DIY LEDs - The write-up

[HeneryH]

Do the math on the voltage drop of each LED and figure out what the remainder is that needs to be dropped by the resistor. 24v - (3.5v * x LEDs) = (Vdrop of Res).

Give the current and the V of the resistor you can calculate the R (V=I*R).

The example uses the ELN-60-24 but you could do the same math with the ELN-60-48.

 

[der_wille_zur_macht]

But if you're using the resistor to control current, why not ditch the ELN alltogether and just use a DC power supply?

 

[Skeptic_07]

yeah but doesn't the driver automatically do that? This past weekend I had a 48D driving 12 XP-G's which has vf around 3.2. I tested the voltage of the string and was around 37 driver does not output the full 48V unless the string needs it. Am I missing something? ohhhhhh... If i have a different vf on one string.. the other strings will get more voltage. is that the reason for the resistor?

 

[der_wille_zur_macht]

If i have a different vf on one string.. the other strings will get more voltage. is that the reason for the resistor?

That's my impression. Seems like a lossy way to account for the difference. You're taking an efficiency hit for the driver to regulate current to the whole array, and then you're taking an efficiency hit on each string for the resistor to burn off power that string doesn't need. IMHO you'd be better off "binning" your strings to get matching voltage drops.

 

[HeneryH]

You guys have way more practical experience than me on this. I just thought is was a straight forward calculation of what the R should be.

 

[der_wille_zur_macht]

Henery, I don't think you were wrong, I just don't agree with that load-balancing approach because it potentially reduces efficiency. And if you haven't picked up on it yet, I'm a freak about efficiency. :lol:

 

[HeneryH]

This is why I posted the drawing; bring out the conversation of pros/cons. If equal, the -24 may be more available than the -48s which seem to have been all bought up (by the reefers?).

 

[der_wille_zur_macht]

The other "con" of course is that max drive current is limited. Many people seem addicted to 700mA or 1000mA current, because it reduces the upfront cost (number of LEDs you have to buy). The 60-24 maxes at 2.5A, so you could do three parallel strings of 6 to be in the current range most people want. So you're doing 18 LEDs per driver.

That diagram showing 8 parallel strings looks really attractive, but that's at 312mA, which is way below what most people want. You'd be more efficient that way, but your upfront cost would be way higher, so the payoff would get pushed out a bit.

 

[Skeptic_07]

That's my impression. Seems like a lossy way to account for the difference. IMHO you'd be better off "binning" your strings to get matching voltage drops.

I guess thats the way I'll be doing it. I'm sure Kcress would argue that that is a bad way to do it because of variations in something or other. I remember once him saying that vf's can vary widely from chip to chip etc.. but really what are the chances? Even if one LED had a little extra vf, say 3.7 instead of 3.5 or something, and the other two strings ended up getting an extra .2V. how damaging would that be to the other strings?

 

[TheFishMan65]

Fuses will only protect the first set of LEDs unless you fuse each LED.

Your voltage is also less of an issue. Take 2.5 amps and the first column (and assume only 3 rows). All three will have the same Vf say 3.5 now the first LEDs may get different currents (but I think they will be close) say .9, 8.33, and 7.66 still 2.5 total. Now they all get tied together and it starts over. The next group may balance out at 3.6 volts and the currents might be 1.2, 1.3 and 0 (it is defective and has no current flow). Good things your are using XPG they can go to 1.5, but you better not loose another one in that column.

So each set of three LED will adjust the current so that they have the same voltage drop. Even if the all the LEDs on a row have a low voltage (for a fixed current) and another row all has high the current will be adjusted between them.

[EDIT]
Don't know why the have a resistor either.

 

[Skeptic_07]

also, wouldn't the resistors just cause the driver to output more voltage?

 

[der_wille_zur_macht]

Skeptic, it's an easy problem to solve. Set up a "test station" with a constant current driver that can power a few LEDs at a time at some reasonable current, while allowing you to probe each individual LED with a multimeter. Turn the test array on, test the voltage drop across each LED, and write them all down. Then, arrange your LEDs into groups such that the total voltage drop for all groups is as close as possible. That's what I meant about "binning" your own LEDs. It should take an hour or two max, and it's cheap insurance if you're running parallel strings.

.2v CAN be quite significant (like 100mA!!!), but I'm not sure you'd see variation that high unless you randomly stacked things up in the worst possible way.

 

[TheFishMan65]

I agree skeptic I think a resitor would just increase the voltage. And yes dwzm binning will reduce some variance. The ones with the same drop should go in the same column. But we have still not solved the problems if they start to go out. Fusing each one would just cause the whole thing to go out. You would end up with all the fuses associated with one column going except the one with the BAD LED. Identifying it is easy that way.

 

[Skeptic_07]

DWZM: Now i see what you mean by "binning". Sounds like a fun project for this evening. I already have a buckpuck wired to a wall wart from an old experiment.

Fishman: I think i understand what you are saying but my wallet tells me that i cannot afford to buy a fuse holder for each LED. fuses are no problem at like .30 a pop but the holders are a couple bucks each. I'll just stick with one fuse per string.

 

[TheFishMan65]

I think we get into trouble by calling them strings since there is no string it is really a matrix which I why I have been trying to refer to columns and rows.

I would not bother with fuses. You are right it is too expensive. But since each column get tied back together you keep getting back to 2.5 amps for the next column. So you can not really protect anything but the first column.

You might consider only running two rows at half the maximum current each. If you loose an LED in a row the other will get the full current, but it is still within range. You get some benefit of buying fewer driver but no risk of losing an LED if the heat sink can handle the LED at full current. Three rows at a third of the maximum current would also work.

 

[Skeptic_07]

now i'm really confused

 

[laverda]

Henry
I am also planning to go a smiler route. I am not sure the connections between strings is really needed. I will have to reread my notes on that. If I remember corectly it is supposed to help protect the strings, but it seams to me it makes the fuses useless. If a fuse blows, it looks to me like it is only going to take one LED out of the circut when it is wired this way.
My guess the resistors are to adjust for any variation in Vf in each string. The resistors might not be needed in some strings and would vary in resistance for the others to get the same total Vf.
I will be measuring the Vf on all my LEDs and try to match the strings as closely as possible.

 

[TheFishMan65]

I don't mean to confuse people. Tell me where and I will try and explain.

 

[TheFishMan65]

laverda,

You are correct about fuses in this approach. Resistors would work in a string formation as you say, but not in a matrix formation. If you measure Vf (at the current you plan to run) then ideally each string should add up to the same total voltage.

Word about resistors: watch how much you are dropping. You may need power resistors. If you ran at 1 amp and had a 1 volt difference in strings. That is a watt of power. No dinky resistor for you.

 

[Skeptic_07]

This is how i understand it: Running in parallel means that I would connect negative to positive to a few strings (just like in series) and then wire them all to one driver. Your comment about a matrix is really what confuses me. When you are running parallel strings off of one of these meanwells, and one LED goes out, the entire string goes out. All of the load from that string is instantly transferred to the other strings, probably blowing all the LEDs on them in just a few seconds. So if i'm running 3 strings off 2.5A each one will draw .833A. If one blows, the other two will get 1.25A, probably only for a short time, but still enough to damage the LED's, i think. if there is a 1A fuse on each string, they'll pop and LEDs will be saved.