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I have a common 12V led strip (SMD5630) connected to my car battery directly. I have noticed that it gets really hot after a couple of minutes so I'd like to reduce the current it takes (now it's 1A) to get less heat, longer lifespan and less power consumption (of course in exchange of some brightness). How to do it properly?

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  • \$\begingroup\$ Start with the data sheet and see what it says about controlling the current/reducing the supply voltage. \$\endgroup\$ – Andy aka Jun 10 '18 at 12:06
  • \$\begingroup\$ Put a few diodes in series. Every diode will lower the voltage some 0.7 - 0.9 V (depending on the type). "12V car battery" usually means 14.4 V and even more sometimes. I'd use 3-4 1n4007 for starters. (I assume the strip is powered by DC only, no PWM or any other sorcery) \$\endgroup\$ – Sredni Vashtar Jun 10 '18 at 12:11
  • \$\begingroup\$ Define where and how hot? Wire or LED? burning finger? Can you attach to heat conducting substrate ( metal?) \$\endgroup\$ – Sunnyskyguy EE75 Jun 10 '18 at 13:11
  • \$\begingroup\$ @Andyaka Unfortunately, there is no data sheet, it's a chinesse low cost led strip \$\endgroup\$ – Ivan Jun 10 '18 at 13:14
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    \$\begingroup\$ Many now use cheaper film substrates and not polycarbonate (Kapton) and all plastics have excellent thermal insulation. 60'C is normal for these designs. Any metal substrate will easily spread heat but limited by Rca resistance of plastic film susbtrates. \$\endgroup\$ – Sunnyskyguy EE75 Jun 10 '18 at 13:50
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The other solution that are mentioned do work but in my opinion the optimum solution to get the current/brightness that you want is to use a regulated 12 V (or less if you want) no matter the car battery voltage is.

You can get such a proper regulated 12 V by using a boost/buck converter module. They're not expensive and here's an example.

enter image description here

Note the screw on the blue potmeter, you can use that to get 12 V (or whatever voltage you need) at the output.

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  • \$\begingroup\$ Sorry for my lack of knowledge, but this will control the voltage, but what about the intensity? Does it will be the same? \$\endgroup\$ – Ivan Jun 10 '18 at 13:20
  • \$\begingroup\$ The voltage defines the current that flows through a resistor-diode line. The current defines the intensity. Basics! \$\endgroup\$ – Marcus Müller Jun 10 '18 at 13:32
  • \$\begingroup\$ Ok, thanks. I've reading about this and I've found another module XL4015 that it's just a step-down (and easier to get from my country). I think that booth will work well, really? (ebay.es/itm/…) \$\endgroup\$ – Ivan Jun 10 '18 at 13:52
  • \$\begingroup\$ No, a step-down only module will not work properly as it needs to have a higher input voltage, like 14 V to get 12 V out. Then if your battery is 12 V (which it often is) then the LEDs would get maybe 10 V and might be quite dim. You can buy the module I suggest in your country as well, I just searched for "XL6009" and voila: ebay.es/itm/… \$\endgroup\$ – Bimpelrekkie Jun 10 '18 at 18:41
  • \$\begingroup\$ Well, it cames from China, so I will have to wait maybe one month :) but it's ok ;) Thanks! \$\endgroup\$ – Ivan Jun 10 '18 at 21:25
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Did you measure the current of 1A? I would suggest to go for 1N54xx series diode to be on safer side.. It should work if the LED array is just passive..network of LEDs and resistors. Also, by default these diodes can withstand large pulses too expected from car battery.

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