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First time on here hope I have given sufficient information. I am trying to power 100x 5W high power multi chip LEDs and 8x 3W high power LEDs. I would like to do so off of 120v Ac house plug. The thing is that there is 6 colors and each color group has its own NPN transistor so it can be control with a microprocessor to dim or turn off/on. There is two heat sinks each with 50x 5W LEDs and 4x 3W LEDs.

On each heat sink there is:

24x red1
6x red2
4x red3
10x blue1
5x blue2
5x white

The led specs.

red1 650-660nm 4-6v 700mA 225-270Lm

red2 620-625nm 4-6v 700mA 225-270Lm

red3 730-740nm 3.5-3.8v 700mA 20-30Lm

blue1 450-455nm 6-8v 700mA 90-135Lm

blue2 460-465nm 6-8v 700mA 90-135Lm

white 15,000-20,000k 6-8v 700mA 500-550Lm

I know that I need to wire the individual color groups in a parallel and series configuration to raise or lower the forward voltage and/or current so I can have a practical power supply. The thing is I don't know how to figure the best configuration to gain the most optimal power. It's not that I do not get the math calculations. It's that the complexity of compounding factors is throwing me for a loop. The factors being:

  1. it pulls more power then the wall plug can give so then do i use a step up converter after the transformer and bridge rectifier?

  2. or what power supply should i use?

  3. or will i have to use more then one power supply to get the right power rating for each color group?

  4. or can I group all the color groups so that i can give all the transistors the same power?

  5. all with keeping low cost and energy efficiency and keeping the power supply small as possible.

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    \$\begingroup\$ You need to use the editing tools provided by site. One hint : giving double space adds a new line. \$\endgroup\$ – User323693 Jan 27 '17 at 1:17
  • \$\begingroup\$ I see. format look okay on my phone hope this is better. \$\endgroup\$ – Lucky hex Jan 27 '17 at 2:23
  • \$\begingroup\$ You have many questions and you could edit the last paragraph so that all questions come in a different line.. What is the rating of your power supply adapter you have? Can you share the datasheet if LED modules you have got? \$\endgroup\$ – User323693 Jan 27 '17 at 2:41
  • \$\begingroup\$ The LED load you seem to have is in excess of 500 W, which is quite substantial. The LED's seem to have a very large range of Vf, which is a concern so some part number may help. Do you need to control the brightness (ie PWM or constant current drive)? Or do you just need them to be on when powered? \$\endgroup\$ – Jack Creasey Jan 27 '17 at 3:28
  • \$\begingroup\$ Except for Red3, it seems like they would all be nicely powered from 30~40 Volts. \$\endgroup\$ – Passerby Jan 27 '17 at 5:10
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Stop. Just stop.

You need to learn about driving LEDs. Apparently what you want to do is

schematic

simulate this circuit – Schematic created using CircuitLab

where the number of LEDs will depend on the voltage of V1 and the color of the LEDs in a string. Am I correct?

If so, you are in big trouble. What will happen is that the LEDs will light up to some brightness. Very likely, the current being drawn will be too great, and you will kill one or more LEDs. You must keep in mind that the data sheets mention a range of values for things like forward voltage, and you cannot count on getting a bunch of "typical" units - not unless you're willing to fix things after they go badly wrong. But let's say that you've gotten lucky, and the LEDs are drawing just the right amount of current. What then?

Well, the LEDs will start to heat up. When they do, two things will happen. First, they will get dimmer. Second, they will begin to draw more current at the fixed voltage. This will cause them to get even hotter, which will cause them to draw even more current, which will cause them to get even hotter... and eventually one LED will die. If you are lucky, it will fail open circuit, which will save the others even though the string will be dead. If you are not lucky, it will fail shorted, and the situation will get much worse very quickly, and a second will fail. If that one fails shorted, a third will die very quickly indeed, and the process will continue until all are dead or the power supply decides it's had enough and shuts down. The increase in LED temperature is called "thermal runaway", and the system failure is called "firecracker mode".

So, how do you avoid this? Simple, you don't connect the LEDs directly to a voltage source. The simplest way is to add a series limiting resistor to the LED string, like so

schematic

simulate this circuit

How big a resistor, you ask? At a minimum, I'd recommend a resistor which drops 1/4 of the power supply voltage, although I'm conservative in these things. So, if you were planning to drive a string of LEDs at 2 amps from a 36 volt supply, you could go with 4.5 ohms, which will produce a 9 volt drop. The value is not super-critical, so you could get away with a 5 ohm resistor, which is a common resistance value. Of course, then you'd only have 27 volts available for your LEDs, but that's tough. Also, 2 amps at 9 volts is 18 watts, so you would need at least a 20-watt resistor, and a bigger unit would be better (Did I mention I'm conservative on these things? Did I also mention that I've flown rocket payloads, including a satellite, and my boxes have never failed? There is probably a connection there.).

This is obviously inefficient. The alternative is to use a switch-mode current source, which will turn the LED on and off very quickly, and get a proper average value. It will do so with a relatively small voltage drop, which means you can use more of your voltage for the LEDs. It's pretty apparent that you are not at the stage of being able to design this sort of circuit, so I really don't recommend that you try it just yet. You can buy commercial LED drivers which do operate this way, and if you're willing to take your chances you can get them very cheaply on eBay. Keep in mind that the cheap ones come directly from unknown companies in China, and the old saying, "caveat emptor" ("buyer beware") applies. It is your choice, though.

But let's say you've decided to go ahead with the DIY approach. Now you need to size your heat sink. You have stated that you want to mount a total of 262 watts worth of LED on each heat sink. If we make the generous assumption that the luminous efficiency of the LEDs is 20%, then we can expect to dissipate 80% of the input power as heat, or 210 watts. Additionally, we can expect to lose 1/3 of the LED power to the series limiting resistor, or another 83 watts. Total heat that the heat sink must handle is 290 watts. If you want the LED temperatures to be no more than 30 degrees C above ambient (which will put them up at about 130 degrees Fahrenheit), you will need a heat sink with a thermal resistance of 0.1 degrees/watt, since 0.1 times 300 is 30. Actually, you will almost certainly need a separate heat sink for each LED, since you'll want to space the LEDs out more than a single heat sink will allow. You can get such heat sinks on eBay, and they will run you about 1 dollar each, or maybe a bit more. You will also need to provide heat sink for each limiting resistor. If you do try to go with a single heat sink for each 262 watts of LED power, you'll find that it will run you somewhere near 100 bucks or more, and will be big and heavy, AND you will absolutely need a cooling fan for each heat sink.

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Based on the numbers, a 36V 16 Amp supply (or 3x 5 Amp supplies) would work nicely.

Break up Red1 and Red2 into 6 led series strings. 5.5V * 6 = 33V

Blue1, Blue2, White into 5 led strings. 6.5V * 5 = 32.5V

Red3 could be done in a single string of 8 series leds. 3.8V * 8 = 30.4V
You could use a 1n4007 diode or 2 to bring this up more. 30.4V + 0.7V + 0.7V = 31.8V

Of course this only applies if you don't need to control each 4 led set of Red3 separately.

If you use a constant voltage supply, with resistors, you'd need 19~20 3~5 Watt power resistors of 2~4 ohms. Each string is slightly different voltage after all.

You could control some of these sets from the same transistor. Like Blue 2 is only 1.4 Amps. Many transistors can do this, or a N-Channel Mosfet. But Red 1 is 8 strings, so 5.6 Amps. A Beefier transistor would be needed, or multiple smaller transistors.

You could control multiple transistors from a single GPIO, assuming that the base current needed for the transistors are within your GPIO's capacity. A mosfet may be simpler.

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  • \$\begingroup\$ i am a little lost. The math you did is just for one heat sink so don't i have to double that to run two heat sinks? and don't i need a power supply for each grouping of LED's? then when you said Break up Red1 and Red2 into 6 led series strings did you mean 5 led series strings? i ask because i want to have control of each color independently so i would not want to mix one color in with another. Then my last confusion is way did you not go with the max Vf of the lights does that not take away from there full potential? \$\endgroup\$ – Lucky hex Jan 29 '17 at 3:27
  • \$\begingroup\$ @luckyhex nope. My math was for all of them. The heat sink isn't relevant here. By red1 and red2, I mean you would wire six Red1 leds in series for an ideal forward voltage that works with 36V. Repeat for the rest of the Red1 leds. Separately do the same for red 2. You will end up with 8 strings of red1 (6*8 = 48 leds), and 8 of red2, and can control each string independently or as a group. I do not mean mix red1 and red2. \$\endgroup\$ – Passerby Jan 29 '17 at 4:49
  • \$\begingroup\$ @luckyhex as for the max Vf, I did that to simplify your circuit. Otherwise you will need multiple voltage regulators, a higher source voltage and likely higher heat issues. You may not notice the difference in brightness even then. But of course this is a basic setup and you have to play around with the actual values. You also have to figure that the MOSFET you use will have its on Resistance Drain - source, and will affect this a bit. \$\endgroup\$ – Passerby Jan 29 '17 at 4:51

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