Designing and Building a LED Fixture

[Hoppy]

Originally Posted by kev82

Yeah I think the remaining would stay lit - but they would probably be given the extra current/voltage.

This is pretty much the problem with parallel strings - if a LED dies in either string, the rest of them need to take on the extra electricity. Unless you're using a current mirror.

I'm not an electrician so I'm really bad at explaining this stuff, but basically you want to avoid parallel strings and aim for series if you can. Or go parallel and have a current mirror installed.

A crude schematic of how I have the LEDs hooked up is:


There are two strings of 12 LEDs in series, each with its own driver, with the two strings in parallel. This was necessary to fit the 48 volt power supply I have. A failure in one of the strings will not affect the other string. And, logic tells me if one of a group of things wired in series fails, the others lose their power too, since the same current flows through all of them. If one LED fails so that it has zero voltage drop, the only effect of that is to make the driver dissipate the extra voltage, as heat. But, I don't know if LEDs fail so that they have zero voltage drop. In any case the current flowing through that series string will remain the same, controlled by the driver.

I guess the question is: do LEDs fail so the voltage drop across them goes to zero or near zero, as the current continues through them, with zero or near zero electrical resistance? I will google and see if I can find the answer, but if anyone knows, please tell me.

[malaybiswas]

if a LED fails, both current and voltage drop across it is 0.

did you find out what the "pop" was? Are the solders intact for the 3 LEDs? Remember that the LEDs will heat up pretty fast, so if there is any weak solder, that can pop out.

[Regloh]

I think you may have created a short across those LED's. The fact that they are right in sequence may be just coincidence. This does mean that the remaining LEDs in that line are taking a higher voltage, so be careful.

[Hoppy]

Originally Posted by Regloh

I think you may have created a short across those LED's. The fact that they are right in sequence may be just coincidence. This does mean that the remaining LEDs in that line are taking a higher voltage, so be careful.

It doesn't change the voltage seen by the other LEDs, since they all run at the voltage they choose. All you can change is the current going through them. But, the driver does have to get rid of the excess voltage, so it can cause a failure of the driver.

I think it is a short too, but multimeter readings of resistance don't show any short. Unfortunately, when the light is on, the glare is so intense I can't see to use the multimeter to measure the voltage drop on the three that are not lit. I might be able to somehow attach the probes to the LED's then turn on the light, but my cheap little multimeter doesn't have alligator clips for that.

I read through the Wikipedia article about LEDs and there are a very few failure modes that short out the LED. But, it looks like they would then show a short when checked with an ohmmeter, with the power off, and they don't.

[Hoppy]

Originally Posted by malaybiswas

if a LED fails, both current and voltage drop across it is 0.

did you find out what the "pop" was? Are the solders intact for the 3 LEDs? Remember that the LEDs will heat up pretty fast, so if there is any weak solder, that can pop out.

I can't find any soldering problem with those 3 LEDs, and if I did, wouldn't that result in an open circuit, so none of the LEDs in that series would light up? And, so far I can't find anything that would have caused the "pop", which wasn't at all loud.

I'm still thinking about what the best way to trouble shoot this is - given that all I can get to is the LED side of the heat sink, without disconnecting the power leads to the LEDs.

[Wasserpest]

Is there a simple way to verify that they actually "popped"? Not sure what their normal voltage is, about 2 or 3V? Can you connect a power source (1.5v battery, or 2?) to them and see if they still light or are truly fubar? If they are dead, you'll probably have to redo things, right? Somehow reduce the current?

[691175002]

LEDs can fail both open and closed, frankly failing closed is better since that way the entire string wont die.

Don't worry about the driver, assuming it is a switching type it should not be dissipating much heat to begin with. Most electronics are rated to 70C or more anyways. If you can touch it without getting burns its fine (I wouldn't touch 48V while its on though).

I'm thinking short as well, mainly because the chances of three sequential LEDs all dying at the same time but not taking out any other ones in the string is near zero.

[Regloh]

Actually, are the three dead ones at the end of a string? If that is the case you are looking for one little short circuit around the first one. I think you had mentioned that you grounded the negatives to the heat sink? That could explain it. That short would be cutting the 3 LED's out of the circuit. This may be really good news, as the LED's may still be viable... if I'm right.

[Hoppy]

Originally Posted by Regloh

Actually, are the three dead ones at the end of a string? If that is the case you are looking for one little short circuit around the first one. I think you had mentioned that you grounded the negatives to the heat sink? That could explain it. That short would be cutting the 3 LED's out of the circuit. This may be really good news, as the LED's may still be viable... if I'm right.

The three dead ones are the first 3 at the + end of the string, where the + lead connects. This means those have the highest voltage above ground on them, which argues for a short to ground, because the heat sink is grounded. The negative or, more accurately, the zero voltage end of the string isn't at ground potential, but is above ground, since the current then flows through the switching transistor in the driver, and through about a .75 or 1.5 ohm resistor. It would help a lot if I measured the voltage drop across the 3 dead ones, with the power on. If it is much less than 9 to 12 volts, the LEDs are shorted internally. (I think.) But, I need to solder leads to the multimeter probes and to the LED pads to do that.

[Regloh]

Eh... you just need to grow a third hand

[Hoppy]

Originally Posted by Regloh

Eh... you just need to grow a third hand

Will regular solder work for attaching my third hand

[airborne_r6]

Originally Posted by Hoppy

Will regular solder work for attaching my third hand

As long as it is lead free to avoid long term health problems it should be just fine.

[Hoppy]

I have a new clue! A few minutes ago I went out to see if I could get a quick PAR reading of the intensity, in the garage, in the air. I plugged it in, it flashed on, then off. Now, both series strings are totally off. It looks to me like they were running at too high a current, and, since my 48 volt supply is a 1.45 amp supply, I probably exceeded that and it shut down - or failed. This because both strings failed simultaneously, and there isn't any other spot I can think of where a single failure shuts the whole thing off.

This is beginning to annoy me. (But, just beginning.)

[kev82]

Don't lose faith just yet - Pioneers are bound to run into troubles

[CardBoardBoxProcessor]

so what exactly is needed? a DC converter.. then some resistor? and then just string them all together?