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Gadget Freak Case #228: Super LED Flashlight Hits 3,000 Lumens
10/19/2012

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John Duffy's super LED flashlight is almost three times as powerful as xenon car headlights.
John Duffy's super LED flashlight is almost three times as powerful as xenon car headlights.

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laserdudephil
User Rank
Iron
Re: Go John!
laserdudephil   10/22/2012 5:00:17 PM
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@Rudy,

     I'm sorry but you're way off.  You're thinking in terms of a fixed voltage.  An n watt resistor is rated for n watts no matter what's next to it (except for a heat gun.)

laserdudephil
User Rank
Iron
Re: Nice work John
laserdudephil   10/22/2012 5:08:45 PM
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Thanks John, I've just checked it out.  Did you notice that they only have line voltage drivers for them?  Trust me on this, once you start playing with constant current switching power supplies, you'll never look back at those burning hot power hungry resistors and linear regulators.  ...and in spite of what I said earlier, do take  your time and get a proper BSEE from an accredited university.  There are all too many people out there who would take advantage of your ingenuity to the detriment of your youth and education.  Best, -Phil

dbell5
User Rank
Gold
Re: Go John!
dbell5   10/22/2012 5:16:19 PM
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@John, regarding the resistor power dissipation:

You didn't include a schematic of your light, so I had to infer from your math bit.

It appears that you are wiring the three 10W LEDs in series, and estimating 9.5V for the set, or ~3.2V each. That sounds correct, so far.

Powering that off of 12V, you need to drop (12 - 9.5) or 2.5V, as you showed.

Then, you place the LEDs in series with a power resistor, and switched 12V battery.

If all the above is true, then the (composite) resistor will see a 2.5V drop, and pass 2.5A. P=IE, for 6.25 Watts total dissipated, or ~1.6W per resistor.

If the 40W of resistor was getting to 200°C, then there it seems like must have been more than 3A in the circuit. If you were dissipating 40 Watts in 1 Ohm, then: P = I^2R, and I = sqrt(40), or 2 sqrt(10), or 6.3 Amps.

+ o------/\/\/\----->|---->|---->|------o -  Right?

Dave

 

John Duffy
User Rank
Gold
Re: Nice work John
John Duffy   10/22/2012 5:49:35 PM
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@laserdudephil, thanks, but I generally, (except on DX), can only really buy stuff from stores, and constant current switching supplies arent exactly cheap or common.  I just used resistors becasuse radioshack carries them for about $1 each. 

 

As for the power dissipation of the resistors, they are sinking a lot more than 1.6W each, and it is definately not much more than 3A, or else the LEDs would burn out.  Also, I'm not sure if heat is solely dependent on wattage, I believe current is generally more important (not sure though...) since high voltage at low current (estimated 100,000V@~20mA) should be about 2000W, but the resistor didn't break a sweat.   

laserdudephil
User Rank
Iron
Re: Go John!
laserdudephil   10/22/2012 5:57:52 PM
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@dbel5

It's actually even less than that, and it's 90W of power handling.  The Cree XM-L datasheet puts 2.5A at 3.25V.  (12-(3.25*3))V^2/1Ohm=5W.  5W/9=1/2W per.  Better get those resistors some airflow John.  You got them wrapped in insulation or something?

laserdudephil
User Rank
Iron
Re: Nice work John
laserdudephil   10/22/2012 6:02:29 PM
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Naw, the power being dissipated IS the heat being given off.  Remember that law of thermodynamics?  Energy shall neither be created nor destroyed.  Here elecrtrical energy (the power x the time it is applied to the resistor) is entirely converted to heat.

dbell5
User Rank
Gold
Power dissipation
dbell5   10/22/2012 6:38:37 PM
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(Let's give this thread its own title!)

"it is definately not much more than 3A, or else the LEDs would burn out"

I wonder. They can probably take dobule the nominal current for a while, and BTW, put out considerably over 1000 lumens each while doing so.

This could be explained by looking at a few assumptions. Your battery voltage is probably actually more like 12.6 V or even a bit higher, at full charge, which will drive up both the current the LEDs draw and the resulting power. The I/V curve for these devices increases pretty sharply! And - the LEDs are in series, correct?

As Phil commented, power = heat, and is not dependent upon either voltage or current more than the other. In the case of the 100 kV at 20 mA load, what's the source? It could be that what you are driving with has limited current capbility, and the terminal voltage drops drastically under a 20 mA load. Otherwise, you would be dissipating 5 Megawatts!!

 

Dave

Leigh
User Rank
Silver
Re: Go John!
Leigh   10/22/2012 6:57:59 PM
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@Rudy. DBell5 is correct.

RudySchneider
User Rank
Iron
Re: Go John!
RudySchneider   10/22/2012 9:29:25 PM
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Dave, John, et al ---

My humble apologies.  I just cranked through the numbers (Using E^2/R, 90W across 1-ohm means voltage of ~9.48V, and current of ~9.48V; break that into thirds, carry the zero, hold yer tongue right, etc...) and John (along with the rest of you) is correct!  Somehow, 30 years as a BSEE and all, I let my mind take over before doing the (full) analysis.

John Duffy
User Rank
Gold
Re: Go John!
John Duffy   10/22/2012 11:30:50 PM
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No need to apoligize.  It's not really a big deal.  Either way, it gave me an exceuse to stop doing homework every couple of minutes. 

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