unknown white LEDs, power question

Using Arduino LEDs and Multiplexing
#1

I bought a light bulb, with 6 of these strips and it was powered from a simple screw in base. supposed to replace a simple light bulb.

it died.. partially. I put it on the range hood and the switch has a high/low, I think it is half wave rectified. not sure.

so, I took apart the lamp and salvaged the LED strips.

when I power it with 12 volts, just the slightest of dots appears in all the LEDs. when I power 4 of the LEDs, they are very bright.

I have never played with these, but I would like to make something out of it.

any suggestions on how to feed this ? |500x372

#2

Put a 100-500 ohm resister in series, and power the strip with 12v. Then test the power draw (amps), and make adjustments.

I can't see it (maybe on the back of the unit), but I suspect they have resisters to drop the amps already.

#3

[quote author=jack wp date=1442882438 link=msg=2406419] I can't see it (maybe on the back of the unit), but I suspect they have resisters to drop the amps already. [/quote] Most unlikely. These are luminaires, not indicators; as so often pointed out, the idea is not to waste power.

I gather most of these lamps use all the LEDs in series (24 LEDs - about 72 volts in this case) fed through a bridge rectifier with a capacitor in series with the input. Also a series resistor but it will be on the AC side of the bridge and has only a protective and inrush-limiting function.

Using a capacitor (vastly cheaper than a switchmode dropper) means that the working voltage will be specified as a narrow range, not a "universal" 100-240 V.

#4

Take a look at the NUD4011. It's a current limiting, LED driver for mains supplies. There would be no required resistors in the LED string.

#5

Chagrin:
Take a look at the NUD4011. It’s a current limiting, LED driver for mains supplies. There would be no required resistors in the LED string.

It is in fact, nothing more than a constant-current “resistor” requiring a heatsink.

And of course, US supply mains only.

#6

Paul__B: It is in fact, nothing more than a constant-current "resistor" requiring a heatsink.

And of course, US supply mains only.

Did you mean to say constant-current "regulator" . It seems to be a semiconductor, not a resistor.

#7

more pics|478x500

#8

So there you are - series capacitor provides reactive current limiting, feeds bridge rectifier, 51 ohm resistor limits turn-on surge (but not continuing current) to 2 Amps which is in fact, not applied to the LEDs because the smoothing electrolytic absorbs the surge.

#9

5 of 2835 size LEDs like those wired in series need larger voltage to turn them on. They have a Vf of ~ 3.2V, similar to these http://classic.superbrightleds.com/moreinfo/surface-mount-smd/cool-white-2835-smd-led/1843/4456/# 5 x 3.2 = 16V. 5 lit up will be really bright. I would use say 18V and 33 ohm current limit resistor for 60mA. Control them with PWM and a N-channel MOSFET, AOI-514 would be good, 50 cents at digikey.com http://www.digikey.com/product-search/en?vendor=0&keywords=aoi514 for brightness control. Don't run at high brightness until you get a feel for how warm they get.

#10

Paul__B: So there you are - series capacitor provides reactive current limiting, feeds bridge rectifier, 51 ohm resistor limits turn-on surge (but not continuing current) to 2 Amps which is in fact, not applied to the LEDs because the smoothing electrolytic absorbs the surge.

the original lamp is starting to make sense. it was 6 of these 5 LED strips and a plate of 6 more. fed from 120VAC. I put it on the range hood that has a high/low light switch. I believe that all the High/Low does is to use a half wave rectifier. as that is the easiest way to reduce an AC light output level. albeit fixed reduction in output. fed with half the AC wave, the circuit burnt up. the circuit failed to the point that when fed 120VAC, the LED's would be so dim as not only have a faint yellow dot of a glow, if you looked hard enough.

#11

CrossRoads: 5 of 2835 size LEDs like those wired in series need larger voltage to turn them on. They have a Vf of ~ 3.2V, similar to these http://classic.superbrightleds.com/moreinfo/surface-mount-smd/cool-white-2835-smd-led/1843/4456/# 5 x 3.2 = 16V. 5 lit up will be really bright. I would use say 18V and 33 ohm current limit resistor for 60mA. Control them with PWM and a N-channel MOSFET, AOI-514 would be good, 50 cents at digikey.com http://www.digikey.com/product-search/en?vendor=0&keywords=aoi514 for brightness control. Don't run at high brightness until you get a feel for how warm they get.

thanks, this was exactly what I needed. since 4 of the lights work with 12v, I can remove one from the board, add an FET and power and get a useable light.

Paul__B: So there you are - series capacitor provides reactive current limiting, feeds bridge rectifier, 51 ohm resistor limits turn-on surge (but not continuing current) to 2 Amps which is in fact, not applied to the LEDs because the smoothing electrolytic absorbs the surge.

just learning about how to post a photo into the forum and not just a link. but, here are two links http://i.imgur.com/FURsRoh.jpg and http://i.imgur.com/cjzpb94.jpg just shows the cap values and the other resistor. since I was not trying to repair the light, the old driving circuit was not particularly important. this was an e-bay special light, non-dimming, got 42 LED's for a couple bux.

#12

dave-in-nj: just learning about how to post a photo into the forum and not just a link. but, here are two links |500x467 and |500x395

Put in your .jpg link, highlight it and use the little "screen" icon to mark it as a photo. I then highlight the whole and centre it.

If you put a diode in series with this, you have a diode in series with a capacitor. The capacitor charges up, then nothing else happens.

#13