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Despite my Computer Systems Engineering degree, I haven't played with electronics for over a decade, and I'd like to get back into it.

I have a project idea, and I think I've got all the design leg work done. However before I commit money to this project, I would like a second opinion. My assumption is that I'm completely out of touch and no longer know what I am doing, so please take a look.

I believe the digital parts of the project to mostly be a nonissue. My biggest problem is powering 2048 LEDs that require that I deliver 640W over 5V, so a 128A current draw.

I've written up what I intend to do (and how I came upon the Idea) here.

In short, and in terms of power supply: I intend to have 512 LEDs, each drawing max. 60mA in parallel. Thus a total of 32A @ 5V. I will then connect 4 of these these in series. Thus 32A @ 20V achieving my 640W supply for 2048 LEDs. Thus avoiding DC-DC converters and yet more heat.

I plan to deal with connecting the logic parts of the circuit to the floating grounds using some form of Digital Isolator IC.

I need to ensure that each of the floating 5V power rails stay at 5V. Remember each LED has its own PWM and so current draw will fluctuate very rapidly. After some searching I landed on the idea of using a Zener diode to provide the base current to a power transistors . I need to look at possible transistors. I think I will probably pepper more then one across each rail to keep current and heat (in any one area) down.

Some capacitors would probably also help.

Is this a reasonable solution? Or is there a better way? Should I arrange the circuit differently somehow? e.g. dividing through 40V (8 x 256 LED strips) to keep the current down even further? Can I get 40V from a ATX PSU or will I have to use some other supply?

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    \$\begingroup\$ Like the link says: "Given each 1mx1m tile has 2048 LED it is probably also a house heating project." \$\endgroup\$ – Federico Russo Aug 31 '12 at 11:15
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    \$\begingroup\$ -1 for hiding important information in a link. See Dave Tweed's comments to my answer. \$\endgroup\$ – Olin Lathrop Sep 1 '12 at 12:34
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I don't know how you came up with needing 5 V, but it sounds like a bad idea. You have a big efficiency problem, so spending a bit more on power electronics will make things easier and cheaper in the long run.

First, I would not bus around power as low as 5 V because that will require too much current. Having a roughly 48 V bus sounds like a much better idea. Each panel can then locally convert that to the specific voltages needed to run the LEDs and the electronics. That also gets around the problem of the bus voltage drooping from the power supply to the panels and between panels. The buck supplies on each panel can tolerate significant variations in the "48 V" power bus. And, because of the lower current there will be less variation in the first place.

Look at what voltages the LEDs need. Red and green will be near 2 V, but blue over 3. Red and green are probably close enough so that you can use one power voltage for both of them. Green has the higher voltage, usually about 2.1 V, so make a little more than that. You want it only high enough so that you can put enough of a resistor there to have the current be reasonably predictable despite variation in the LEDs. Maybe 2.5 V is a reasonable tradeoff. Red LEDs usually drop a bit under 2 V, so the regulation for red will be slightly better. Either way, this is still way better than 5 V. For the same LED brightness, just switching to 2.5 V instead of 5 V will save half the power.

Blue usually requires significantly more voltage, like over 3 V. Make a separate supply for blue. It should be a few 100 mV above the LED voltage, just like for the red and green LEDs.

48 V is a common voltage for off the shelf power supplies, and is the limit you are usually allowed before you get into legal regulations. There are various buck converter chips out there, or if you're clever you can maybe have a existing micro handle the buck conversions. Either way, these are readily available blocks you can use in your circuit.

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    \$\begingroup\$ you've got the wrong end of the stick. He's not building this from discrete LEDs, he's using an OEM "strip" of addressable LEDs that has integrated PWM drivers. That's where the 5V power requirement comes from. \$\endgroup\$ – Dave Tweed Aug 31 '12 at 13:26
  • \$\begingroup\$ @DaveTweed: I don't see that. He even talks about using 40 V possibly, and seems to have flexibility as to how many LEDs per strip and how the strips would be arranged. If what you are saying is true, then he needs to be a lot more clear. \$\endgroup\$ – Olin Lathrop Aug 31 '12 at 20:01
  • \$\begingroup\$ if you read the page at the link the OP provided, you'll see that he's talking about various series/parallel combinations of entire strips. There are 32 LEDs + controllers on each strip. \$\endgroup\$ – Dave Tweed Aug 31 '12 at 22:37
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To answer your question, putting the strips in series, with shunt regulators across each group is a really bad idea. It basically means that you'll always be dissipating the full 640W, even when all of the LEDs are dark! Whatever power doesn't go to the LEDs gets wasted in the shunt regulators. This is a lot worse than the ~10% loss you'll get with switching regulators.

Since you're already spending $1600 for the LED strips for one panel, it doesn't make sense to try to save a few bucks by repurposing a surplus computer power supply. Invest in a proper primary power supply (48VDC @ 15A would be a good choice) and use DC-DC converters locally ("point-of-load") to power smaller groups of LED strips. This way, they can all share a common ground and you don't need to worry about isolating the digital signals.

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If they're common color LEDs, like red, they'll drop a lot less than 5 V. 2 V is a typical value for a red LED. So you don't place 4 of them in series for a 20 V supply, but more like 10 (just estimating now, we'll adjust the figures later). Then 1 such a string will consume 20 V \$\times\$ 60 mA = 1.2 W. You would need 2048 / 10 = 205 of those, so that's a total of 205 \$\times\$ 1.2 W = 250 W, that's 58 % less than the 600 W you estimated.

So the 250 W is our first estimation, but we'll have to make some corrections: if we have ten 2 V LEDs in series on a 20 V supply we don't have anything left for a current-controlling resistor. We remove one LED in each string, so that we have 2 V left for the resistor. To get 60 mA at 2 V we need a 33 Ω resistor. The resistor will only dissipate 120 mW, so a standard 1/4 W type will do. We'll have to add a couple of strings: 2048 / 9 = 228 strings instead of 205. So our total power will rise to 228 \$\times\$ 1.2 W = 275 W, so that's still a more than 50 % saving.

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    \$\begingroup\$ Don't forget each LED is individually addressable. \$\endgroup\$ – Codism Aug 31 '12 at 17:07
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Floating power rails will drive you mad as you have to get the control signals to that level as well. Your best bet I think is to find a source of cheap mains-5V supplies at e.g. 10A or 20A (Dealextreme show these for about $20) and connect them to suitable copper busbars to drive a 20A group. There's no reason why the power supply division should be the same as the data signal divison. Just connect all the grounds together (but not the 5V rails).

I see you've also already noticed that an Atmega is under-resourced for controlling something this size. Try a Pi instead.

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I don't believe that pre-wired addressable RGB strips can be run in series as you're thinking, since your data lines will still be using low voltage signaling. If such an arrangement is not specifically allowed by the datasheet of the LED strip I would expect it would not work. However, you can avoid the need for putting LED strings in series by simply using proper high-current 5V power supplies.

For example, two 350W 5V supplies, for example as sold here for $33 each, would work well (note I am not affiliated with this company and have no experience with their products, but it looks right for your application): https://www.ledsupply.com/power-supplies/mean-well-lrs-enclosed?gclid=CjwKCAjw8uLcBRACEiwAaL6MSbrrKr8ewAflfgFXISZHQ__uRMUrJhc8oraVp3Nt5HXx_7eDfX6J5xoCoywQAvD_BwE

Do note that you'll still want to divide your LED strips into shorter sections and run individual power wiring to each, since the wiring within the strip can't support enough current for the number you're planning. The datasheet for your LED strips should contain the maximum number you're able to run before cutting the strip and splicing in new power wiring.

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