Check out this heat sink

[MoreGooder]

http://www.newark.com/fischer-elektronik/ick-r/heat-sink-for-plcc-19c-w/dp/34M6426

datasheet: http://www.farnell.com/datasheets/7781.pdf

Imagine a 20mm star XPG mounted to that.:twothumbs
Comments?

 

[thepaan]

Whats so special about it? I don't get it.

 

[evilc66]

Looks tiny, and 19C/W?

 

[PMM]

either incorrectly linked or a big fail in the understanding of heatsinks.

 

[jason 77]

mine is bigger than yours! lol

 

[ma_sha1]

The fins need to be exposed to air to dissipate heat,
stick it in flashlight, you then lost the function of what the fins are designed for. Also, your heat transfer surface are cut-to half vs. solid surface, so this doesn't look like a good sink for flashlight.

But if you aren't doing flashlights, you'll likely have something that's constant on for a long time,
then you'll need a fan, as the sink is pretty small for 3-5W led constant on w/o a fan.

 

[zippittydoodah]

Isn't a solid puck pasted on the back of the LED best?,as long as there is a good conductive interface with heat transfer to the outside of the housing which is exposed to the air for cooling?.
Heat sinks come in all types,they don't have to have fins and stuff,they just have to conduct away the heat from the heat sensitive component,yes?.

 

[PMM]

Surface area and ability to radiate heat is important.

Equally big is not always best If you whacked that big heatsink above on a Computer CPU for example it would do worse than the stock heatsinks supplied an area I experimented in some 10+ years ago.

However when wanting to cool power IC's the big bulky heatsinks are better because they run higher temps but equally as a net result radiate that heat away more effeciently so while the object being cooled is hotter more heat is moved making it good for power applications like power controllers and cooling for Hi-fi output stages etc.

C/W is a very important measurement when determing heatsinks its the temperature increase per 1 watt of energy.

So if you have 5 watts of heat you will have a gain of 5 * 19c = 95deg C
however that's ontop of your ambient so if its 25deg C then 25 + 95 =120Deg C

A tad hot


Looking at the XPG data sheet you could not run it at 1amp your looking sub 800ma about 2.5watts to stay in the thermal spec assuming heatsink is open to the air. if enclosed you might as well use an aluminium or copper slug its just a transfer medium.

 

[MoreGooder]

I am trying to find a heatsink with fins that can be installed inside a custom built host. I would expose the fins to ambient air some how.

I do see the points made, however.

It is possible that I could get the same heat dissipation by using a copper slug with a host that is ridged heavily that this heat sink can provide.

My ultimate goal is to design a flood-to-throw host that can handle the heat of an XP-G R2 or better (or hotter). I have some ideas on how to do it, but I want to use as many purchased parts as possible to keep the cost down.

Thanks everyone for your excellent comments! (well, most of them were anyway)

 

[thepaan]

I get my heatsinks from these guys.

http://www.alphanovatech.com/

Or, at least, I find one made by them then google for a place that sells one for cheap.

 

[MoreGooder]

Excellent. THanks!

 

[EndOfTheTunnel]

C/W is a very important measurement when determing heatsinks its the temperature increase per 1 watt of energy.

So if you have 5 watts of heat you will have a gain of 5 * 19c = 95deg C
however that's ontop of your ambient so if its 25deg C then 25 + 95 =120Deg C

A tad hot


Looking at the XPG data sheet you could not run it at 1amp your looking sub 800ma about 2.5watts to stay in the thermal spec assuming heatsink is open to the air. if enclosed you might as well use an aluminium or copper slug its just a transfer medium.

I'm not very familiar with heat transfer math, but I'd like to know more. To calculate the heat energy produced by an LED, wouldn't you have to multiply the power sent to the LED by the LEDs efficiency? So, if you send 10W to an LED that is 75% efficient, for example, wouldn't that equate to roughly 2.5W of energy converted to heat? Wouldn't you use 2.5W in your heat transfer calculations?
I'm not sure how efficient the XP-G is, but if it's above 80%, then wouldn't 5W of power sent to it result in generation of less than 1W of heat?

 

[lilmarvin4064]

not that you would want to use it in flashlight, but if you're looking for a round finned heatsink, something like this is larger and less expensive....

http://www.excesssolutions.com/cgi-bin/item/ES953

 

[AnAppleSnail]

I'm not very familiar with heat transfer math, but I'd like to know more. To calculate the heat energy produced by an LED, wouldn't you have to multiply the power sent to the LED by the LEDs efficiency? So, if you send 10W to an LED that is 75% efficient, for example, wouldn't that equate to roughly 2.5W of energy converted to heat? Wouldn't you use 2.5W in your heat transfer calculations?
I'm not sure how efficient the XP-G is, but if it's above 80%, then wouldn't 5W of power sent to it result in generation of less than 1W of heat?

Yes, but do look at the heat from your driver as well. One approximation for LED heat I've heard is: White light is 330 lumens per watt. So if your LED takes 4 watts and makes 330 lumens, it's shooting out about 1 watt as light, with the other 3 being heat. There's probably better ways to calculate this though.

 

[PMM]

I'm not very familiar with heat transfer math, but I'd like to know more. To calculate the heat energy produced by an LED, wouldn't you have to multiply the power sent to the LED by the LEDs efficiency? So, if you send 10W to an LED that is 75% efficient, for example, wouldn't that equate to roughly 2.5W of energy converted to heat? Wouldn't you use 2.5W in your heat transfer calculations?
I'm not sure how efficient the XP-G is, but if it's above 80%, then wouldn't 5W of power sent to it result in generation of less than 1W of heat?

TBH it does my head in so I just generalise a slight rule of thumb from years ago.

That being 15 / 85

STD bulb = 15 % heat + (5% visible light & remainder radiated Infrared).
LED = 85% heat 15% visible light (No Infrared)

So for an LED sort of the opposite to your calc so assume 85% of your input power needs sinking you can't really go wrong.

 

[thepaan]

"Power dissipated by the LED (Pd) is the product of the forward voltage (Vf) and the forward current (If) of the LED"

If you send 10W to the LED then you use 10W in your heat transfer calculations.