Question on heat sinking

[ahfoong]

I modified a headlamp using a Seoul P4 running off a DX driver (SKU 4735).
It's drawing 1.4a and is pretty hot.
There is still space on the heatsink and I was thinking of adding another LED (a Cree P4) parallel to the Seoul to reduce the amps to 700ma each.
But would the heat generation from the 2 LEDs @ 700ma be equal, lower or more than a single LED at 1400ma?

 

[Lynx_Arc]

I am guessing it will be slightly less heat generated because of the LEDs being more efficient at lower voltages. I doubt it will be enough difference to reduce the heat enough though.

 

[LukeA]

There will be less heat and more light from two LEDs each running at 700mA, but there will be no tremendous difference. Running two LEDs at 700mA is better for each emitter because of the lower junction temps.

But adding LEDs in parallel is not a good idea. One of the LEDs will end up drawing all or almost all of the current. Series is better if it can be swung. If not, two separate drivers would also work.

 

[eebowler]

You're better off taking up the extra space with a heatsink instead of another LED.

BTW, a current draw of 1.4A doesn't mean that the LED is getting 1.4A. If the driver is a boost circuit, current would be less, if the driver is buck (reduce V), current can actually be more than 1.4A. Either way, a current draw that high means a lot of current going to the LED and a lot of heat produced.

 

[TigerhawkT3]

There will be less heat and more light from two LEDs each running at 700mA, but there will be no tremendous difference. Running two LEDs at 700mA is better for each emitter because of the lower junction temps.

But adding LEDs in parallel is not a good idea. One of the LEDs will end up drawing all or almost all of the current. Series is better if it can be swung. If not, two separate drivers would also work.

Yes, parallel is usually not a good idea, but not if the max current is okay for a single emitter in the first place. And about that "1.4A"...

The OP said this driver draws 1.4A. According to "Kyle" (a DX employee), current to the emitter is 350mA on an alk or NiMH, and 850 on a Li-Ion. I'm guessing he's using an alk or a NiMH and that the 1.4A is current draw from the cell, which would mean a 53% efficient driver. That's pretty darn close to the test figures here, which list an efficiency of 55% with a Vin of 1.57V (fresh alk, give or take).



...Aaaaaaand eebowler beat me to it.

 

[ahfoong]

Ok... thanks for all the feedback.
I am actually using a 18650 and the 1.4a measurement was taken between the battery and the driver... so I guess the emitter is not getting 1.4a.
This means adding a parallel emitter is not good... sigh.

 

[TigerhawkT3]

Ok... thanks for all the feedback.
I am actually using a 18650 and the 1.4a measurement was taken between the battery and the driver... so I guess the emitter is not getting 1.4a.
This means adding a parallel emitter is not good... sigh.

That's weird. Looking again at the topic in the second link I posted, Iin is 1.4A when Vin is about 2V. For Li-Ion voltages, Iin shouldn't be over 1A. :thinking:

Strange.

 

[TorchBoy]

According to "Kyle" (a DX employee), ...

I have no reason to think that Kyle is an assumed moniker, and he's the one who does the employing.

It seems to me a not very efficient driver. If you got a more efficient driver you wouldn't have so much heat produced. For example, an 85% efficient driver would produce just half the waste heat that a 70% efficient driver does. I guess using it with an 18650 its efficiency wouldn't be too bad.

 

[VegasF6]

http://www.candlepowerforums.com/vb/showthread.php?t=199215

OP Rizky P. Lots of input on this driver. Do you have the newer blue version? This board isn't running anywhere near spec at those currents and I highly doubt it will last.

You can change the resistor to lower current, or do as Tohuwabohu did and series a schottky diode to lower voltage, thereby lowering current.