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Outdoor lighting project

Discussion in 'LEDs and Optoelectronics' started by Log2, May 16, 2011.

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  1. Log2

    Log2

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    May 16, 2011
    Ok, so as the title describes, I'm looking to build some outdoor lights. Now for some elaboration, I want to make some deck lights using LED's and some rechargeable batteries.

    So far my project is in the R&D state.

    What I'm thinking is using 4 rechargeable batteries in a parallel circuit to keep the voltage at about 1.5V

    Now I know basics of electronics, and this project shouldn't be too complex with a little help and the internet. But what I'm thinking is putting a 1 Watt solar panel at the beginning of the circuit to constantly, or charge the batteries under control with the help of a switch.

    Basically I want to know if there's an easy way to trickle charge, or automatically turn off the charge when the batteries are full? That's my first problem.

    Now my second problem is that the solar panel I have fluctuates from 0 (nigh time) to about 20 Volts. So I'm wondering if I need to make a small voltage regulator circuit to lower it to 5V when I'm charging the batteries? Or maybe even regulate it, then lower it to 2V or 1.5V to make sure the batteries don't over heat.

    Basically what I have right now (I don't have a schematic program, so please excuse the written version) I have the Solar panel as the main power source going to a 200 microfarad electrolytic capacitor in parallel, then to a 7805, followed by a second electrolytic capacitor in parallel, then after that I can throw in a diode or a resistor to lower the voltage in series, then put the batteries next in the circuit in parallel so they keep their voltage. After that I have my lights and that's all easy after that.

    Sorry for the long message, but if you're still reading, then great, I'll sum up. What I need to know, is will that circuit work properly? Also is there a transistor, or a chip I can add in to see when the batteries are fully charged, and make the solar panel stop charging?

    Any help would be very appreciated, thank you
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You would probably be best off with a sealed lead-acid battery as the charging for these is somewhat easier and small battery trickle charger panels are relatively easily obtained.

    An SLA battery would give you 12V and then you'd need an appropriate switchmode LED driver to efficiently drive your LEDs.

    Tell us more about the LEDs? How many? What colour (voltage will differ)? How will you arrange them (series, parallel, etc.)

    With an SLA battery you would also probably want everything to turn off if the voltage falls below about 10.5V
     
  3. Log2

    Log2

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    May 16, 2011
    I was looking into a bit more, and I think I will go to the 12 V application for ease of use. Also I plan on using white, 3.6V Leds. approx 38-50 of them, haven't really looked much into it, it might be less than that if they're brighter than expected.

    Also I'm not sure how I want them to be wired, as far as I can tell, if I have them wired in parallel, I read the article you posted about it, and it looks like parallel would be best in my situation. But if you know of a better way, please let me know.

    Please tell me more about this turning off everything if the voltage dips, I assume the lower the voltage is, the dimmer the lights would get, to some extent, but please elaborate as it's something that is coming to mind as I further research
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    If you're running from a SLA battery, 10.5V marks the point at which the battery is effectively discharged. Discharging it more may result in accelerating the chemical changes which kill lead acid batteries over time.

    You should get a couple of these LEDs and try them first, then figure out how many you'll need.

    The important part is the current they'll run at. If they require 300mA each (1W LEDs) then you may be best off looking at LED drivers. If they are 20mA each, then resistors may be sufficient.

    As a rough guide, you might consider placing 3 in series with a resistor, then duplicating this as many times as required. The problem is that the brightness will vary greatly as the battery discharges.

    Some form of constant current supply will allow you to retain full brightness until the battery is discharged. A switch-mode circuit has the advantage that it is much more efficient and your battery will last longer.

    You need to figure out the battery capacity required to keep the LEDs illuminated as long as you need them, and then determine the size of the solar panel required to charge that during the day. Remember that cloudy/rainy days will reduce charging greatly. You may want your battery and charger capacity to be multiple times your requirement to allow for a string of cloudy days.
     
  5. Log2

    Log2

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    May 16, 2011
    Oh I Wasn't aware that it dies at 10.5V

    What would I need to do to turn it off automatically before that?

    Also putting the LEDs in series, would it look like the schematic below if I wanted multiple lights? I plan on putting a switch before all the lights, but if there's an something I need to turn it off before 10.5V then I might need a circuit that can reset a push switch or something, not sure, any suggestions?

    Also the LEDs I'm buying come with a resistor specifically to hook it up to a 12V battery, so I assume they're only 20mA or less
     

    Attached Files:

  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Every battery technology has its problems. Lead Acid batteries suffer from sulphation. This is accelerated when the battery is discharged too far, or when it is left discharged for too long. Preventing discharge below some voltage (I have suggested 10.5V) will provide some protection and longer life for your battery.

    To automatically turn it off you would require a circuit that constantly monitors the voltage and disconnects the battery when it falls below 10.5V. This would typically be a comparator and some voltage reference, and a couple of resistors. To do the switching you would require a mosfet. Unless you have some skill with electronics, it may be a little tricky to build this up.

    The schematic you show is a series/parallel arrangement of LEDs This would be a suitable arrangement to use.

    I'm interested in the reason you want to pursue solar power. Is it that there is no mains power nearby?

    Will these lights come on automatically, or will you manually switch them on and off. If it is the latter, you could use a switch that automatically turns off after some time (I've seen some mechanical ones that give 10 to 15 minutes. That would make it impossible to leave the lights on, thus mitigating the problem of over-discharging the battery.

    You're using low power LEDs. I presume this is for a visual effect rather than lighting per se?
     
  7. Log2

    Log2

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    0
    May 16, 2011
    I will be using standard LEDs, so yes, it is more of an aesthetic design as opposed to a lighting design, I decided to make my summer project to redo my deck, and figured some floor pot lights would be a nice addition. It has lattice board for it's rail, so the light would be able to reflect off that to give a more warm feeling, also the deck I plan to install them in is only about 150 sq ft. so it's pretty small, and if I put 8-10 lights in the floor, it should light up the deck with a soft glow, nothing too powerful.

    I'm planning on doing 3 LEDs per light, so a total of 30 LEDs max. I only plan on using it about 2 nights a week, for 5 hours max. It will be mounted to a normal switch, so I can turn it off when I leave the deck, and turn it on when I enter.

    As for the reason for going solar. I want a project to build up, I've done a little solar work before, but not much, so I wanted to try it out. I don't want to use an outlet, and I don't really want to drill into my house to connect to a main, at least not for such a small project.


    Now for the problem of the resistor, if I were to put the 3 LEDs in series after the resistor, it would change the resistor needed would it not? And if the LED's take 3.4V (as indicated on the web site I'm purchasing from) and at 20mA each LED, totaling 60mA needed to light them, from a 12V battery, I would need to put a 51 OHM resistor before them correct? or is my math completely wrong...
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Remember that a 12V battery will vary in voltage from perhaps 14.5V down to 10.5V.

    3 LEDs in series will draw the same current, so 20mA for one is 20mA for all, and 20mA total. using a nominal 12V for the supply and 3.4V each, the resistor value is:

    ( 12 - (3.4 x 3) ) / 0.02 = (1.8 / 0.02) = 90 ohms.

    However the current will vary significantly with battery voltage.

    at 14.5V it will be 48 mA
    as 12V it will be 20 mA
    as 10.5V it will be 3.3 mA

    It's not an ideal situation. 48 mA may well damage the LEDs, 3.3mA may be quite dim.

    One practical solution is to get a boost DC/DC converter to boost the battery voltage up to 20V, then a buck DC/DC converter to convert it to 12V. Whilst there are losses involved in this, it will result in a constant 12V output independent of battery voltage.

    I'm pretty sure you can get small DC/DC converters like this on eBay...

    OK, not eBay, but here and here -- you would need 1 of each, and a low ESR capacitor to place between them. AAAGH! actually, they're both step-down (buck), you need one step-up (boost) and one step down.

    The alternative is that you could have 2 LEDs in series (6.8V), use a single buck converter (one of the above) set to (say) 7.5V and a 39 ohm resistor.

    The alternative would be more efficient, but you would not be able to have 3 diodes in each fitting, you could have 2 or 4.

    With a choice of 2 of these DC/DC converters, I'd almost certainly go for the former with it's higher current capability.

    Note that with these, the LEDs will have a constant brightness from the battery fully charged down to almost no charge at all (probably below 10V) so you would need to be careful.
     
  9. Log2

    Log2

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    May 16, 2011
    will an IC like this: Step Down Voltage Regulator

    Will that not work? It takes an input of 12 and drops it to 3.3V or am I way off base here? Anyway that's the site I'm ordering from, and that's the only step-down, or buck I found that would be suitable, seeing as how the LED's operate between 3.2 and 3.4

    Also I was told, on another site, to use 180 ohm resistors if I were going to do strings of 3, like previously discussed. Would that dim the lights too much I guess?

    Also as for your 2 led and a buck idea, I don't mind having to use 2 or 4 lights, I'm ordering 50 LED's in the case some burn out, or don't work from factory, so 4 lights would be possible.

    Also if the lights stay constant (which is what I'd prefer) how would I figure out how long they can last before I have to turn them off? it's a 7AHr battery
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    It might, but that's just the surface mount chip. The links I gave you were for a complete regulator -- just connect up the wires ad go (ok, well, set the voltage first).

    Remember that LEDs are current and not voltage devices. Powering them from a voltage source, you will always need a resistor to limit the current.

    The value of the resistor depends on the supply voltage and the current you require through the LEDs.

    180 ohms will drop 3.6V at 20mA, so for 20mA through your LEDs you would be looking at approximately a 13.8V supply. That is a voltage commonly seen on a reasonably charged 12V battery. It is simply another variation of the problems associated with the battery voltage changing with the state of charge. You simply can't have a resistor value that is optimal for all states of charge.

    It's rare to find any damaged, and with care you shouldn't burn any out, but dtuff happens so it's always best to have a few spare.

    Well, you could measure the battery voltage and turn them off when the voltage gets too low. Alternatively you could build a little circuit that would tell you. You may even be able to incorporate such a circuit into a power on/off arrangement so that the lights automatically turn off when the battery gets low.

    What's your skill with electronics? have you ever built anything involving IC's on veroboard?
     
  11. Log2

    Log2

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    0
    May 16, 2011
    Well as you may have been able to tell from this topic, I'm not the greatest with the theory of the electronics, the application I can do no problem.
    I took a course about 5 years ago in my last year in high school, or secondary school, whatever you want to call it.
    Basically we built circuit boards, designed them and such, stripped the copper off ourselves (unfortunately I don't have the proper acid to do such a thing at home) the only problem with the course, was that there wasn't much theory, sure we learned how to read resistors, and what a transistor is, and the proper way to mount diodes, and electrolytic capacitors, and how to tell the difference in most components, but if we ever worked with IC's we would only learn where pin 1 is and what gates we'd be using, and how to mount it, but never anything to severe, the AND, XAND, NOT stuff is some basics (which I mainly forget now)
    But currently I am a millwright by trade, so I work with a lot of very precise tools, so I have a lot of patience, and am willing to learn what I need to learn.
    So to sum up, I do know how to solder, and test, and how to be a sheep and put the proper parts where they belong, but was never taught how to choose which parts I needed to use, always seemed to confusing as a young lad.

    Either way, I looked at the web site you pointed me to, and the ship to Canada, so I will most likely buy the second one you posted, Step Down But could I not just use that (rather than using a step up as well) and bring the voltage to 10v or something... or would that limit my current to 3mA or whatever you mentioned in a previous post? I'm not fully sure how the current is calculated in the battery, but that buck converter says the max current output is 500mA

    But yes, if possible I would like to make an auto-off type thing, if it's not too expensive that is. I am like most DIYers and prefer to keep costs as low as possible, while the product is semi-professional looking/working, haha.
     
  12. Log2

    Log2

    9
    0
    May 16, 2011
    Ok, (*steve*) so if I have the first buck converter, how would I go about putting that into the circuit to make it so the lights are a constant brightness rather than the opposite. Or would it be better to buy a step converter (which I don't know where to find one, because I looked on that site and couldn't find it) and a buck converter and step it up to 20 then down to 12
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You connect the buck regulator the the 12V battery and set the output voltage to 8V (there will be an adjustment for this). In order to do this you will need to put a load on the regulator. A resistor between 100 ohms and 1K should be fine.

    Then you wire up pairs of LEDs (with appropriate resistors) to the output.

    Double check that you're getting 20mA (or whatever you calculated) to each pair of LEDs and you're done.
     
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