XP-L VS XP-G2 Heat at same power?

recDNA

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Does an XP-L at 2 amps produce more or less heat than an XP-G2 at 2 amps? Thanks.

I would also be curious about heat from triple xpg2 at 3 amps total vs single xpl at 3 amps total.
 
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alpg88

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good question, xpl has bigger die thus more efficiant, so in theory it should be making less heat, especially since 2A is overdrive for xpg2, but still within specs for xpl.
now the bigger question is how much more, that i do not know for sure. but i do not think there will be big difference.
 

SemiMan

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Does an XP-L at 2 amps produce more or less heat than an XP-G2 at 2 amps? Thanks.

I would also be curious about heat from triple xpg2 at 3 amps total vs single xpl at 3 amps total.

Order of magnitude, heat = (power in) - (lumens out)/280 .... roughly ... 280 goes up for low CRI high CCT, down for high CRI warm white
 

RetroTechie

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XP-G2 also has a higher thermal resistance from die -> case (4 o​C/W vs 2.5 o​C/W). Read: same heat dissipation will (usually, depends on mounting / base material etc) give higher working temps for XP-G2 compared to XP-L.
 

recDNA

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How about triple 219b 3 amps (1 amp each) vs one XP-L 5000k also at 3 amps?
 

ShineOnYouCrazyDiamond

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You are missing one very important metric in this discussion - forward voltage (Vf) of the LED. You can't calculate power, and thus heat, from just current. Each of the LEDs mentioned will have a certain Vf at a specific drive current (and, this Vf will vary from LED to LED - not just between different LED types).

Let's look at nominal Vf at a set current. Cree doesn't spec the XP-G2 over 1.5A so let's use that for the example.

Vf XP-G2 @ 1.5A = ~3.13 volts
Vf XP-L @ 1.5A = ~3.05 volts

So at 1.5 amps the LEDs will have to dissipate the following amount of power:

XP-G2 1.5A x 3.13V = 4.70 watts
XP-L 1.5A x 3.05V = 4.56 watts

Just looking at pure watts you can deduce that the XP-L will run a little bit cooler than the XP-G2. But, as mentioned earlier in this thread, you also need to take into account the thermal resistance of the die transfer. The XP-G2 has a higher thermal resistance than the XP-L so that means the XP-L can also dissipate the heat faster than the XP-G2. Now there are too many other variables such as the material the mPCB is bonded too, the type of the junction, the bond between the mPCB and the heatsinking, the amount of heatsinking material, etc, etc, etc. But if you consider all things being equal the XP-L should be run cooler than the XP-G2. I just don't know all the math to be more specific as to how much.

Now if you consider the Nichia 219B triple @ 3Amps vs one XP-L.

N219B @ 1.0 Amps = 3.5 volts. 1x3.5 = 3.5 watts x 3 = 10.5 watts
XP-L @ 3.0 Amps = ~3.33 volts. 3x3.33 = 9.99 watts

So the XP-L will run cooler. But the thermal resistance of the N219B is 7o​C/W so it will be much less efficient at dissipating the heat than the XP-L so the N219B will run significantly hotter.
 

RetroTechie

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Now there are too many other variables such as the material the mPCB is bonded too, the type of the junction, the bond between the mPCB and the heatsinking, the amount of heatsinking material, etc, etc, etc.
Which add up into this thing called thermal resistance, for which the semiconductor die -> solder point value is specified in the datasheet. :) There are methods to measure thermal resistance for other parts in the path from semiconductor die -> hand + flashlight-surrounding air.

One thing of note is that with multiple emitters & the same total heat flow, each emitter's thermal resistance is effectively parallel-ed with the other emitters. Or in other words: all emitter's thermal conductivity gets put in parallel (= added up). Example:

An XP-G2 die dissipating 5W would sit at 20 o​C higher (4 o​C/W) than its solder point temperature. An XP-L die dissipating 5W would sit at 12.5 o​C higher (2.5 o​C/W) than its solder point temperature. Here the XP-L has the edge due to its lower die -> case thermal resistance.

But if that same 5W is distributed over 3 pcs. XP-G2's, each will take 1.67W. Putting each die at 6.7 o​C higher than its solder point temperature. Effectively behaving like a (theoretical!) single XP-G2 with 4/3 = 1.33 o​C/W die -> case thermal resistance. Which gives 3x XP-G2 the edge over an XP-L, due to the lower overall thermal resistance.

Then there's efficiency: even if electric power is known, it's not known exactly what % of that power is radiated out as light, and what % dissipated as heat. So even with a higher die -> case thermal resistance, a higher lm/W LED may still come out on top, simply because of each W put in, a smaller % is dissipated as heat. Possibly up to the point where heat flow * thermal resistance gives a smaller number.

It doesn't end there, though: :hairpull: even though a multi-emitter setup appears preferable, that heat still flows to another part of the flashlight. Which in turn may become the bottleneck in the heat flow's path. 6.7 o​C vs. 12.5 o​C temperature rise makes little difference, if pill -> flashlight body thermal contact is so poor, that the entire pill sits 30 o​C above the flashlight's body temperature. Likewise: if you're dissipating 5W total, that will cause the same temperature rise for each element that's next in the path, since the same heat flow is going through it. Regardless of whether that 5W is dissipated by 1, or 3 emitters. Only exception here is if the heat flow is distributed such across flashlight parts, that a 3-emitter setup actually causes a lower thermal resistance from pill -> flashlight body.

Adding thermal mass (eg. a heavier pill) in itself doesn't make any difference whatsoever for what final temperatures are reached (!). It does change this though: 1) how fast temperatures rise/fall with the same heat flow, and 2) more thermal mass often means better thermal conductivity (lower thermal resistance). But this depends a lot on the light's construction & the route along which most of the heat flows.

For those who think this is all very complicated: thermal resistance, and the related math, very much behaves like voltages, currents, series & parallel resistors in electronic circuits. Thermal mass could be compared to a capacitor in such circuits. So although complicated at first sight, it's very possible to measure, calculate & predict things when the necessary lab work is done. Coming up with a good measurement setup isn't easy, though...

One final note: usually the bottlenecks are closest to where the heat is produced. So in the case of a LED, that's primarily die -> LED case. Next in line is pcb the LED sits on (copper or aluminium for example, and how thick). Followed by the question whether a heat-conducting paste is used between LED pcb and flashlight pill. From there on, things are likely to be less critical. Read: alu or brass pill likely makes a smaller difference for LED temps than the difference between alu or copper LED star.
 

SemiMan

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Actually it is normally the final heatsink (or flashlight) that is the biggest bottleneck
 
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