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I have purchased these following components:

  1. 12V/18Ahr SLA battery

  2. 300W modified sine wave inverter

  3. 50W LED floodlight (NO SENSORS I.E. PIR, AMBIENT, ETC)

  4. 12V relay module

We are currently experiencing a shortage in electrical supply in our country. So in order to keep the property light during the night I decided to make an emergency backup light.

My plan is to connect a fully charged 12V SLA battery to the input of the inverter and then connect the floodlight to the output of the inverter via the 12V relay module. The 12V relay will activate the floodlight when the power from the mains is off due to load-shedding.

My question is, will it be safe to keep the battery connected to the inverter continuously, so the system is autonomous and only turns on and consumes the battery when there is load-shedding?

And I only to intervene when the battery needs to be charged. On this point is it also safe to put a 12V SLA battery charger to charger the battery and still keeping it connected to the inverter?

The LED only runs for 2 hours per day. There is no timer controlling this. The power utility (Eskom, South Africa) which implements the blackout, turns of the power lets say at 20:00pm, the utility then leaves the power off for only 2 hours. They then turn the power back on after that. The reasons for this is because the utility tries to lessen the burden on the power infrastructure by having planned power outages for 2 hours per day. Hence the relay is used to switch the light on when there is a power outage via the mains control signal and then turn the LED off when there is mains power.

enter image description here

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    \$\begingroup\$ You should add a schematic of your plan. You can use the CircuitLab button on the editor toolbar and use the Custom Component for any odd devices. (Double-click to edit.) You should also show your battery charger. A 12 V light might have been a much simpler solution. \$\endgroup\$ – Transistor Feb 2 '20 at 13:09
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    \$\begingroup\$ Unless you really want to build something from scratch (which always is more fun), why not just purchase a standard UPS? \$\endgroup\$ – AnalogKid Feb 2 '20 at 13:33
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    \$\begingroup\$ Your block diagram looks fine. You just need to make sure that the trickle charge is the correct value for long term otherwise you'll cook the battery. \$\endgroup\$ – Transistor Feb 2 '20 at 14:04
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    \$\begingroup\$ If you already have all the components, then why'd you ask the question? There's nothing we can add to your plan. \$\endgroup\$ – Harper - Reinstate Monica Feb 2 '20 at 22:19
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    \$\begingroup\$ You say there's a shortage of power, but you should know that efficiencies, in a chain of devices, multiply, so if thecharger has (i.e.) 90% and the inverter 80%, then the total efficiency would be 72%, which means you're more efficient directly connecting the lights to the mains than letting them run from a battery which needs to be charged from mains, anyway. (something that's already mentioned in the answers) \$\endgroup\$ – a concerned citizen Feb 2 '20 at 22:31
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"That depends."

12V/18Ahr SLA battery should not be discharged deeper than 50% if you want it to last a long time, so I'll consider it a 9Ah battery.

That is only 100 W, you have a 50W light, and the inverter will also add losses, so battery life will be less than two hours.

If a blackout occurs during the day, and goes on into the night, the battery could be drained before it gets dark and you actually need light. So it could be useful to add an ambient light sensor to only turn the light on at night. A PIR motion sensor to only turn it on when there are people around would also increase your battery life. You could also use a PIR sensor to turn on a powerful light, and leave a low-power light on continuously. Note that such mains powered motion/light sensors tend to use capacitive dropper mains supplies which do not go well with modified sine wave power.

The inverter will use power even if the light is off. Probably something like 5-10W if it's a cheap inverter. So if there is a blackout, the battery will drain even if the light is not used... If you don't use sensors this is less of a problem, but your charger will still have to waste power to compensate for the inverter's idle power. So, in your current schematic it would be more efficient to put the relay between the inverter and the battery.

IMO a better option would be to get rid of the inverter and use a 12V LED floodlight instead. These are common and quite cheap due to high market demand (every pimped-up truck has at least one) so look for the terms "12V led bar" or "12V led flood". You can get cheap 12V motion detectors too if you need them.

This will be more efficient as voltage will only be converted once rather than twice, there will be no inverter losses.

You will need a low-voltage cutout to make sure the battery is never over-discharged as that will severely shorten its life. You can make the cutout latching or non-latching. In your case, non-latching is probably better if you want it to work without having to babysit it. "12V battery low voltage disconnect" and sort by price on Amazon gives results.

Even if you use the inverter, it is important to check it has a low-voltage cutout too! If it does not it will kill your battery.

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    \$\begingroup\$ I agree with this plan. \$\endgroup\$ – Tony Stewart EE75 Feb 2 '20 at 16:17
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    \$\begingroup\$ Unfortunatley has I said I already have these components on hand. Going out and purchasing new stuff such as 12V floodlight although its more efficient but it will add additional expenses which I cannot afford. In South Africa these 12V flood lights still offer less lumens than than AC options. Secondly there is no motion sensor, the LED is turned on via the relay, when there is a blackout. \$\endgroup\$ – Joey Feb 2 '20 at 16:30
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    \$\begingroup\$ Please remove the PIR from the answer. This is giving people the impression that my LED as a PIR sensor when it does not. The schematic shows a relay acting as a switch to turn on the light when there is a power failure. \$\endgroup\$ – Joey Feb 2 '20 at 23:40
  • \$\begingroup\$ I think this is a good plan too, though yes, it would help to remove the PIR assumption now that we have more info from OP. \$\endgroup\$ – Harper - Reinstate Monica Feb 2 '20 at 23:46
  • \$\begingroup\$ I have updated the answer. \$\endgroup\$ – bobflux Feb 3 '20 at 1:31
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Better to have the relay on the inverter input so it doesn't have the inverter running when not needed.

The charger arrangement is OK as long as the charger is designed to competently both charge and float a 12V lead acid battery.

The ability to correctly boost charge when charging from depleted state also desirable for full capacity and long battery life.

Having a low voltage battery cutout is very desirable if operation 'flattens' the battery.

Discharging a battery deeply shortens its life and loading it down to very very flat is very bad for it.

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  • \$\begingroup\$ The inverter comes with a low battery protection circuit builtin. \$\endgroup\$ – Joey Feb 2 '20 at 16:57
  • \$\begingroup\$ @Joey If the light is used occasionally then deep dischargemay be acceptable, with some loss of capacity. If use is frequent then, as peufeu says - limiting discharge to a fraction of capacity will very greatly increase lifetime. \$\endgroup\$ – Russell McMahon Feb 2 '20 at 21:41
  • \$\begingroup\$ The light is only used for 2 hours per day. \$\endgroup\$ – Joey Feb 2 '20 at 21:52
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As proposed, it will function, but badly enough to have usability problems.

The inverter has got to go

The weak point is the inverter. You should be staying at native 12V (or 24V DC, if distance is a factor) for several reasons.

  • First, it takes the inverter's parasitic losses out of the picture. Inverters are not magic; they require energy to sit there "spun up". In fact, it would be wise to picture an inverter as a M-G set that is actually spun up. You wouldn't leave that spinning for no reason, would you?

  • Second, your battery is way, way, way too small to burn that light any length of time. I get where the battery says "18AH" but that is only relevant to extremely rare use e.g. in a UPS. Lead-acid batteries have a serious problem being deep discharged. If you regularly dip them to 50% it will greatly shorten their life. For a reasonable life in a frequent-use senario, lead-acid better dip only to 70% full. So you now have a 5.4AH battery, giving 64.8 watt-hours, so your pack will barely make it through the night carrying the inverter's parasitic load, let alone any time of lighting.

    • If you're serious, we really should have a conversation about battery packs.
  • Therefore the only way this dog hunts is if you use a motion sensor to minimize lights-on time. DC motion sensors are half the price of AC motion sensors. Simply because motion sensors are low-voltage critters, and AC motion sensors need to have an internal 230->12 power supply in them, and that's not free.

    • Further, simpler AC motion sensors (i.e. series-wired) do not play well with inverted power.
    • In any case, you would need to keep the inverter spun up all night to keep the motion sensor powered up, which is nuts.

    • Motion sensors also should not have their power cut off, ever. That is because they also contain day-night sensors. They are not human eyes; they do not auto-adjust; they see true solarization. That, at any given spot, can vary dramatically depending on shade, moon, and night lighting. Therefore the sensors need to be powered up for at least 1 24-hour period so they can calibrate to the difference between day and night. Otherwise they will work in broad daylight, or refuse to turn on under a full moon/street lighting /etc.

    • Low voltage DC motion sensors are the way to go, therefore. Their parasitic load is negligible. Most sensors that accept 12V are also labeled for 24V.

About that relay

Why don't you want this light coming on when the power is on? Does your yard not need lighting then? I bet it does, and I bet there's another light already. That is silly, there don't need to be 2 lights unless you want to do a high/low thing. But then you mentioned an "On when power on" relay! That's a good use for that relay.

Other than that, I'd skip the relay altogether, reduce system complexity and simply allow the light to work even when the power is on. Make sure your power supply is big enough to both top up the battery and carry this load. Which, to your original question, is fine, as long as it's capable of 3-stage charging.

If you really want the relay to knock out the light with AC power on, then put the relay after the DC motion sensor, i.e. interrupting power to the light (NOT the motion sensor) if AC mains is on. At that point, post-relay, if you really, really want to have the inverter power a 120VAC light, then that is fine. The inverter only "spins up" when the motion sensor commands the light on, so you aren't wasting energy on its parasitic load the rest of the time.

If carrying 12V long distances is an issue, use 24V, or put the whole package out there at the light.

About those lights

It's hard to find bright, useful 12V yard lighting in the normal places you look for yard lighting. You have to look elsewhere. I've found happy choices in automotive / off-road. I use a 42W 8-degree spotlight (12V of course) that is as bright as a locomotive headlight. They also make them in 30 degree and 60 degree spot. So work out your angles and see what you really need; often, far less than you think.

The most important thing about yard lighting with LED is aiming it properly. Usually people are replacing metal-halide lights, which cast light in a sphere - with reflectors they cast light in a very wide wedge of almost 180 degrees. We had seven of them lighting up a yard; they burned 3300 watts. Seriously. And they mostly lit up the sky... trees... our neighbor's bedrooms... I redesigned the lighting system to use LED spot lighting, and cut the draw down to 300 watts simply by aiming the lights properly. Mind you, metal halide is already near 100 lumens/watt so there was no lumen efficiency gain. It's all about the aiming.

enter image description here

For a wall-pack light, the presence of the wall is not a surprise, so there is simply zero excuse for throwing any light at the wall. LED fixtures should be lensed so all the light misses the wall and most gets thrown beyond the here-overbright sidewalk. This involves about a 45 degree wedge of light, but that is easy for LED. These here 40W lights could be 15W and cover the yard as well.

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  • \$\begingroup\$ Thank you for this answer, its really eye-opening! The light will only be lit for 2 hours per day. Not kept on the entire night. \$\endgroup\$ – Joey Feb 2 '20 at 23:20
  • \$\begingroup\$ Just for the record I did not mention nothing about any motion sensor. This was introduced by a user NOT me, so please delete that from your answer. Secondly, yes there is another light that is switched on during normal mains supply. I am not interfering with this light as it is connected to mains directly and I am NOT an electrician thats wants to fiddle with the DB board. So I am making a seperate system \$\endgroup\$ – Joey Feb 2 '20 at 23:41
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    \$\begingroup\$ @Joey As a mains person, I applaud your caution and respect for mains power. However you not only want to use an inverter, you want to add a relay past it. You do understand: the output of an inverter is mains power. And must be treated with equal caution, not to mention Code compliance and use of mains-rated parts. That's why I'm trying to keep you in the low-voltage world as much as possible; that's pretty safe to work with and you're not confined to mains-rated parts. I'm well aware the light you intend to use isn't an AC PIR; I am suggesting a LV PIR. \$\endgroup\$ – Harper - Reinstate Monica Feb 2 '20 at 23:43
  • \$\begingroup\$ I have experience with an inverter but only small loads of 60W. I am thankful you are providing me with a much safer option, but I am aware of certain safety measures that needs to followed, hence I wanted a second opinion if the system described in the pic is safe. \$\endgroup\$ – Joey Feb 2 '20 at 23:53

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