Linear LED driver - avoiding a hot resistor? - Page 1
Forum section: electronics / projects

author: Artlav
author_rank: Frequent Contributor
timestamp: February 24, 2014, 06:53:39 pm
content: Hello.I want to power an 80W LED, which needs 2A at about 38-40V, from a 43V power supply.Sounds fairly simple - LM350, a resistor, and a couple of capacitors - a linear constant current source:However, there is a little problem here.The resistor gets pretty hot, having to generate 1.2V of feedback voltage for 2A of current.Certainly, this can easily be solved by using a large enough resistor, but a big LED is hot enough as it is, without the extra help.So, is there a cooler way to drive an LED in this scenario?I.e. a 2-3A linear regulator with lower feedback voltage than LM350?

author: Marco
author_rank: Super Contributor
timestamp: February 24, 2014, 07:25:39 pm
content: Quote from: Artlav on February 24, 2014, 06:53:39 pmWhat the resistor doesn't burn the regulator will ...

author: Artlav
author_rank: Frequent Contributor
timestamp: February 24, 2014, 07:36:46 pm
content: Quote from: Marco on February 24, 2014, 07:25:39 pmThere are plenty of regulator heatsinks, there are no resistor heatsinks.Even if it does not improve the efficiency, it should at least make heat management simpler.

author: SeanB
author_rank: Super Contributor
timestamp: February 24, 2014, 08:01:30 pm
content: No room for a 5W resistor? Place it next to the heatsink and use a small amount of thermal epoxy ( or a silpad) and it will run cooler. Alternatively clamp it to the heatsink, you do get clamps for that, or you can make them easily enough with a small length of aluminium strip and either long screws or a bending brace.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 25, 2014, 12:00:05 am
content: Quote from: Artlav on February 24, 2014, 07:36:46 pmNo resistor heatsinks? That's news to mehttp://www.digikey.com/product-detail/en/PWR220T-20-R750F/PWR220T-20-R750F-ND/2192696(At a glance, Digikey doesn't have a 0.68 ohm in stock, unfortunately. Lots of kinds to pick from though.)Otherwise, if you're having trouble getting the heat out at all, your only other choice is a switching regulator.Tim

author: mariush
author_rank: Super Contributor
timestamp: February 25, 2014, 12:31:10 am
content: Or, you know, just get 7w cement covered resistors and put some thermal paste between it and the heatsink and use two screws and some metal bar to keep the resistor pressed against the heatsink.http://www.digikey.com/product-detail/en/2-1623789-4/A103699-ND/2055566http://www.digikey.com/product-detail/en/PSP700KB-0R68/YAG1122-ND/4135291or put two 1.3-1.5 ohm in parallel :http://www.digikey.com/product-detail/en/SQP10AJB-1R3/1.3W-10-ND/18714

author: dannyf
author_rank: Super Contributor
timestamp: February 25, 2014, 12:46:43 am
content: QuoteSure.1) use a smps;2) use a linear regulator with low Vref.3) use a linear led driver with low Vref.4) use some diodes/bjt + mosfet / bjt as regulator....

author: dannyf
author_rank: Super Contributor
timestamp: February 25, 2014, 12:48:19 am
content: A typical bjt's b-e junction, for example, as a voltage drop of 0.7v -> that will cut your dissipation significantly.Having said that, using a linear regulator here is really inefficient.

author: Artlav
author_rank: Frequent Contributor
timestamp: February 25, 2014, 06:56:51 am
content: Quote from: dannyf on February 25, 2014, 12:46:43 amAnd that is the essence of my question - what linear regulator/special driver have a lover Vref?You can't exactly search them on digikey by this parameter.Quote from: dannyf on February 25, 2014, 12:48:19 amHm?44V, 2A input = 88W; 39V, 2A output = 78W.78/88=88% efficiency.It's not exactly simple to get better level of efficiency out of an SMPS, especially in a non-standard range like that.Add to that a 44V->12V converter to power the SMPS itself, and you total with a whole lot of complexity for a very tiny gain.Quote from: SeanB on February 24, 2014, 08:01:30 pmEssentially...The rest of the driver is the size of two 5W resistors, and the place for the resistor is at a somewhat wrong angle to to stick one to the heatsink.So, i'm looking for simpler ways.

author: timelessbeing
author_rank: Frequent Contributor
timestamp: February 25, 2014, 08:08:40 am
content: http://www.ebay.com/itm/151172379115

author: Artlav
author_rank: Frequent Contributor
timestamp: February 25, 2014, 08:34:33 am
content: Quote from: timelessbeing on February 25, 2014, 08:08:40 amHm.Input Voltage: 100V-240V, 50HZ/60HZOutput Voltage: 30V-36VFits neither input, nor output requirements.More to the point, an assembled LED floodlight - driver, case, LED and all - cost much less than it's components, yet i didn't buy one.That's a hint about my purposes, ain't it?

author: dannyf
author_rank: Super Contributor
timestamp: February 25, 2014, 12:03:11 pm
content: QuoteIt was offered right there in the post.QuoteSure. But you can search for Vout, and think about the relationship between Vout and Vref.

author: Artlav
author_rank: Frequent Contributor
timestamp: February 26, 2014, 09:05:17 am
content: Quote from: dannyf on February 25, 2014, 12:03:11 pmOk, i'm officially dumb.However, it's no big loss, since i still can't quite find anything that would fit.There are ones with low dropout, but their feedback voltage is still 1.2V.And there is LP38853, which kind-of, sort-of looks like it might work if the LED won't get hot enough to go too far down.All in all , the path of least resistance appears to include a bigger resistor stuck to the heatsink.Thanks all.Quote from: dannyf on February 25, 2014, 12:03:11 pmDiscrete transistor regulator of some sort?Can't quite see what you mean.

author: muvideo
author_rank: Frequent Contributor
timestamp: February 26, 2014, 03:27:37 pm
content: Quote from: AcHmed99 on February 26, 2014, 11:51:04 amIn your circuit if Q2 is coupled to heatsink, the current will decrease of about 0.3 to 0.4% per °CBut a decent protection need to be more sharp...

author: NiHaoMike
author_rank: Super Contributor
timestamp: February 26, 2014, 03:46:05 pm
content: Use several resistors instead of one. Or use thermal epoxy to attach a heatsink.

author: dannyf
author_rank: Super Contributor
timestamp: February 26, 2014, 04:34:55 pm
content: QuoteDepending on what you are protecting against. Q2 sets the current for the whole thing. The whole set up is immune to Vbe decline in Q1 but does not protect against Vfwd decline in the diode (but that's irrelevant here).So if anything, you should NOT thermally couple Q2 to Q1 - if you do that, the current will actually decline (2mv / 700mv per C) when Q1/Q2 heat up.

author: muvideo
author_rank: Frequent Contributor
timestamp: February 26, 2014, 05:33:02 pm
content: Quote from: AcHmed99 on February 26, 2014, 05:26:27 pmSorry, I should have said "the circuit you posted"as for the right approach for the op problem, I said nothing about

author: muvideo
author_rank: Frequent Contributor
timestamp: February 26, 2014, 06:02:54 pm
content: Quote from: AcHmed99 on February 26, 2014, 05:37:20 pmOk, so now I can say something about:I like the minimalistic approachAnyway, more seriously, a 40V led array will have widely varying forward voltage,the voltage will depend on the particular unit and on the temperature, temperaturealone will probably change forward voltage around 1-2V from room temp to full power.So I agree with you that the current regulation with such low headroom, better besmarter than a pair of transistors.

author: Artlav
author_rank: Frequent Contributor
timestamp: February 26, 2014, 08:20:31 pm
content: The LED is a nice Cree CXA3050 series one, about $30 equivalent each, i got two of them more or less randomly, of different colour temperature.I wanted to see what modern high-power LEDs look like, how hard is it to keep them cool, how to drive them, etc - get familiar with the technology.If it was to survive all that, and be viable, then it would end up as a light in a room (outshining everything already there).There aren't exactly ready-made drivers to fit it's specs around here, i'll have to order one online, and i hate doing that.Therefore, i hook it up to a tweaked-down 45V power supply (which are found in abundance everywhere), with a simple current regulator.The reason for simplicity is - there is pretty much nothing that can go wrong with that LM350 circuit, unless i screw it up (or rather not screw it tightly enough to the radiator).My trust in things i make is proportional to their simplicity - a constant current buck converter is not that hard to make, but i would trust it much less than a simple, stupid linear one.I've already verified that the cooling is sufficient with a huge safety margin, so no need for temperature monitoring.So, that's all there is to it.

author: dannyf
author_rank: Super Contributor
timestamp: February 26, 2014, 10:12:18 pm
content: QuoteSMPS can be easily repurposed to drive those leds, efficiently as well, in a fashion highly similar to what you are doing with a linear regulator.