Theory question. . Series a 1W & 5W luxeon

This is a theoretical question. If [for some reason] someone wanted to series a 1 watt luxeon with a 5 watt luxeon and hook them up to a 12 or 15 volt power source, would you want the 1 watt luxeon to be at the positive lead or at the negative lead, . . . .or would it matter WHAT the position was?

+ wire_+ lead of 1W_- lead of 1W_+ lead of 5W_- lead of 5W_-wire

+ wire_+ lead of 5W_- lead of 5W_+ lead of 1W_- lead of 1W_-wire

It would not matter either way. But either way the 1W is going to get pretty toasty and the 5W not so much. This is because the same amount of current will flow through the 1W as the 5W, wheras the 1W should get half as much current. (I don't know what that current would turn out to be).

A good way to fix that would be to add a second 1W (if you have one) in parallel with the first, and then have that parallel pair in series with the 5W.

That would make very good use of all three LED's and a 12V supply. If it's 15V, probably you'll need to have some resistance as well.

Good luck.

Interesting. I guess that would be true about the 1W getting toasty, since it would be sharing 12 volts with the 5W, thus about 6 volts. Perhaps 2 paralleled 1W's would be a better idea.

This is a truly spooky direct drive idea, at least to me. In addition to the current differences cited, I can see the whole lot going chain reaction style. If the first to go 'failed to short' the other parts, already stressed, would follow in half a heartbeat.

A second 1W (or an appropriate resistor, say 10 ohms) seems a minimum, as I believe some sort of current control. A series resistor at the minimum, something a bit 'smarter' still better.

I vote a minimum total resistance of say four ohms for 12.0 Volts, twice that for 15. Remember, this is for twice the 'normal' current draw.

Or so I see it.

Cheers.

Doug Owen

Hm.... I don't see the problem, though it's not great. Yes, the same amount of current will go thru the 1W as the 5W - the amount of current that would ordinarily go thru a 1W. The 5W would be underdriven, most likely.

Assuming a 1W with Vf of 3.5V and a 5W with Vf of 7.0V you're talking about 10.5V nominal driven by a 12.0V supply. Worst case for the 1W is that it is overdriven by 12-10.5 = 1.5V which is pretty tame compared to what some folks are doing. As far as calculating current goes, this would be the equivalent of running the 3.5V 1W by itself from a 5V supply, the 5W LS doesn't even need to be considered. No matter what, it will be limited to whatever the 1W can stand, which is certainly going to be less than what the 5W can conduct.

Understand that a Vf of 3.5 Volts is not the value you need to use, since this includes the voltage drop across the internal one ohm, .35 volts at 350 mA. The true Vf of the part (as measured at the junction) is therefore 3.15 in your example. Adding the 1.5 volts above to this .35 makes a total of 1.85 Volts across that one ohm, 1.85 amps, over *five times* the "absolute maximum" spec from the maker. IMO you've a prefect right to expect problems. This is what, 1.85 Amps times 5 Volts or 9.25 Watts, a modest overload on a nominal 1 Watt (actually closer to 1.25) device.

Simply driving a typical 1W Luxeon with a stiff 5 Volts is sucking around for a nearly instant failure. No matter what your heatsink.....

Proper current regulation is the way to go.

Or so I see it.

Cheers.

Doug Owen