LED lighting on a budget!

[kcress]

I'd use "multi-stranded" for everything. Don't use solid for anything. The idea here is to prevent stress and strain on the LED's star pads.

The sizes are very accurately controlled and get smaller with the larger number. 22 is much smaller than 16, etc.
Do not be fooled by the insulation. Some wire has thick insulation ans some thin making you think the wire is bigger or smaller.

 

[ronald7410]

I'd use "multi-stranded" for everything. Don't use solid for anything. The idea here is to prevent stress and strain on the LED's star pads.

The sizes are very accurately controlled and get smaller with the larger number. 22 is much smaller than 16, etc.
Do not be fooled by the insulation. Some wire has thick insulation ans some thin making you think the wire is bigger or smaller.

Got it! One last thing, if the diagram would change from series to parallel then the Drivers would be so far off from 3.5v in parallel, Whats the math to get a high voltage Driver (48v) to 3.5v? Is this overkill on most resistors?

 

[kcress]

??? I don't understand. You would never want to run all your LEDs in parallel.

If you have 48V driver you divide that by 3.3V to see how many you can run in series.

48/3.4 = 14 So you can run 12 to maybe 14.

If you want to run two strings you divide the driver's current ability by 2. If that is a reasonable amount of current then you can add resistors and fuses and run two strings in parallel.

 

[ronsonol]

I wanted to make light for my fuge - I used an old laptop power supply, it outputs 18.69 volts with load, up to 3.5amps. I have two bridgelux es leds on individual heatsink/fans. They run from 12.8-13.5 all the way up to 2 amps. So I added some 2amp fuses for protection as well.

I want to run the leds at 13 volts where they will draw about 1.2 amps.
I choose a couple of 25W adjustable resistors up to 5ohms...and started adjusting until the voltage across the resistor/ohms equalled about 1.2 amps. It turned out that i needed the nearly full 5 ohms to get this reading, so in the end I removed the adjustable poles and just running it end to end as a fixed value, which I read at 4.9 ohms.

At this setting I measure 6.05v across both resistors, so divided by 4.9 ohms gives me about 1.23amps. Is the 6.05v the amount the resistor is dropping from the full 18.69 volts? If that is true, then the leds are running at 18.69-6.05 = 12.64 volts...but according to the datasheet they shouldn't be anywhere near 1.2amps at that voltage.

One thing that worries me is the heat coming off these resistors, it is so hot I can't even touch it for a second, and you can feel the heat from 8 inches away...do i need to get larger 50w ones? Its done up in a small project box...where I will add a small fan.

Another note on efficiency...my killawatt shows 52 watts....seems high for 2 leds running 13 volts at 1.2amps. Or do I lose some because of the resistors, and the watt usage is high because it is really 18.69v at 1.2amps.

 

[Floyd R Turbo]

Well, you say that you wan the LEDs to run at 13V and draw 1.2A. You are getting 18.69V-6.05V (your math/thinking is correct) = 12.64V which is close to 13V, so what is the amperage supposed to be at that Vf?

I'm just looking at this from an EE standpoint, I'm don't have very much working knowledge about LEDs, but the standard formulas apply here, which is V=I*R and P=I*V.

If your PS output is 18.69V and up to 3.5A (I assume it is constant-voltage then?) and your P=52W, you will have some loss inside the PS but not that much (10% maybe, or 5W) so that leave 46W+/- on the output side.

EDIT: my math was totally hosed there. Sorry. I deleted the last part of that post due to sheer embarrassment. Brain is tired, been working this weekend...

 

[ronsonol]

Thanks. The specs on the datasheet indicate at 12.6 volts they should pull about 900ma...far off the 1.23 I read.

Also, the PS is constant voltage...and I am running only one LED in the series. The power splits to a (2) 2-pos terminal blocks. The (+) 2-pos block - each side goes to fuse and a resistor then the the single LED.

 

[Floyd R Turbo]

I'm not fully following your description and the math is screwing with me. Can you take a pic or draw a schematic and post it?

 

[ronsonol]

Here is the power distribution box. Each string goes to only 1 led.

 

[ronsonol]

I refuse to post a picture of the LEDs themselves also out of sheer embarrassment.
I forgot to tin the stars before I used thermal epoxy to attach them to the heatsink...I was lucky to get solder on it at all...but its ugly =)

 

[ronsonol]

Reading the datasheet (BXRA-C1202) again it says min. 11.8v typ. 12.8v and max. 13.8v for 1050ma...so its likely with the big range things are running as the math would seem.
18.69 with 4.9ohm 25W resistor...6.05 drop/4.9ohm = 1.23 amps.

So the concern still being the heat coming off these resistors. The math on the first page figures 6.05vx1.23a needs a 7.45W..double that for 15W resistor.

 

[kcress]

Thanks for the clear info ronsonol. It's nice for a change.

A couple of points. You say you have 6.05V 'dropped' across the 4.9ohm resistor. From that ALL we know is that you are indeed running:
6.05V/4.9ohm = 1.23A thru that circuit.

If your supply is indeed 18.69V loaded then yes! You are running 18.69V-6.05V = 12.64V thru your LED. But why math it? Measure it. Measure across the LED. You could actually be dropping a lot of voltage in your wires or possibly thru a crappy solder joint.

As for you question on the LED Vf verse LED current. The Vf is ALL about the LED's temperature. Higher temperature demands more current at the same voltage. If your LED is running hot you will see a large deviation from what the table states. Hopefully you have an excellent heatsink setup. Why? Because the math sez you're dumping:
12.64V x 1.23A = 15.5W into the LED. Some people have problems getting 3W away from their LEDs let alone five times that. This is definitely not a U-channel build. You need something like a CPU heatsink.

Now as for your resistors. You're dumping:
1.23 x 1.23 x 4.9 = 7.41W off of each one.

Your 25W resistors are just fine for that if they're in air that's room temperature. Putting them in a plastic box like that is not going to cut it, as you're probably starting to suspect. That's a house fire in the making. You should absolutely mount them in a non-flammable container. That means metal.

It also needs to be open... Very. Open. Get a metal bud box. That's those boxes that slide together. One side is purely a large U that slides over its mate that's also a U but has all the edges bent into little lips. They allow you to screw thru the top U so that the two halves make a closed box.
They are always aluminum sheetmetal. If you don't know what I'm talking about, ask. Anyway, get a generous sized one and mount the resistors in it using the proper technique for those power resistors. That's some sort of sheetmetal fingers that only stick into the ends of the ceramic tubes.

You cannot cover any of the outer body like your black straps are doing. Once you have them happily mounted pitch the cover in the trash. Replace the cover with any kind of large metal mesh bent in the same simple U as the discarded lid.

Screw it down just like the lid was. Make sure the mesh has the largest holes you can find that fingers won't fit thru. Window screen size will not work!! Expanded metal is a good size. Then mount your box somewhere safe that has air flow. It's also important that your two resistors are side by side. Don't have one mounted above the other.

Things will be much cooler and sustainable after this improvement.:dance:

 

[philshel]

A question for Kcress or any of the other electronics experts

A question for Kcress or any of the other electronics experts

When running parallel strings, do the resistors serve a purpose other than a place to measure the strings amperage? Reading all the dozens of various threads, that is the only purpose that ever seems to be mentioned. Would the resistors also help equalize the individual strings current draw or am I barking up the wrong tree?

 

[TheFishMan65]

IMHO they might help some, but not much. Say you have 2 strings at 700 ma. Over time they go to 600 and 800. With 1 ohm resistor the voltage will change 0.1 volts. Not a lot of change to help with balancing. And if you are using smaller resistors there will be even less change. A larger resistor will help more, but then you are wasting power.

I don't have a lot of hands one experience, but this is what the math tells me. So if I am wrong hopefully some one will correct me.

 

[ronsonol]

kress, thank you for the feedback - exactly what i needed to hear. I will modify things accordingly before I put it in use. Also, your assumption of the heatsink is dead on, as I am indeed using two CPU heatsinks and fans to cool the LEDs.

My build for my display tank is underway using 2 HLG Meanwells, that I have also mounted inside an enclosed wodden box to mount on the wall. The hinges provide a small gap around the perimeter of the door. I have two fans installed...do the same concerns apply as in this budget build?

 

[kcress]

When running parallel strings, do the resistors serve a purpose other than a place to measure the strings amperage? Reading all the dozens of various threads, that is the only purpose that ever seems to be mentioned. Would the resistors also help equalize the individual strings current draw or am I barking up the wrong tree?

Fish nailed the answer. Again, yes, but not enough to matter.

kress, thank you for the feedback - exactly what i needed to hear. I will modify things accordingly before I put it in use. Also, your assumption of the heatsink is dead on, as I am indeed using two CPU heatsinks and fans to cool the LEDs.

Your welcome and whew.. Good to hear the heat sinks are good. Maybe you should try a temp gun on the LEDs.

My build for my display tank is underway using 2 HLG Meanwells, that I have also mounted inside an enclosed wodden box to mount on the wall. The hinges provide a small gap around the perimeter of the door. I have two fans installed...do the same concerns apply as in this budget build?

You always have to be very mindful of electronics mounted in or around wooden boxes. How many wooden, power electronics filled appliances have you seen? :worried: There's a reason for that.:eek2:

If you're going to use wood it's best to line it with metal first. That way a roasting hot spot is going to be half or 1/3 as hot because it will necessarily conduct its heat off to the surrounding areas 'spreading the hot' to perhaps 'very warm'. And of course it's fire proof.

Your clean layout there will have something like 4.2W for the resistors and 2 x 60W x 0.2 = 24W call it 25Watts

You see how much heat is coming off your two boxed resistors that are dissipating 15W? Add 10W more and you will be in the ball park. Not much heat but if it's bottled up - a lot of temperature.

I'd add a fan or two. But! Fans fail. You should make sure that your setup doesn't overheat in the absence of running fans. Make sure you mount the box so convection will still occur. Mount it and run it with the fans off and the cover on. Open it and check after 5 minutes. If it seems fine check again in 30 minutes. If that's fine check in 2 hours and 4 hours. The temp should remain below hot, say about 105F anywhere on the wood. If it does you should be fine. Add your really quiet fan(s) to extend the life of the components.

Remember to use fans to extend lifetimes where needed. But use fans and temperature switches in cases where fans are required to prevent fires. (The temp switch must turn off the equipment being cooled not the fan!)



BTW: Here's the aforementioned Bud Box type I referred to:
http://www.budind.com/view/Small+Metal+Electronics+Enclosures/Miniboxes

 

[ronsonol]

kress will do, and many thanks for this post and all your other posts that have been so helpful in understanding parallel strings.

For the heck of it, because I had to disassemble everything in the enclosure anyways - I removed the 4.9 ohm resistors and replaced with 1.0 ohm 5 watt, and tried a 12.03v power supply with a load I had kicking around...these leds pull 475ma at 11.53v.

 

[philshel]

Originally Posted by philshel
When running parallel strings, do the resistors serve a purpose other than a place to measure the strings amperage? Reading all the dozens of various threads, that is the only purpose that ever seems to be mentioned. Would the resistors also help equalize the individual strings current draw or am I barking up the wrong tree?

Fish nailed the answer. Again, yes, but not enough to matter.

Hmmmm. So if I understand this correctly, if the only way to balance the LEDs is to physically move them from string to string and then they can change over time, the resistors are pretty much useless. I am currently using four Mean Wells for a fixture of 48 LEDs, it is easy but I hate the clutter with all the drivers and cords. Going parallel when possible seems like a better way but only if you can efficiently and permanently balance the strings. Is there a method?

 

[kcress]

philshel; The resistors are NOT useless. They allow you to measure the individual string current in situ any time you want to. This is as compared to getting out a soldering iron or some other scheme to interrupt the strings and insert an ammeter each time. The resistors are a superior solution in my view.

Generally your strings will stay balanced. Occasionally one might not... Do you want an easy way to check once in a great while so you could detect and correct a diverging issue, or do you just want to wait until one string fails then do a postmortem?

A commercial string likely has no way to check this drift. They just run until something fails. Or, they tie every single LED to every other LED and watch the slow individual failure of areas. Look at signal lights. You will see areas of them go dark. They have no way to check for these drifts and correct them. Of course, it's cost prohibitive to do so in that setting anyway, but it's assumed a room in your house and access to your fixture shouldn't be cost prohibitive.

It is better to go parallel. Using half as many ELNs and checking for drift once in a while is a better way. Yes, I agree, it has a little more "clutter".

 

[philshel]

Ok, thanks for the succinct explanation. I was hoping for a more automated way of balancing the strings.

 

[reefmusic]

Thanks. The specs on the datasheet indicate at 12.6 volts they should pull about 900ma...far off the 1.23 I read.

You shouldn't try to read actual current off a spec sheet graph or anything like that. The voltage drop change across the LED for a wide range of currents is small, making it hard to read it that way. Instead, subtract the typical voltage drop of the LED from the loaded voltage output of your constant voltage power supply, and use this difference voltage to calculate the size resistor that will cause the desired current to flow. Because the resistor/LED is a series circuit, all current that flows the the resistor must also flow through the LED.

In short, assume a (near) constant voltage drop on the LED and use this to size the resistor. This will usually be near enough when you actually measure the current.