current, and temperature.
SW2 could be used to test the lamps.
Good point about the charger capacitance holding the relay open.
I'm going off'f 12v relay idea now, maybe I'll Plan A it on mains. Fewer components.
And no-one has a problem with an AC relay being energised for weeks / decades at a time? (which was my initial concern - hence the resistor and switch)

 

According to Ian0 (I think) the holding voltage of the coil is much lower than the trigger voltage, and I'm keen to reduce both operating current, and temperature

I believe latching voltage is typically 70% of operating voltage and 30% is the dropout voltage. But those have all been on DC relays. I don't know if you need to worry about that on AC relays, nor do I know what the latching or dropping voltages are. That's what spec sheets are for. They'll give you far more accurate information than anything I've offered.

 

A resistor will improve the sensitivity to voltage but most of the relay's I've messed with seem to have a 70% pull in and a 30% drop out.

OK, it wasn't Ian0, it would appear to be "ThePanMan" who said it first. And he seems to have the same numbers as I do. Maybe I'm smarter than I think. Nah!

 

Tonyr1084 - that's why R1's a variable, just turn it until it works!! IMHO spec sheets are only a starting point anyway.
And I think Ian0 suggested the dropped holding coil voltage (something about the magnetic flux circuit being more efficient) but he didn't give me any figures. I was worried about high prolonged coil currents and any medium value resistor should work (100 - 1000 ohm).

 

If you wish - the resistor is fine. Just be sure to calculate for the wattage it must dissipate.

Mains voltage of 240VAC, a relay coil resistance of 200Ω means there's 1.2 amps (and I'm making up the coil resistance for sake of explanation). At 240VAC and 1.2 amps a resistor would have to dissipate 288 watts. That's a HUGE resistor. You're not going to get away with a 1/4 watt resistor. Keep in mind I don't know what the coil resistance would be. It could be 2000 ohms. In that case you're still dissipating 28.8 watts.


All I'm saying is that you have to do the math. The data sheet is where you'll get the correct information from. It'll likely tell you how much current the coil draws. Hard facts are far better than someone making wildly wild guesses as to what the resistance may be. I think you're going to find it hard to find a small variable resistor that can handle that much wattage.

But if that's the way you want to go - then the math is your best friend. You should be able to calculate the correct resistance for the set drop-out voltage you seek. Just keep this in mind, if you want the coil to run cooler you have to move that heat somewhere else. A resistor will run the relay cooler, but you're still going to dissipate all that heat. Either in the coil or in the resistor. Or both. There's still waste heat.

AC is not my best subject, but I think the coil is far better suited for handling the AC than a resistor. Reactance (the actual measurement of the coil load placed on the line) is what is used to define it's - um - resistance (wrong term for sure); but maybe you're picking up on my point. The resistor is just going to get hot. But again, if that's the way you want to go.

 

For some AC or DC relays it is possible to adjust the spring tension so that the dropout voltage can be changed. But you might need to buy an antique relay to find that capability intentionally provided. For charging the batteries you really only want to use the float voltage to avoid damaging them.
As for the DC relay heating, there are intermitant types that will not last an hour,there are also those made for constant operation. They have a higher resistance and a lot more turns of wire in the coil.

 

As for the DC relay heating, there are intermittent types that will not last an hour, there are also those made for constant operation. They have a higher resistance and a lot more turns of wire in the coil.

There you have it. Spoken from someone with far more experience than what I have. The only relays I've ever messed with have been automotive type relays like the horn relay. I suppose the fuel pump relay must be different from a horn relay.

Years ago (1970's) I had a horn get stuck on. Short in the steering column. The only way to silence the horn was to cut the horn wire from the relay. When I got home (15 minutes about) opened the hood and the horn relay was hot as hell. Pulled the plug (couldn't do that on the road, needed a tool to unlatch the catch). Traced out the problem. Was the after-marked steering wheel I installed. Accidentally pinched the horn wire, which eventually pierced the insulation and caused the malfunction.

Nuf bout dat. I'd still opt for an AC relay on mains because it has no chance of causing problems with the smart charger. I also don't care for having to "Reset" the relay after a power failure. If you have 20 units mounted on walls then you need to go around and either climb a ladder or push a reset button with a stick on all 20 units. If you miss one - the battery will be drained or the charger will try to carry the load, which may lead to premature failure. And you won't know it until the next power failure.

 

Excellent point, Tonyr1084. Checking the specs now . . . . . . . . .
Coil Power 2W Coil Voltage 230VAC Coil Resistance 8.2 kΩ Switching Current 10A
This is a typical relay from RS components (the main UK warehouse is only a few miles from me)
So:
230VAC / 8.2kΩ = 28mA
230VAC x 0.028A = 6.44W

OK so I'm confused. Either my maths is wrong or the specs are wrong. Probably my maths. So, what am I getting wrong?

I don't want a 6watt coil constantly energised (waste heat and current) so a Vdrop resistor - lets say 1.8kΩ, rounds it up to 10kΩ coil+resistor ;
230/10k = .023A
230x.023 = 5.3W
So, using the resistor (10watt is plenty, it seems) we save a whole watt, and the power will be dissipated partly by the R itself, and partly by the coil, keeping it cooler. Less heat and current, as I wanted.

Does my Math look big in this?

 

. I don't know if you need to worry about that on AC relays, nor do I know what the latching or dropping voltages are. That's what spec sheets are for. They'll give you far more accurate information than anything I've offered.

An AC relay should Never be ran lower than the rated voltage, the only advantage an AC relay has is at pull-in time, due to the high in-rush, after that steps have to be taken to keep it retained, i.e. shaded pole etc.

 

An AC relay should Never be ran lower than the rated voltage, the only advantage an AC relay has is at pull-in time, due to the high in-rush, after that steps have to be taken to keep it retained, i.e. shaded pole etc.

A highly respected member.

The point I was making about wattage is that you're going to have to deal with heat no matter how you look at it. 6 watts is still 6 watts. Whether it's 2 watts in the coil and 4 watts in the resistor or 4 watts in the coil and 2 watts in the resistor - you still have the same amount of waste heat. And introducing extra components means introduction of potential failure points.

At this point I don't think I can add anything further to this conversation. I'll let others with more experience direct you. I still favor the relay (without the resistor) for a design (if I were to design such a thing).

 

Cheers Tonyr1084. Slight disagreement tho' - adding a Vdrop resistor reduces the current, therefore also the wattage, less waste heat.
But only slightly. And if AC relays need the full supply, (or whatever MaxHeadRoom, was saying), then that idea is no good.
I suppose I'll have to learn to love a 6watt coil.
But why did the specs say it was 2W?

Britain's in a heat wave so I'm going home early today. Catch ya on the other side peeps.

 

Britain's in a heat wave so I'm going home early today. Catch ya on the other side peeps.

OT: I did see that in the news, also that due to global warming, UK is now able to grow grapes for the wine ind, And now in France, it is too hot for it!!

 

Excellent point, Tonyr1084. Checking the specs now . . . . . . . . .
Coil Power 2W Coil Voltage 230VAC Coil Resistance 8.2 kΩ Switching Current 10A
This is a typical relay from RS components (the main UK warehouse is only a few miles from me)
So:
230VAC / 8.2kΩ = 28mA
230VAC x 0.028A = 6.44W

OK so I'm confused. Either my maths is wrong or the specs are wrong. Probably my maths. So, what am I getting wrong?

I don't want a 6watt coil constantly energised (waste heat and current) so a Vdrop resistor - lets say 1.8kΩ, rounds it up to 10kΩ coil+resistor ;
230/10k = .023A
230x.023 = 5.3W
So, using the resistor (10watt is plenty, it seems) we save a whole watt, and the power will be dissipated partly by the R itself, and partly by the coil, keeping it cooler. Less heat and current, as I wanted.

Does my Math look big in this?

I wonder about Max and the statement about full voltage for the AC coil relay. That has not been my experience with SMALL relays. Power relays, (contactors) usually should have close to full voltage so as to maintain contact pressure to avoid contact melting and welding.
At some lower voltage the smaller control relays will start to buzz, that is where my diode scheme fits in. Operating an AC relay controlled by a type 2050 thyratron there was sometimes a tendency to buzz just a bit, The diode solved that problem completely.
So you may need to experiment a bit with the series resistor. But certainly a low enough value so that the relay will pull in when power returns. Otherwise, one momentary dropout, the emergency lights come on and stay on and the batteries run dead. Bad show all around.

 

I suppose I'll have to learn to love a 6watt coil.
But why did the specs say it was 2W?

As you can see, at 230 VAC relay with coil resistance 8.2 k consumes 2 W with current 15.42 mA.

 

The current limit for a AC relay coil is the Inductive Reactance, the DC version is the resistance of the coil.
If the armature of an AC relay does not pull in or 'chatters' the current goes up.

 

The units are not the same, or even similar, and so the calculations must not be the same, or even similar. So somewhere there is an error. With a current of 2.677 microamps (2.677 x10-6amps) the voltage will be huge to get 2 watts. Thus my claim that some place there is an error.

 

The units are not the same, or even similar, and so the calculations must not be the same, or even similar. So somewhere there is an error. With a current of 2.677 microamps (2.677 x10-6amps) the voltage will be huge to get 2 watts. Thus my claim that some place there is an error.

Really?



"The average value is zero over one complete cycle,
as the positive average area would be cancelled
by the negative average area ( VAVG – (-VAVG ) )
in the sum of the two areas, thus resulting in zero
average voltage over one complete cycle of a sinusoid."
https://www.electronics-tutorials.ws/accircuits/average-voltage.html

 

Once again I suggest that the two-diode arrangement to provide more constant magnetization is worth trying. The current does not drop to zero, while the power is less, the frequency remains the same. In addition, experience has shown that it is effective at avoiding buzz and chatter.

 

Once again I suggest that the two-diode arrangement to provide more constant magnetization is worth trying. The current does not drop to zero, while the power is less, the frequency remains the same. In addition, experience has shown that it is effective at avoiding buzz and chatter.

Mechanical force diagram: