0
\$\begingroup\$

Prerequisite: For a home-decor project, we wish to work with different LED strips and even a larger LED chip as a central lighting source by attaching everything to a powerful, high quality 12v CC/CV DC PSU. (eg., Mean Well HLG series).

  • Each portion of LED strips, depending on chips / resistors, will draw different currents.

  • For the central lighting, we wish to use a CC / CV DC booster to drive a high-power LED.

Given that the total power consumption from all components is well below 90% of PSU capacity;

  • Would it be safe to connect all components in parallel and use the single PSU to power them?

Edit:
This question is not a duplicate on this site. It is irrelevant to, for example, the following:
Can I safely and efficiently connect batteries parallel to a DC source? or:
Choosing power supply, how to get the voltage and current ratings?

\$\endgroup\$
15
  • 1
    \$\begingroup\$ shouldn't be a problem, just add appropriate fuses with maximum allowed currents. also make sure you're cables are rated for the currents you're drawing. \$\endgroup\$
    – Flying Swissman
    Commented Dec 1, 2014 at 18:42
  • 1
    \$\begingroup\$ Can you give more information about the LED strips and the high-power LED chip? If they all work at 12V DC input and they have integrated constant-current sources, you should be good. If you are directly driving an LED or string of LEDs without resistors, which requires a constant current, you may be in trouble (especially with multiple different constant-current components on the bus). \$\endgroup\$
    – Justin
    Commented Dec 1, 2014 at 18:50
  • 1
    \$\begingroup\$ @Justin he said he's limiting the current with resistors \$\endgroup\$
    – Flying Swissman
    Commented Dec 1, 2014 at 18:53
  • 1
    \$\begingroup\$ @FlyingSwissman - It should be fine then. It's not clear in the question whether the high-power LED is current limited (well, I'm not really sure where the "CC / CV DC booster" fits in...) \$\endgroup\$
    – Justin
    Commented Dec 1, 2014 at 18:58
  • 1
    \$\begingroup\$ @Justin oh sorry missed that. \$\endgroup\$
    – Flying Swissman
    Commented Dec 1, 2014 at 19:05

1 Answer 1

Reset to default
5
\$\begingroup\$

Yes. The power supply will make 12 V as long as you don't exceed its current rating, which you say you know you won't exceed. Otherwise, 12 V is 12 V, regardless of how much current someone else is drawing from it.

A related issue you should think about carefully is voltage drop in the wires. The power supply will do a good job of making 12 V at it's output, but that's not the same as 12 V at the other end of a 18 guage wire across the room, especially when something else is also trying to draw power from the remote end of the wire.

Think of the wires as series resistors and plan out your system accordingly. The resistance of various sizes of copper wires is well documented out there. Remember that you have two wires, + and -, so the total series resistance seen at the end of a cable will be twice that of one wire.

You can trade off losses in the wire, voltage drop at the end of the wire, number of units fed from a single cable, and thickness of the conductors. There is no single magic answers. It is up to you to pick the best tradeoff, where "best" is measured by a weighting of factors that only you can decide.

\$\endgroup\$
2
  • \$\begingroup\$ Whoever downvoted this, please explain what you think is wrong, misleading, or badly written. I have re-read this and still think it is correct. Silent downvotes do the site a disservice since we have no way to know what to fix, or whether the misconception is on your end. \$\endgroup\$
    – Olin Lathrop
    Commented Dec 2, 2014 at 14:54
  • \$\begingroup\$ +1 +4 years on to make up for stupid downvote. And +1 anyway :-) \$\endgroup\$
    – Russell McMahon ♦
    Commented May 20, 2019 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.