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I am planning on powering a set of RGB LED strips (I am leading towards APA102 https://cdn-shop.adafruit.com/datasheets/APA102.pdf). I know that, at full brightness, each LED requires 60mA. So, at 30LEDs/m, that is 1.8A/m, or (5V x 1.8A =) 9W/m.

I will be using 13m, and so I need a (13*9W=) 117W power supply that can provide 23.4A at 5V.

My question is this. Can I use a 12V 10A (=120W) power supply, and 5 LM1084 voltage regulators in parallel to covert the 12V to 5V? Each LM1084 is rated for 5A output. http://www.ti.com/lit/an/snva558/snva558.pdf

Please excuse my ignorance. Perhaps my reasoning is off, or thee is a glaring safety hazard here. Please feel free to offer any helpful advice at all.

Thanks in advance.

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  • \$\begingroup\$ Where is the link to the APA102? Put it in your question where it belongs - not in the comments. \$\endgroup\$ – Transistor Jul 2 '17 at 19:16
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If I understood you correctly you basically want to put an LM1084 every around 13m/5=2.6m, so each one of them will need to supply a current of 2.6m*1.8A/m = 4.7A (I just took your numbers, I didn't check if they are correct).

The thing with the LDOs, like the LM1084, is that you have to be very careful regarding the maximum power dissipation they can afford due to their low efficiency. The reason is that simply the excess power that is not needed on each of them will be turned to heat. That is equal to \$(Vin-Vout)*Iout = (12V-5V)*4.7A = 32.9W\$! All this power will be turned to heat, which is enormous!

Check the chapter 10.3 ("Thermal considerations") of the datasheet on this very important issue.

Basically you have to find out if this 33W of heat can be dissipated in the package. If not, consider using a heat sink. If even with the heat sink the dissipation is still too high, you'll have to find and use a switching regulator instead, where the efficiencies achieved are usually around 80% to 90%.

UPDATE: Of course another thing is that your input power supply, the 12V/10A will not be able to provide enough current! The LDOs need (almost) so much current at their input as they provide at their output, meaning the 5 LM1084 will need a total current equal to 5*4.7A=23.5A, which is higher than the 10A of your power source.

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  • \$\begingroup\$ That is exactly the kind of thing I was hoping I didn't miss. Are you saying the LM1084 uses the same current on input as output, but the difference in power goes to heat? If that is the case, then my reasoning is flawed, in that the 23.4A needed by the LEDs (5V) actually requires the power supply to provide 23.4A at 12V to the regulators? Is this correct? If that is the case, there is no point to this kind of setup. Oops, noob question... \$\endgroup\$ – Fed Jul 2 '17 at 20:27
  • \$\begingroup\$ Oh yeah, that's also true! I just concentrated on the power dissipation thing. I'll update the answer. Check Equation (4) of the datasheet, page 21. Input current equals output current plus the copmaratively very small quiescent current. \$\endgroup\$ – nickagian Jul 2 '17 at 21:06
  • \$\begingroup\$ Thanks for the clarification. I guess the simplest, and safest, solution is a dedicated 5v 150w power supply. \$\endgroup\$ – Fed Jul 2 '17 at 21:17

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