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So, this is my first question here in this forum and I don't know if I'm doing this right. Then in fisrt place, I'd like to apologize myself if I'm doing something wrong or if my english is a little ununderstandable. I also don't know much about electronics, but even so I want to do this project and because of that I'd be extremely grateful if someone could help me.

What I'm trying to do is a 16x16 (= 256) led panel with 5W leds that need a 700 mA current each. The leds have differents wavelenght and beacuse of that, I want to divide these leds in two differents groups so that I can control the brightness of each group independently with its correspondent dimmer. The main point is, I want that when the dimmers allows the maximal current flows, each led must bright with the full intensity.

So, having said that, I'd like to ask 3 questions:

  • Would I need to use something else than a heatsink, transformer, the leds, and the dimmers to do this? (of course things like wires and plugs are already implicit).

  • Wich transformer suld I use ?

  • How should I set those things to not lose any brightness intensity of each led ?

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  • \$\begingroup\$ I'm not sure, but if its just a simple sum of the individual consumption of each led, then it would be 256 (leds) * 5W = 1280W \$\endgroup\$ – Lucas Sievers Mar 26 at 3:29
  • \$\begingroup\$ (Sorry I posted the wrong quantity of leds, now its right :D) \$\endgroup\$ – Lucas Sievers Mar 26 at 3:43
  • \$\begingroup\$ lol ... that is why i asked about the power consumption \$\endgroup\$ – jsotola Mar 26 at 8:56
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The most common solution to driving large numbers of LEDs at (approximately) equal brightness is to use constant current drivers. A common size driver is 250W. If you used that size, you would set up about 5 series circuits, each consisting of about 50 LEDs. You would then choose a driver with 700mA current and 360V maximum voltage. This is a common voltage/current range from vendors such as Meanwell (e.g. HLG-240H) and many others. Dimming is usually included. Alternatively, if you wanted lower voltage, you could put two parallel strings of 25 LEDs and use 1400mA.

It should go without saying, but take care that your wiring is properly isolated. High voltages keep the current reasonable, but they can be very dangerous if you come into contact with them.

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  • \$\begingroup\$ Thanks for your answer, you were very clear and I think that you solved my problems, but I want to confirm if I really have understand, just to be sure: If I use 2x the current, and then put 2 series of 25 LEDs in parallel, then I can cut the voltage in a half. I can do that because if the two 25 LEDs series are the same then the current will be equally divided and I will have 700 mA for each, its that right ? \$\endgroup\$ – Lucas Sievers Mar 26 at 14:16
  • \$\begingroup\$ I forgot to say that the foward voltage of the LEDs are 6-7v, to know the voltage that I will need, I must just sum the foward voltage of each led that are in serie (25*7V = 175 V), or just use W/I = V ? are those two options the same ? I hope that I'm not abusing of your good will, I swear that's the last question, will I need then to use 5 LEDs drivers (like this one that you did mention), 1 for each 2x 25 series (if I want to lower the voltage in a half) ? \$\endgroup\$ – Lucas Sievers Mar 26 at 14:16
  • \$\begingroup\$ Putting all of the LEDs in series will give the most consistent brightness. If the LEDs are all the same model and from the same batch, putting two parallel strings of 25 each will still be very consistent because variation will average out across the 25 diodes resulting in very close to 700mA per string. Do not put different models of LED in parallel unless you want a fire. You need to pick drivers with both enough power and enough voltage for the strings you're connecting to them. \$\endgroup\$ – user1850479 2 days ago

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