dissipated power of sst-90 @9A

[lightliker]

Has anybody measured the dissipated power of a SST-90 led running at 9amps?
Thinking of making a triple SST-90 flashlight with dc-fan cooling and asking myself how many watts I have to get rid of from 3 x SST-90.

 

[xul]

From
http://www.luminus.com/products/datasheets/PDS-001342_Luminus_SST-90, Product Datasheet.pdf


Vf max at 3.15 A is 3.9 v so 12w
Vf typ at 9A is 3.87 v so 34w.


but see the current derating curve.

 

[xul]

SST-90 notes:


Understand completely the meaning of each of the 8 footnotes on page 10.


Send for the app note mentioned in footnotes 4 & 5 of pg 10. I can't find my copy of it.
If you want higher brightness in exchange for shorter lifetime you need some numbers to figure your tradeoff.


It looks like you get slightly more than 14,000 hours if you run it at Tj = 150C.


You can run it at 9A with an ambient of 55C with a junction to ambient thermal resistance of less than 1.8 C/W.
See note 2 on pg. 11.


On pg 12 I don't see how the Vf can be TBD if they've actually built any of these.

And what's this
'Special design considerations must be observed for operation under 1A. Please contact Luminus for further information.'
about?

Pg. 11 tells how to use the LED as its own thermistor but the interpretation of these coefficients depends if the RGB LEDs are wired in series or parallel. The Vf of the red LED seems to say they can't be in parallel, but a series connection doesn't make sense either. Seems like these guys have some 'splainin' to do.

 

[lightliker]

Thanks for responding people!
So it seems that I have to get rid of about 105 watt.
assuming a thermal resistance of total 2,0 C/watt (0,64 for the junction to LED-body plus 1,36 for the cooling body) and a maximum T-junction of 100 C, I can use the thermal difference between 30 C (t-ambient) and the cooling finns (at about 77 C) to transfer the heat to the air around the flashlight.
deltaT = 47K
Power = 0,034kW
mass of air = 1,2
calculated airvolume will be about 1,8 m3/h times 3 (3 leds) = 5.4m3/h
This should be possible for a small fan :thumbsup:
Can anybody correct me if I'm wrong?

 

[Justin Case]

If you are going to use a driver, you also need to consider the waste heat generated by the driver. To deliver roughly 100W to three SST-90s, a 90% efficient driver will generate about 10W of additional waste heat.

Also, not all of the approx 35W of power drawn by an SST-90 at 9A drive ends up as waste heat. Some of it actually products light. LEDs are probably about 20%-30% efficient, so of the 35W, about 70%-80% is waste heat.

 

[lightliker]

So this levels the heat produced by the leds.....
0,8 x 105 watt plus 10 watt for the driver will be about 90-100 watt, not a mission impossible for 3 fans to cool away

 

[MikeAusC]

. . . . LEDs are probably about 20%-30% efficient, so of the 35W, about 70%-80% is waste heat.

Yes, these figures have been confirmed in my tests - although I expect the SST-90 running at full power will be closer towards the 80% lost as heat.

 

[LFP11]

1lm is about 0.0015W (683lm=1W). If you use 9A current total lumens are about 3x2000lm=6000lm. This is same as 9W, not more.

 

[MikeAusC]

1lm is about 0.0015W (683lm=1W). If you use 9A current total lumens are about 3x2000lm=6000lm. This is same as 9W, not more.

Blindly using formulae without understanding their meaning is dangerous.

683 Lumen = 1 Watt defines the relationship between Luminous Flux and Watts of RADIANT Power ONLY AT 555 nm.

Unless you're talking about a Green LED, this is meaningless.

Unless you know the relationship between Electrical Input Watts and Radiant Output Watts, this formula is useless to to indicate heat output.

 

[LFP11]

I have use this from Wikipedia:

ideal 5800 K black-body, truncated to 400–700 nm (ideal "white" source),Luminous efficacy of radiation
(lm/W)=251, Luminous efficiency=37%

and my idea was that led is similar.

 

[LFP11]

Thanks for responding people!
So it seems that I have to get rid of about 105 watt.
assuming a thermal resistance of total 2,0 C/watt (0,64 for the junction to LED-body plus 1,36 for the cooling body) and a maximum T-junction of 100 C, I can use the thermal difference between 30 C (t-ambient) and the cooling finns (at about 77 C) to transfer the heat to the air around the flashlight.
deltaT = 47K
Power = 0,034kW
mass of air = 1,2
calculated airvolume will be about 1,8 m3/h times 3 (3 leds) = 5.4m3/h
This should be possible for a small fan :thumbsup:
Can anybody correct me if I'm wrong?

1.8m3/h=0.5dm3/s.
0.5dm3/s*1.2g/dm3*1J/Kdm3*47K=28.2W.
If you want delta T=47K then airflow is 34W/(1.2*1*47)=0.6dm3/s=>2.2m3/h, total 6.6m3/h
Can anybody correct me.

 

[MikeAusC]

I have use this from Wikipedia:
deal 5800 K black-body, truncated to 400–700 nm (ideal "white" source),Luminous efficacy of radiation (lm/W)=251, Luminous efficiency=37%

and my idea was that led is similar.

But you need to know what this means - Luminous Efficacy of RADIATION = how good the eye is at converting OPTICAL watts to units of visibity. 555nm radiant energy produces lots of lumens, 755nm radiation produces zero lumens.

The OP is interested in Heat Output Watts, - the eye is NOT sensitive to HEAT i.e energy at less than 700nm - this Luminous Efficacy info is totally irrelevant.

 

[xul]

For fans, try
http://www.google.com/cse?cx=partne...59-1&q=fan&sa=Search#gsc.tab=0&gsc.q=fan heat

 

[LFP11]

But you need to know what this means - Luminous Efficacy of RADIATION = how good the eye is at converting OPTICAL watts to units of visibity. 555nm radiant energy produces lots of lumens, 755nm radiation produces zero lumens.

The OP is interested in Heat Output Watts, - the eye is NOT sensitive to HEAT i.e energy at less than 700nm - this Luminous Efficacy info is totally irrelevant.

You are right. Led emit invisible wavelenght too.