I need to build a high current 12 to 5 volt regulator.

This will power neopixels, which draw 60mA each, and we have like 800 of them, so around 48A when all of them are on at full brightness and white.

I was tempted to use linear regulators in parallel (and actually witnessed them work) but I have read here there’s an unbalancing issue, where current will not be shared equally between regulators.

Also, I have seen the high current application using a transistor, but I’m afraid it won’t be enough for such a high load.

The main power supply is a 12V 120Ah battery, so I can’t use the regular power supplies that are used for this purpose.

I also need for them to have their short circuit protection. The main neopixel circuit will be made by students who might make mistakes and it would be great to preserve the regulator and not have it blow up.

Is it ok to use linear regulators in parallel in this case? Which IC would be most appropriate? I was thinking a couple 7805 might work but maybe there are some better ones. Do you have another solution for this? Thank you very much!

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    \$\begingroup\$ Using 5v LEDs when you have a 12v supply does not make sense. Buy the 12v version. But no, driving 48A from a linear regulator is not realistic. \$\endgroup\$ Commented Sep 23, 2023 at 22:07
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    \$\begingroup\$ Why use regulars in parallel? Just use one regulator for every 100 devices or whatever works out most economical for the parts you can find. \$\endgroup\$
    – The Photon
    Commented Sep 23, 2023 at 23:36
  • \$\begingroup\$ @user1850479 Yes, i know. But they already have the 5V pixels. I would rather not have them buy another 800 pixels and have a reliable way of powering 5V circuits. \$\endgroup\$
    – Juan010
    Commented Sep 24, 2023 at 14:41
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    \$\begingroup\$ @ThePhoton I wanted to have a single output for simplicity. This is not a professional environment, it is a high school, and I know if I leave a box with multiple 5V outputs, the first thing they will do is wire them in parallel. \$\endgroup\$
    – Juan010
    Commented Sep 24, 2023 at 14:45
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    \$\begingroup\$ Can you put two 5V strings of LEDs in series so you just need to drop 2V, or two stacked 5V LDO regulators (if their current draws are the same, and maybe with optocouplers for data connections if those need a common ground)? Oh, you need something that's robust for inexperienced and/or curious people who might wire them incorrectly. \$\endgroup\$ Commented Sep 24, 2023 at 15:22

3 Answers 3


Using linear regulators would be nuts! 5/12 of your power will go to the LEDs. 7/12 will be wasted as heat and will be a major headache to dissipate (using massive heatsinks). If this was 1970 you might not have had good options (but you wouldn't have had white LEDs either). Parallel connection would require some means of load balancing.

These days there is no good reason not to use an efficient DC-DC switching regulator or buck converter. Any decent one will be short-circuit proof.

There is probably no need to power all the LEDs from the one supply. It will be simpler and safer to split the load onto separate supplies (5 A each maximum, for example), connect all the grounds together at one main "star" point, connect all the 12 V inputs to the battery (with protection) and connect the groups of LEDs to their group regulator. (Check the NeoPixel datasheet to check that they can handle 5 V signal on the data lines if there is no power.)

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    \$\begingroup\$ Another advantage of multiple regulators is that they can be distributed around the array and keep the brightness of all the LEDs approximately the same. If you have a single 48A supply the LEDs will get progressively dimmer the further they are from the supply due to losses in the wiring. If you have multiple regulators then the bulk of the wiring losses will take place up stream. The regulator furthest from the battery might only see 10V at its input, but will still give 5V out. \$\endgroup\$ Commented Sep 24, 2023 at 1:41
  • \$\begingroup\$ Thank you for your answer. I have thought of making groups, but for simplicity, I wanted to leave a single 5V output. When I said I have witnessed one working, I mean I have seen around 10 linear regulators connected in parallel using a steel tube as a dissipator. Keep in mind I intended to build this and leave it where it will be needed year after year, even when I'm not there, and there's no guarantee there will be someone who even knows how to differentiate voltage from current. \$\endgroup\$
    – Juan010
    Commented Sep 24, 2023 at 14:19
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    \$\begingroup\$ I understand my application is pretty uncommon, it's for a float parade locally where every high school takes part in. The org that makes it happen provides each one of us with two 120Ah 12V batteries to power it for a full hour parade, and they charge them before each one of them. The short circuit protection is needed because 13-17 year old students are the ones to solder those LEDs, and where a shortcircuit happen in any of the 5V circuits, it would be ideal for the power supply to shut off. \$\endgroup\$
    – Juan010
    Commented Sep 24, 2023 at 14:24
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    \$\begingroup\$ "... where a shortcircuit happen in any of the 5V circuits, it would be ideal for the power supply to shut off." In your solution the short circuit has to handle (effectively) unlimited current from the battery. This is dangerous and brings you into fire territory. My recommendation is the splitting up of the regulators into circuits that can only provide 3 - 5 A max. \$\endgroup\$
    – Transistor
    Commented Sep 24, 2023 at 15:35
  • \$\begingroup\$ To put some numbers to it: OP needs 240W at 5V (48A). A linear regulator would need 576V at 12V, losing 336W of waste heat in the process. Toasty! \$\endgroup\$
    – Alexander
    Commented Sep 24, 2023 at 21:23

Welcome. As previous answers have explained, reducing a 12V supply down to 5V by dropping 7V across a series element (in this case, the linear regulator) will result in most of the power delivered by the battery being wasted as heat. Using the numbers you provided:
Current required by the load = 800 * 60mA = 48A
Power delivered by battery = 12V * 48A = 576W
Power consumed by the load = 5V * 48A = 240W
Power lost as heat in linear regulator = 576W - 240W = 336W

Not only will this be impractical to build, but it will also drain the battery very fast. You didn't mention the type of battery, so for purposes of this discussion let's assume it is lead-acid, most lead-acid batteries are designed to be discharged at what is called "the 10-hour discharge rate", or "C10 rate", which simply means: "C10 rate" current (amps) = Amp-Hours / 10 hours.
So a 120Ah battery has a C10 rate current (10-hour discharge rate) of:
120Ah / 10 = 12A. As the discharge current exceeds this value, the actual charge the battery will supply reduces; meaning: the battery will deliver significantly less than 120Ah.

A 120Ah battery being discharged at 48A will not last long before needing a recharge, and its lifecycle will be reduced compared to a battery discharged at its C10 rate (12A, in this case).

Here are some suggestions:

Option #1 Single PSU
1x Switch-mode power supply (buck converter buck), rated at:
Vin=12V, Vout=5V, Pout=240W (48A @5V).
This is not a common device, but I did find one:

The problem with having just one big supply is that it is quite capable of destroying (burning up) any of the 800 NeoPixels should they suffer some kind of internal fault.

Option #2 Multiple PSUs
Several PSUs of lower power, eg:
10 x Switch-mode power supply (buck converter buck), each rated at:
Vin=12V, Vout=5V, Pout=24.0W (4.8A @5V).
This is much more common and easier to source, here is just one example:

Compared with option #1, this has a far lower risk of burning up any of the loads. Also, if one of the loads fails, its effect will be limited to just that PSU (eg: that PSU may go into current-limit mode), the remaining PSUs will continue operating as normal.

  • 1
    \$\begingroup\$ Thank you for your answer. I have not provided enough information about this. This is for a kind of local float parade, and we are given 2 fully charged 120Ah batteries before each hour-long parade (not necessarily the same ones each time). You may say that it would be irresponsible to diminish their battery life, but it would be the least of the misuses those batteries face. I do agree multiple PSUs might be the way to go. \$\endgroup\$
    – Juan010
    Commented Sep 24, 2023 at 14:33
  • \$\begingroup\$ @Juan010 There is another way: stick to the one battery, but have several regulated 5V outputs each with a current limit high enough to supply the expected load, but low enough that will ensure that (a) the faulty has at least some chance of not being vaporised, and (b) the power dissipation in any one individual component (including the 5V regulators), is not too high. should any "load" go short-circuit. \$\endgroup\$ Commented Sep 28, 2023 at 3:03

Linear regulators will dissipate an enormous amount of heat. For every 5 W of power delivered to the pixels, 7 watts must be dissipated into the air with a heatsink. Obviously, switching buck regulators would be much more efficient and cost effective.

If you are committed to the idea of linear regulators, do not operate IC regulators in parallel. Instead, have individual regulators for groups of pixels. 10 pixels need 600 mA, well within the range of a 7805, but it still will need a heatsink. The fewer pixels per IC, the smaller the heatsink.

There are other linear regulator IC's rated for 3 A and 5 A, but the heat load problem is the same. No matter where you move it, each pixel causes 0.42 W of heat in its regulator.

  • \$\begingroup\$ Thank you for your answer. I'm not that commited to using linear regulators, but i would prefer to have a single output for convenience. This is a power supply that will be used long after I'm gone, year after year, by high school students. I'm not that devoted to efficiency, more than reliability that the thing will not suddenly blow up or stop working had anything gone wrong in the wiring of the pixels. \$\endgroup\$
    – Juan010
    Commented Sep 24, 2023 at 14:37
  • \$\begingroup\$ This is for a local float parade, and we are provided with two fully charged batteries before each hour long parade. I have seen these linear regulators being used with a square pipe as a dissipator, so no much regard for efficiency. I guess the obvious solution would be using 12V pixels, but this school already has 5V pixels. \$\endgroup\$
    – Juan010
    Commented Sep 24, 2023 at 14:40

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