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I got some high lumen LEDs. When I power them singularly they are indeed very bright.

I connected 52 of them in parallel in parallel and plugged to the same power source and the brightness dropped significantly.

The LEDs are 3.2v,1W. The battery pack (a controller for christmas lights) has 3 AA batteries. How can I fix the issue? Shall I add more batteries in parallel?

These are the specs given on the shop:

  • Matériel: Base d'alliage d'aluminium
  • Type: Perles de LED, SMD
  • Couleur de la lumière: blanc chaud, blanc (facultatif)
  • Longueur d'onde: 380 ~ 840nm
  • Puissance: 3W
  • Température de couleur: 3000 ~ 3200K, 6000 ~ 6500K
  • Tension: 3.2 ~ 3.4 V
  • Courant: 350mA
  • Luminosité: 100 ~ 110lm

enter image description here

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    \$\begingroup\$ Yes. A single AA battery can only supply a few watts. You need a more powerful supply. One way to do this is use more powerful batteries, or use more of them in parallel. \$\endgroup\$ – polwel May 1 at 18:03
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    \$\begingroup\$ @algiogia: How are you regulating the current to the LEDs? Connecting them directly to a battery pack is a very good way to damage them. \$\endgroup\$ – JRE May 1 at 18:05
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    \$\begingroup\$ Please link to the exact LED type datasheet. They rarely are specified to 5V but your LED might have a driver. Also 52 pieces of 1W LEDs need 52 watts of power, or more than 10 amps of current. And 3 AA batteries cannot provide 5V, and cannot provide 10A. \$\endgroup\$ – Justme May 1 at 18:07
  • \$\begingroup\$ @justme I don't have the datasheet, just the description given in the online shop \$\endgroup\$ – algiogia May 1 at 19:16
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    \$\begingroup\$ @algogia: What exactly are you using for the supply? Is it just the three batteries in series, or is there more to it? The proper way to drive a power LED is to regulate the current, not the voltage. In the operating region that produces useful light levels, the voltage is a weak function of current, or conversely, the current is a very strong function of voltage. So to control power, you must control current. Batteries are not good for doing that. With 52, you probably care about what you're doing, so you'll want a regulator (even if it's only a resistance). \$\endgroup\$ – FrontRanger May 1 at 19:41
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You need to completely re-think your power source. If each LED takes 350mA, and you have 52 of them, then that's 18.2A to power all of them. There's no way you're going to get that from 3 AA cells.

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First, those specs are not correct and probably copy-paste from a different product, so you will have to do some testing to figure out what you bought. I would probably treat the current they give as the absolute max and run them lower, maybe a lot lower if you want them to last. Use one resistor per diode to set the current.

Second, you'll want to get batteries with an actual datasheet, and ideally ones rated for relatively fast discharge. My suggestion would be to use lithium ion batteries in parallel.

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  • \$\begingroup\$ Thanks. I found that with 3v they don't even light up, that's why I put 3 AA in series. Can I use a single resistor for the set? The space is limited \$\endgroup\$ – algiogia May 1 at 19:54
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    \$\begingroup\$ @algiogia using a single resistor isn’t recommended since it’s likely that one LED will ‘hog’ the current; as the LED heats up the forward voltage drop reduces and so the imbalance tends to get worse and worse. It’s preferable to use one resistor per LED, although commercial designs often use a resistor for each two or three LEDs. Also be aware that the LEDs can run at 3W only if mounted on a heat sink or aluminium-cored PCB. Otherwise you’ll need to run them at reduced power (so they aren’t too hot to touch). \$\endgroup\$ – Frog May 1 at 20:52
  • \$\begingroup\$ 350ma is the spec that really matters, but that's a very common spec so I believe it. If the others are off, not a big deal, the specs I see are in the "typical" range. I think the battery is the least of the problems here. OP has other more crucial lessons to learn... and once those are learnt, the battery issue will be easy to solve. \$\endgroup\$ – Harper - Reinstate Monica May 1 at 22:06
  • \$\begingroup\$ @algiogia If you put them in parallel each must have their own resistor. \$\endgroup\$ – user1850479 May 1 at 22:44

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