LED and constant current supplies

[[email protected]]

I've got a bit of a mystery. I've ordered a couple of these:

<http://www.dealextreme.com/details.dx/sku.4669>

along with a constant current supply:

<http://www.dealextreme.com/details.dx/sku.13553>

The only tidbit of information I can find is that you drive them with
300 ma @1.7V.

I've asked a couple of questions on their forums on how best to drive
2 of these from that one supply; I'm getting conflicting answers.
About half the answers say to drive them in series and the other half
in parallel.

My limited understanding says that you drive LEDs in series so that
each gets the same current.

With a constant current supply, my gut feeling says that driving these
in series should be OK. The PS uses a PT4105 chip, which is a
constant current LED driver. I've looked over the datasheet but alas
I'm not an electronics engineer... :-(

<http://www.chipcatalog.com/TI/PT4105.htm>

Also, the vendor says I can drive it with 12-24 V; the datasheet says
24V.

I have one PS and two LEDs on the way. How do I start once they get
here?

Thanks,

--Yan

 

[ehsjr]

I've got a bit of a mystery. I've ordered a couple of these:

<http://www.dealextreme.com/details.dx/sku.4669>

along with a constant current supply:

<http://www.dealextreme.com/details.dx/sku.13553>

The only tidbit of information I can find is that you drive them with
300 ma @1.7V.

I've asked a couple of questions on their forums on how best to drive
2 of these from that one supply; I'm getting conflicting answers.
About half the answers say to drive them in series and the other half
in parallel.

My limited understanding says that you drive LEDs in series so that
each gets the same current.

With a constant current supply, my gut feeling says that driving these
in series should be OK. The PS uses a PT4105 chip, which is a
constant current LED driver. I've looked over the datasheet but alas
I'm not an electronics engineer... :-(

<http://www.chipcatalog.com/TI/PT4105.htm>

Also, the vendor says I can drive it with 12-24 V; the datasheet says
24V.

I have one PS and two LEDs on the way. How do I start once they get
here?

Thanks,

--Yan

Put the LEDs in _series_

Do NOT put them in parallel.

Ed

 

[IanM]

Put the LEDs in _series_

Do NOT put them in parallel.

Ed

If the LEDS are going to be left on for more than a few seconds at a
time, they also should be mounted on a heatsink if you dont want to fry
them.

 

[IanM]

How much of a heatsink should there be? Most commercial applications I've seen only have the circuit board tracks to sink heat.

Not a lot If you looked at the OP's link, the LED dies are supplied
mounted on what appears to be an aluminium backed circuit board. Each
LED is dissipating about 1/2 a watt. They will probably survive for a
while albeit running rather warm in free air, but if fully or partially
enclosed in a non-metallic enclosure I would expect them to fail
prematurely due to overheating. If they are mounted to an aluminium
or copper bar or as you say a circuit board with heavy copper tracks and
there is a reasonable extra surface area to dissipate the heat they'll
do fine. The key thing is to monitor the temperature rise, If they are
running more than 30 deg C over ambient they should probably have a
bigger heatsink.

 

[[email protected]]

Not a lot  If you looked at the OP's link, the LED dies are supplied
mounted on what appears to be an aluminium backed circuit board.  Each
LED is dissipating about 1/2 a watt.  They will probably survive for a
while albeit running rather warm in free air, but if fully or partially
enclosed in a non-metallic enclosure I would expect them to fail
prematurely due to overheating.    If they are mounted to an aluminium
or copper bar or as you say a circuit board with heavy copper tracks and
there is a reasonable extra surface area to dissipate the heat they'll
do fine.  The key thing is to monitor the temperature rise, If they are
running more than 30 deg C over ambient they should probably have a
bigger heatsink.- Hide quoted text -

Well, I plan to use a salvaged aluminum external hard drive enclosure;
that way I get a big heat sink area + a 12VDC supply. And it's free,
or nearly so.

If that's not enough, I will add a couple of old CPU heat sinks and
some cpu heat sink goop.

 

[IanM]

Well, I plan to use a salvaged aluminum external hard drive enclosure;
that way I get a big heat sink area + a 12VDC supply. And it's free,
or nearly so.

If that's not enough, I will add a couple of old CPU heat sinks and
some cpu heat sink goop.

I seem to remember a rule of thumb for LED heatsinking that was
something like 30 sq inches of Aluminium surface in free air per watt
(rather less with a fan). I'm pretty sure a drive case is big enough.
You should be good to go with just a smear of heatsink grease on the
back and three screws engaging the notches (dont forget a fiber washer
on each under the head so they dont short the LED) or a dab of thermally
conductive epoxy to mount them. If you can keep your finger on them
without getting burnt, they're cool enough.

 

[Don Klipstein]

That is a HUGE area, those figures must be wrong.

Yes, that is huge. I do OK with 3 square inches per watt, sometimes
less.

In fact, a 1 watt Luxeon "Star" in free air (both sides of the MCPCB
exposed, or if the "star" is mounted on a heatsink barely big enough to
mount on) *usually* runs adequately cool in a 25 degree C ambient.

===============================================

Area of blackbody radiator (anodized aluminum is close enough) to remove
1 watt when at 55 degrees C and in a 35 degree C ambient:

69 square cm, 10.7 square inches.

Area of blackbody radiator to remove
1 watt when at 70 degrees C and in a 35 degree C ambient:

36.4 square cm, 5.64 square inches.

These figures are assuming no convection or conduction - only radiation,
but perfect ability to do so.

My experience indicates that heat transfer from a heatsink by convection
varies greatly, but normally well exceeds what radiation can do.

Keep in mind that effective area for radiation from a finned heatsink is
close to that of a metal box whose overal dimensions are the same as those
of the heatsink. But fins do help with convection as long as they are
vertical.

- Don Klipstein ([email protected])

 

[IanM]

Yes, that is huge. I do OK with 3 square inches per watt, sometimes
less.

In fact, a 1 watt Luxeon "Star" in free air (both sides of the MCPCB
exposed, or if the "star" is mounted on a heatsink barely big enough to
mount on) *usually* runs adequately cool in a 25 degree C ambient.

===============================================

Area of blackbody radiator (anodized aluminum is close enough) to remove
1 watt when at 55 degrees C and in a 35 degree C ambient:

69 square cm, 10.7 square inches.

Area of blackbody radiator to remove
1 watt when at 70 degrees C and in a 35 degree C ambient:

36.4 square cm, 5.64 square inches.

These figures are assuming no convection or conduction - only radiation,
but perfect ability to do so.

My experience indicates that heat transfer from a heatsink by convection
varies greatly, but normally well exceeds what radiation can do.

Keep in mind that effective area for radiation from a finned heatsink is
close to that of a metal box whose overal dimensions are the same as those
of the heatsink. But fins do help with convection as long as they are
vertical.

- Don Klipstein ([email protected])

Thanks for the correction. I forget where I got that from, but obviously
someone has slipped a decimal place (probably me :-( ). My first
reaction for something like this would usually be see if one of those
little twisted fin 30 deg C/watt TO220 heatsinks could handle it.

 

[neon]

THEY got you confused . all of these calculation at 70c is immaterial since ambient temperature is 30-35. you have to worry about temperature rise from there. These leds lexon are hot and brilliant and power hogs. they are tiny so the heat will accumulate there unless you get rid of it. Distibution is the answer. And yes a fan is a clamsy solution but a fan even if it blows hot air is better then a passive heatswink.

 

[Don Klipstein]

Thanks for the correction. I forget where I got that from, but obviously
someone has slipped a decimal place (probably me :-( ). My first
reaction for something like this would usually be see if one of those
little twisted fin 30 deg C/watt TO220 heatsinks could handle it.

Sounds reasonably OK:

OK, add that 30 degrees C per watt to the thermal resistance within the
LED. That figure is anywhere from 5.5 to 23 degrees C per watt for the
various Luxeons, 5.5 to 20 for any of the various blue, green or white
Luxeons.

That makes for a total of 35.5 to 50 degrees C per watt for a white
Luxeon or as high as 53 for a yellow, orange or red Luxeon, depending on
which one.

At 350 mA, power dissipation is usually 1.1 to 1.25 watts, but in a bad
case can be as high as 1.4 watts (somewhat less for yellow, orange or
red). It is not really good engineering practice to absolutely rely on
some of the heat being removed in the form of light as opposed to being
removed by the heatsink - the heatsink still has to work if the LED is on
the low side in light output, dimmed by temperature rise, dimmed by age,
dimmed by dirt on it, or experiences light being reflected back into it.

So depending on which Luxeon, whether its voltage drop is on the low
side or on the high side or in between, and your luck, at 350 mA the
junction temperature can be anywhere from 39 to 70 degrees C warmer than
the ambient - as little as 29 degrees C warmer or a bit less if efficiency
at producing light is really good.

Depending on which Luxeon, maximum juntion temperature for 50,000 hour
life expectancy is anywhere from 90 to 120 C.

So, a Luxeon I may being operated conservatively on that heatsink at 350
MA but would be at 250 mA. On the other hand, a Luxeon K2 (variant with
TFFC) should work nice and fine and conservatively on that heatsink at 450
mA. The Luxeon I will work at least reasonably OK on that heatsink at 350
mA, and the K2/TFFC will work at least reasonably OK on that heatsink at
550 mA.

- Don Klipstein ([email protected])