# do i need a heatsink?



## poolman966554 (Oct 28, 2010)

Hi everyone! i have purchased a few 1 watt leds. I have only toyed with small 5mm 20ma leds in the past.. Here are the specs for the led.

[FONT=Arial, Helvetica, sans-serif]Emitted Color : Blue
DC Forward Voltage (VF): 3.6 ~ 4.0 Vdc
DC Forward Current (IF): 300 ~ 400 mA
Reverse Voltage (VR): 5V
Reverse Current (Ii): 5uA
Temperature Co-Efficient: 0.04nm / Degree C
Viewing Angle: 140 Degree
Intensity Luminous (Iv): 40 ~ 50 LM
Wavelength: 460nm- 470nm









[/FONT]
Im very new to this, and have no idea how bright these are, or how much heat they will produce.

As i said in the past, i have used 5mm leds. The blue ones were 3.2v - 3.8v maximum, and the power supply was from a pc. I just 
soldered 4 leds in a series and connected them to the 12v rail on the power supply.. Will i be able to do the same with these?
Btw, power supply is not in a pc, and is a 400w supply with 16a available on 12v rail.

Thanks!


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## mds82 (Oct 28, 2010)

In short, yes you will need a hearsink on these. You can just a strip of aluminum, as long as they are mounted to something using thermally conductive glue


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## AnAppleSnail (Oct 28, 2010)

poolman966554 said:


> Emitted Color : Blue
> DC Forward Voltage (VF): 3.6 ~ 4.0 Vdc
> DC Forward Current (IF): 300 ~ 400 mA
> Reverse Voltage (VR): 5V
> ...



Each LED is rated for 300-400 mA, at 3.6-4v DC. Heat equals Watts in - Watts out. With LEDs, some of the power (watts) becomes light, but if you ignore that then you have a decent safety margin.

Power = Watts = V*I
We've got 2 voltages and 2 amperages. LEDs tend to need more voltage at higher amps, but we'll do 'em all.

3.6v * 300 mA = 1.08W
3.6v * 400 mA = 1.44W
4v * 300 mA = 1.2W
4v * 400 mA = 1.6W

So I'd expect about 1-1.5W of heat if you're running these things at 300-400 mA.

Voltage: If you have 12.0 VDC between those rails, you'll be running these LEDs at 3.0v each - they're likely to be dim. If you tried 3 LEDs, you'd be running them at 4.0v, which for the LEDs is likely to be screaming hot.

Consider adding a resistor. The way to calculate the value you want is: Pick a drive current (350 mA, say). Make the voltages add up. For 3 LEDs:

V(supply) - V(resistor) - 3* V(LED) = 0
12V - I * R - 3* V(LED) = 0

The voltage of the LED depends on current - I'd guess it's about 3.7v, but that's a WAG.

12V - .35A * R - 3*3.7 = 0
12V - 11.1 = .35*R
.9V = .35R
2.57 ohms = R. Make sure it's rated for at the very minimum 350 mA.


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## poolman966554 (Oct 29, 2010)

thanks for the responses, and the breakdown!:thumbsup:
I just read "each watt of LED power needs about 9 square inches of surface area open to free air for cooling." 

As for using aluminum strip, a 3" square piece will work? How thick should it be?


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## Yoda4561 (Oct 30, 2010)

That seems to be extremely conservative. The average surefire 6p has around 17 square inches of aluminum exposed to air, LED's are regularly driven at 5 watts and this is just a single XR-E or XPG, and 10-15 watts seems to be the upper limit. This is also with a relatively poor thermal path compared to an LED directly mounted to a heatsink. The surefire 6p example can be rounded down to an easy 3 square inches per watt, which should be well within the margin of safety, but the heatsinks will get fairly hot to the touch. I suggest using 1/8 inch aluminum, or whatever thickness of U channel you can find close to that. 1/4 is better of course, but 1/8 is easier to cut with tin snips or a hacksaw to make fins. If you have any old computer heatsinks laying around these will work great too.


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## AnAppleSnail (Oct 31, 2010)

Yoda4561 said:


> That seems to be extremely conservative. The average surefire 6p has around 17 square inches of aluminum exposed to air, LED's are regularly driven at 5 watts and this is just a single XR-E or XPG, and 10-15 watts seems to be the upper limit. This is also with a relatively poor thermal path compared to an LED directly mounted to a heatsink. The surefire 6p example can be rounded down to an easy 3 square inches per watt, which should be well within the margin of safety, but the heatsinks will get fairly hot to the touch. I suggest using 1/8 inch aluminum, or whatever thickness of U channel you can find close to that. 1/4 is better of course, but 1/8 is easier to cut with tin snips or a hacksaw to make fins. If you have any old computer heatsinks laying around these will work great too.



Usually those lights you mention are exposed to hands, which are far better than air at removing heat.


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## poolman966554 (Oct 31, 2010)

I have tested the power supply voltage with my multimeter, and i got 12.22v constant. Instead of a 2.57 ohm resistor that AnAppleSnail suggested at 12v, i will now need a 3.3ohm 1/2watts resistor correct at 12.22v correct? If so, are these the right ones? 
http://cgi.ebay.com/10-pack-1-2-wat...408?pt=LH_DefaultDomain_0&hash=item255937ed50

How much heat will these resistors make? Any idea on what average temps these resistors should run at? Basically im trying to figure out if they will get hot enough to burn you.


Also i have this piece of aluminum alloy..





Its about 12" x 16" x 1/8" thick. Is this appropriate heat sink material? If it is, how would one go about making the most efficient heat sink?

Would cutting a square chunk off and mounting the led to the center of it be the way to go? Or would cutting slots in it like this be better?





Its only 1/8" so im not so sure it would help alot, since im not creating much more surface area. Im very open to suggestions if anyone wants to share their method of fabrication and mounting!

If any of the above is wrong, would you kindly guide me in the right direction?? Thanks!


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## Watashi (Oct 31, 2010)

As for making the heat sink out of the sheet of aluminum. You could cut 4 1.5" squares and 3 .5" squares and sandwich them together with thermal glue and then mount your LED assembly to the aluminum stack. You could make the stack as tall as needed to dissipate heat.


EDIT: If you are going to use SMD LED's then you probably want to go with a SMT resistor, you can find a huge selection here


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## poolman966554 (Nov 1, 2010)

Thanks Watashi! Thats a great idea!

I am still a leary of buying any resistors. The website Watashi mentioned has many options for the resistors, and im not sure what really pertains to me in my exact scenario. I am the first to admit this is still very new to me!

To give you a clue, I just figured out what smd stood for!

This website http://led.linear1.org/led.wiz says i need a 3.3 Ohm 1 watt resistor. I have no idea even what kind of resistor to use, let alone what tolerance or temperature coefficient or voltage rating they should be!


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## Watashi (Nov 1, 2010)

(Input voltage-Vf*#of LEDs)/what mA you want to drive the LED=Ohms resistor

Same formula as above but multiply (V-Vf)*mA for the wattage of resistor needed.

You have a range of 3.6 to 4.0Vf, most LED's have a typical (Typ.) Vf, yours will probably be 3.8. The max LED's you can use with 12v is 3 LED's.

Assuming 3 LED's per string and your current of 350mA your formulas then would be:

12.22-11.4/.350=2.34Ohm Resistor (get the next bigger available)
12.22-11.4*.350=.287w Resistor (use a 1/2 watt)
Also recommend getting 1% tolerance or lower.
And I believe Thick Film are the type of SMT Resistor you want.

I'm 99% sure this math is right.


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