LED Heat question

[Axkiker]

How do you all estimate the amount of heat an LED will produce.

I understand that wattage is the unit of measure for power output however im unsure how to translate that into something I can relate to.


For instance if you are running a 35W led, does that produce as much heat a 35W incan bulb..


Im sure someone hear can share a long formula that I probably wont understand... however ill try LOL

 

[Steve K]

the quick answer is that we usually run calculations based on thermal resistance of the semiconductor device and heatsink.

For instance, I was doing a quick calc on an audio amp by National Semiconductor. I estimate it might dissipate up to 36 watts. The amplifier package has a thermal resistance of 1 degree C per watt from the junction to the case. If I pick a heatsink that also has a thermal resistance of 1 degree C per watt, that means that there will be a temperature differential from the junction to the air of 2 degrees C for every watt dissipated. At 36 watts, there is a 72 degrees C rise. If the air is 25C, then the semiconductor junction will be at 97C (almost hot enough to boil water on).

you might want to look at this wiki entry:
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics

For your calculations, I'd just assume that all the power put into a LED ends up as heat.

As far as "does it produce as much heat as a 35W incandescent?", the answer is no, because a LED turns more of the power into light, so less is wasted as heat. If the question is "for a given amount of light produced....", then the LED will produce much less heat since it needs less power to produce xxx amount of light.

Steve K.

 

[Axkiker]

the quick answer is that we usually run calculations based on thermal resistance of the semiconductor device and heatsink.

For instance, I was doing a quick calc on an audio amp by National Semiconductor. I estimate it might dissipate up to 36 watts. The amplifier package has a thermal resistance of 1 degree C per watt from the junction to the case. If I pick a heatsink that also has a thermal resistance of 1 degree C per watt, that means that there will be a temperature differential from the junction to the air of 2 degrees C for every watt dissipated. At 36 watts, there is a 72 degrees C rise. If the air is 25C, then the semiconductor junction will be at 97C (almost hot enough to boil water on).

you might want to look at this wiki entry:
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics

For your calculations, I'd just assume that all the power put into a LED ends up as heat.

As far as "does it produce as much heat as a 35W incandescent?", the answer is no, because a LED turns more of the power into light, so less is wasted as heat. If the question is "for a given amount of light produced....", then the LED will produce much less heat since it needs less power to produce xxx amount of light.

Steve K.

Okay good info.

What if you are not using a traditional heat sink that you can get thermal resistance info from. Such as a piece of aluminum... Is there a formula you can use to calculate that based on the size of aluminum etc

Thanks

 

[AnAppleSnail]

Okay good info.

What if you are not using a traditional heat sink that you can get thermal resistance info from. Such as a piece of aluminum... Is there a formula you can use to calculate that based on the size of aluminum etc

Thanks

With a known-wattage heat source (soldering iron?) and an accurate thermometer you could measure this. Suppose I put my 15-watt rat shack soldering iron on the heatsink and let things sit for a bit - then I measure the air temperature at 30 degrees, the heatsink at 80, and the iron at 110. Then I know that the iron-to-the-heatsink connection is (40C/15W) 2.67 C/watt. And since the heatsink can't "store" heat once it's at steady-state, then it must be (30C/15W) 2 C/watt.

Solid metal is a darn good conductor, so you can generally assume that one piece of metal is the same temperature.

LED datasheets give a C/watt - the XP-G, for example is 6 C/watt. So if I run it at 1.5A (3.5v Vf) then it's 5.25W. The diode junction will be 31.5C hotter than the heatsink. If my heatsink is the one above, then running this LED in the same air gives the following steady-state die temperature.

Air: 30C
Heatsink: 2C/watt, 5.25 watts = 30C + 5.25W*2 C/W = 30C + 10.5C = 40.5C
XP-G: 6 C/watt, 5.25 watts = 40.5C + 5.25W * 6 C/W = 40.5C + 31.5C = 72C

 

[vaska]

So if I run it at 1.5A (3.5v Vf) then it's 5.25W. The diode junction will be 31.5C hotter than the heatsink.

You are a bit mistaken: in this case 5.25W of electric power means 1.575W turned into light and 3.675W of heat. So the temperature gradient is 22.05C.

 

[blasterman]

If all else fails, you can always use the Design Notes on Bridgelux's site regarding heat sink size and their large LEDs.

They break a lot of this down into some basic rules, and rough guides for heat-sink size.

 

[Axkiker]

Good deal

Thanks

 

[Axkiker]

Wow I just read on the bridgelux site it recommends 10 sq in for every watt.

Wow I have some heat to dissipate.