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I have a 12v SLA battery from an old UPS and two 10w led. I'm new to these stuff. LED specs

I connected one LED directly to the battery and in about 15mins it go hot and now one row doesn't work. I didn't know that we had to limit current to leds. So my question is, What is the most efficient way to drive this 10w led using a 12v SLA ? Adding just a resistor will waste about 2watts right ?

As I'm new to electronics, please draw me a circuit schematic :)

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    \$\begingroup\$ The LED may in fact have been drawing the right amount of current, but if it wasn't on a big enough heatsink, it'll cook. 10W is a bit of heat and you'll need a reasonable heatsink (something the size of your hand should be a good starting point). But the best solution is to get a LED driver, you can either build one or get an off the shelf one (there are plenty for sale on a popular auction house) \$\endgroup\$
    – Sam
    May 4, 2016 at 8:16
  • \$\begingroup\$ Well here is a photo of the heatsink that I used. Room temp is about 30c, so maybe that means I need a bigger heatsink ? (12v SLA for size comparison.) imgur.com/a/sUmPW \$\endgroup\$
    – IshanFdo1
    May 4, 2016 at 14:30
  • \$\begingroup\$ That looks pretty good for 10W (you could probably get to 30+ with a fan). The problem with LEDs is that it only takes a tiny bit of extea voltage (above normal) for their current draw (and hence their power) to shoot right up. It would drop your efficiency by 10% or so, but adding a 1+ Watt, 1 ohm resistor may do the trick, even 0.5 ohms might do it as the LED voltage is so close to the battery voltage. A proper LED driver would be ideal, but a single resistor is a pretty cheap & quick mod, just be aware a 1W resistor will get HOT when burning 1W, that's normal. \$\endgroup\$
    – Sam
    May 5, 2016 at 0:07
  • \$\begingroup\$ Meh, it's about how much higher the Battery voltage is compared with the leds, 1 ohm at 1A is one volt, which would take you down from the 12.6V of the battery down to 11.6V at the leds, led current drops off like a brick with only small changes in voltage so that might be all you need. Given how cheap resistors are, maybe try a couple of values and work your way down. If you can get a cheap current meter or multimeter and measure the current that'd give you a pretty solid idea of which resistor works best. Just be aware that without a led driver, they'll get dim quickly as the battery drains. \$\endgroup\$
    – Sam
    May 5, 2016 at 1:28
  • \$\begingroup\$ According to ledcalc.com, If I type in 12.5v as battery voltage and 900ma as led current. it gives 1.2ohm resistor. and 1w will be dissipated by the resistor, I found cheap 3w resistors, So tinyurl.com/hwpz4sy 1.2ohm will do the trick right ? \$\endgroup\$
    – IshanFdo1
    May 5, 2016 at 1:37

2 Answers 2

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A switching converter with current limiting setting is what you need, don't go building one yourself as layout of switched converters is critical.

Luckily cheap modules can be bought that do everything you need, get one like this: http://www.ebay.nl/itm/DC-DC-LM2596-HV-S-60V-3A-Constant-Current-Voltage-CC-CV-Step-Down-Buck-Module-/281858238205?hash=item41a00f66fd:g:7nEAAOSwEgVWSYuw there's an adjustment for voltage and current, allowing you to set it to 12 V, 1 A which is exactly what these LED modules need.

Also: these LED modules need cooling mount them on a heatsink !!!

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  • \$\begingroup\$ If the module gets hot, how much power does the module dissipate ? \$\endgroup\$
    – IshanFdo1
    May 4, 2016 at 7:47
  • \$\begingroup\$ if it's built by a competent engineer, a 10W LED driver shouldn't be dissipating more than a watt or so, at those levels, the driver probably doesn't need a heatsink, but the LED itself surely does \$\endgroup\$
    – Sam
    May 4, 2016 at 8:18
  • \$\begingroup\$ Indeed, I was not clear about what needs cooling, added it now. It's the LED module which needs cooling. The switched converter will dissipate almost nothing (because it is a switching converter). \$\endgroup\$
    – Bimpelrekkie
    May 4, 2016 at 8:36
  • \$\begingroup\$ I doubt that a buck converter will (success)fully drive a 12V LED from a 12-13.5V SLA battery. deltaV is often 2V or more... \$\endgroup\$
    – JimmyB
    May 4, 2016 at 13:53
  • \$\begingroup\$ ah yes, I use a heatsink for the LED but still it gets pretty hot in about 5mins after directly connecting to the 12v sla battery. ( cant hold the heatsink for more than 2seconds hot, room temp is about 30c" \$\endgroup\$
    – IshanFdo1
    May 4, 2016 at 14:07
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The LED could take 1A at a terminal voltage of 9V according to the specification. This means you need to drop 3V at 1A to guarantee it operates successfully from a 12V power supply. That's probably more like a 4 watt resistor. But your LED is throwing away (as heat) about 7 watts so another 3 watts isn't that big a deal.

However, you can use something like this: -

enter image description here

It will take as few as two LEDs in series - it is an active switching current limiting circuit and will have an efficiency of about 90%. The above is just an example - try searching for "boost current limiter for LED string".

Here's another one: -

enter image description here

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