Which resistor for 3 XPEs on 12V?

I am trying to work out the cheapest way to run clusters of 3 XPE LEDs from 12v. Given that I need many such assemblies, the cost of using a board rules it out, preferably I would like to simply use an inline resistor given that the 12v is already coming from a regulated output.

Normally to work out the resistor to use to get a given voltage over a load I would assume that voltage is dissipated evenly throughout the circuit, but in this case, I don't know what the resistance is. In any case, I think that's the wrong way to be going about it.

I want the 12v stepped down to 11.1v (3 * 3.7v) for the LED load. In order to calculate the resistor needed, do I have to get a multimeter out and measure the resistance of each cluster? Or am I doing it wrong?

You should use a multimeter to measure the forward voltage and current of each string. Then you can calculate the resistance you want for each string.

There's a danger running LED's in series, one could run away and burn out.

That said, in theory you want to take the .9 V (12-11.1) and divide it by the current you want. This will give you the proper resistance.

example: To drop .9 V at .35 A would be a 2.6 ohm resistor.

EngrPaul said:

There's a danger running LED's in series, one could run away and burn out.

That said, in theory you want to take the .9 V (12-11.1) and divide it by the current you want. This will give you the proper resistance.

example: To drop .9 V at .35 A would be a 2.6 ohm resistor.

Click to expand...

Ah! Of course!

Now, can you tell me more about this runaway LED scenario? How would it happen? Should I perhaps use a 12v+ 1A current regulated driver instead.

Basically, it's thermal runaway. The LED with the highest resistance will have the higher voltage. There will be more heat dissipated in this LED, then the increased temperature leads to more resistance... then a higher voltage. The higher voltage then leads to higher power dissipated, and therefore more heat. The temperature of the emitter gets out of control. I may not be explaining it perfectly, but you could do a search to find somebody who can explain it better.

You are better off driving each LED in parallel with it's own dropping resistor. The only problem is, this is inefficient because most of what you put into the system is wasted as heat.

Perhaps you could get a regulator that will drop the 12V to 5V. Then you can put the LED's safely in a parallel arrangement with 3 resistors, on for each LED. And you'll only be shedding a fraction of the power in the resistors.

EngrPaul said:

Perhaps you could get a regulator that will drop the 12V to 5V. Then you can put the LED's safely in a parallel arrangement with 3 resistors, on for each LED. And you'll only be shedding a fraction of the power in the resistors.

Click to expand...

I have thought about using a driver. There are 12v capable 1A drivers, which would be ideal for a tri-LED cluster. However, cost per unit is an issue, and a simple series with inline resistor is the cheapest way to build them, especially when I factor in build time per unit.

The eventual application is to replace the bulbs in a MR16 socket. I don't know how much heat the sockets will be able to dissipate, however, I think that if I use three emitters and underdrive them by upping the value of the resistor, I can keep heat in check. I'm confident that three emitters under driven will be a worthwhile replacement for a 50W halogen. I have done a test with the following unit:



That is just a pair of XPE units mounted to a heat sink with arctic silver and a few tiny globs of epoxy. I do not have a multimeter that goes past 100mA so I can't measure the current it draws, but when direct driven from 2 series CR123As it is very bright indeed. I have 8 MR16 sockets in my room, and I think that were each socket as bright as this unit, the room may even be too bright.

The only other issue I have is colour tint. The above LEDs are far too cold for a bedroom. I was thinking of using three of these LEDs in the assembly:

http://www.dealextreme.com/details.dx/sku.13612

Being 1w emitters, I don't think that heat will be a problem if I use a decent heat sink. Three of those per socket gives me a total of 24 emitters in the room, which is 1,200 lumens or so of diffused 3000K light, which is more than a 100w incan bulb. This would be a 24w energy draw (compared to 400w currently).

Also, driving those emitters with ~10.8v @ ~340mA would be easy compared to trying to push 11.1v @ 1A through a set of XPEs. It would be nowhere near as bright, but cost would be far lower as would power draw.

Thoughts?

I would defer to testing. I would monitor the individual voltage drops, while using decreasing values of the resistor until you hit the current you want.

Thermal runaway might not be a problem if it's quickly moderated by good heat sinking. Also, binning your emitters with similar Vf could help too.

Those be XR-E stars on that heatsink, not XP-E

Tekno_Cowboy said:

Those be XR-E stars on that heatsink, not XP-E

Click to expand...

Indeed. Thanks for the correction.

EngrPaul said:

Basically, it's thermal runaway. The LED with the highest resistance will have the higher voltage. There will be more heat dissipated in this LED, then the increased temperature leads to more resistance... then a higher voltage. The higher voltage then leads to higher power dissipated, and therefore more heat. The temperature of the emitter gets out of control. I may not be explaining it perfectly, but you could do a search to find somebody who can explain it better.

You are better off driving each LED in parallel with it's own dropping resistor. The only problem is, this is inefficient because most of what you put into the system is wasted as heat.

Perhaps you could get a regulator that will drop the 12V to 5V. Then you can put the LED's safely in a parallel arrangement with 3 resistors, on for each LED. And you'll only be shedding a fraction of the power in the resistors.

Click to expand...

This is not correct. Thermal runaway between the leds is not a problem with leds in series the leds will see the same curent. It is a potential problem with Leds in parallel. As the led heats up its forward voltage DROPS so it gets more curent hence it heats up more and you get a runaway senario.

The OP will not have this issue as they are talking about series leds. The size of resistor will depend upon the leds and the desired curent. 0.5A with a vf of say 3.5 * 3 = 10.5V so 1.5v so a 3 Ohm resistor rated at 0.75W. The problem is the next set of 3 leds may have diffent vfs and you find they need a 2 Ohm and the next set need a 4 Ohm... And then the tempreture effect comes in and as the leds heat up the VF Drops hence the curent will rise. This is why you need a driver realy.

Ifor

ifor powell said:

This is not correct. Thermal runaway between the leds is not a problem with leds in series the leds will see the same curent.

Click to expand...

I've sucessfully proven otherwise while making fixed lighting

Heat is a product of current and voltage, not current alone. Emitter temperature is a result of heat vs. thermal conductiveness.

I'm not as qualified as others to proscribe your setup. I'll just share my experience... :tinfoil:

I'm no engineer and I'm sure others will chip in but I think you should revise your theory. I'm with Ifor on this one. The fact that your circuit didn't fail is not a proof of theory; it only proves that the LEDs you started with were well balanced or that your are lucky.