To heat up the body used a BD438 , a little but powerful transistor, 36 W. The reflector is out to access the pill with the wires.
The body was hold in a vise thermal isolated and heated at various currents...the pic below at 280 mA 3.2 V =.896W...the rise temperature was minimal , around 5 C, ambient here 29C
Now what happen at 1,28 A 3,76 V=4,9W , the rise is 36C , too much, in an XR-E the junction temperature is at 104 C.
Below a pic heated at 2,7W perfectly stabilized at 56C even one hour later
Given the measures made the thermal resistance of this body is around 10 C/W ..
Can manage 3W easily and I guess a little more in winter time.
An XM-L in this body has a total thermal resistance of 13,5 C/W at 1A 3W could give 352 Lumen with a rise of 40C ...
I have 2 P31 , one RC-G2 and other in the way,one MTE F15 and 2 C78 ....all aluminium, forgot to say, the pill of this P31 is filled with melt solder , I know that alloys aren't good thermal transmiters but is better than air
Very nice test, arenat. I'm assuming this represents worse case scenario in which the torch is left on high while set down. I wonder how much impact the torch being held would change the outcome. The little P31 might be able to sustain 5w biologically cooled.
What you are assuming is correct Match. Here I'm follow the most conservative path. In fact this torch days ago has had a XP-G R5 at 1,4A and meanwhile hot never pass I guess 50 C after running 5-6 minutes.That was 5W. The thermal contact was bad due the reflector/star height issue and a lot of light were missing but anyway the output was impresive in a dark enviroment, it lighting up all in front 12 meters wide.
I'm planing an XM-L at around 4-5W for this one, the big die maybe match better than the XP-G one which look more like a flame all unfocus...plan B an XR-E R2 or XR-E Q4 warm at 1A or a little more because I'm using the 3 modes drivers...
I own a decent one and don't work for small things close range, , you never know what is measuring also the type of material change the reading. Those are aimed at enough area stuff, engines parts ,freezer, oven and the like...
The proper way to measure the emitter temperature is attaching a K thermocouple to the ceramic substrate near as possible the solder pad... could be glued, soldered, press with a bolt and a plastic piece to isolate it, even hold by hand .
On a side note, is it really that simple? Comparing heat from a resisor to heat from a led at about same wattage? I would guess the resistor is prone to emitt more heat than a led... or not?
A conventional lightbulb has the "nice" feature to emit 3% energy as light the rest is heat...
I have one thatlooks a lot like that, but other colors and a laser switch button and C/F display switch (did cost me $35 in a local store). It has a distance Spot Ratio: 8:1 "Yours" has a better (narrower) ratio 12:1
I just tested on a WF-502B with XP-G at 1.4Amp - Ouch - 49 celsius, very plausible.
I was so upset about the broken lens that forgot completely the Ultrafire P10...
This thing is so lightweight that seem weaker than the P31 in this aspect. Well, the greater area do his job and the numbers are better .Around 8,8 C/W , ambient today 30,5 C.
Sadly for the record this one has weight added to the pill, a few grs but given the lightness of this thing is around 20% of the total
No, is not that simple but the hard part is in the datasheet, that parameter is called thermal resistance. If I'm not wrong, since I'm not much into the matter just read a bit about , knowing the thermal resistance of the emitter in question,XR-E 8 C/W, XP-G 6 C/W , XM-L 2,5 C/W at more efficient the lower the number. As say, knowing the thermal resistance of both, emitter and body is possible predict with reasonable certainty how hot an arrange could go at a given current.
That is good to keep the led in the cold area where the thermal loss is minimal and if not at least to know what can expect really.
You're quite right. The thermal resistance is equivalent to electrical resistance R. If you call the temperature difference between two sides for V and the heat that flows from the hot to the "cold" side for I you can use Ohms law V = R * I to calculate.
V is degrees (celcius and Kelvin), I is Watt and the unit of R is degr. / Watt.
The easiest is to think of a wall with different temperatures on the two sides. It has a thermal resistance R and there is a heat flow through the wall (I Watt). Hope you can use this (for understanding)
Many thanks sixty545,but I knew that, look at the first post.
The Ultrafire R5 at 1A as came from factory could give 283 Lumen.
That's the result of working close at 80C where you get 90% of 250% minus 7% blocked by the lens...that at an ambient T of 25 which is average in many locations.
At 1,4 A the junction could reach 105 C easily and that mean 80% , 20% loss....
Sorry, it was not my intention to patronize you, I must read the thread more carefully. Perhaps my explanation can be of help to others. I very much appreciate your work and are going to read about the method right away. I just did'nt had the time before.