Lux I requires an 18ohm, 3watt resistor @ 9volts ??

Need to verify with the experts... Is this correct ?

I'm trying to use a 9 volt battery to power a 1 watt Lux Emitter. At 350ma of current and 2.95 forward volts...

I used a resistance calculator. It says I have to use a 2.2 Watt (3) resistor @ 18ohms !!

Does that sound right.? The 3 watt resistors are huge compared to a 1/2 watt even !

Is there any alternative ?

Thanks,
Bob

Sounds about right. Just use P=VI... Where V= the amount of the voltage that the resistor must drop (in your case, about 6V). And I=350mA. So P=6*.35, which gives a power of about 2.1W.

Alternatives? Put a few smaller resistors is series - make sure they all add up to your target of 18ohm.

Rather than resistors in series wasting all that power, put two led's in series so the power turns into twice as much light. Calculate the series resistance by adding together both vf's and subtracting from 9v. Better yet, use a Downboy regulator (or similar) instead of resistors.

I hope you aren't trying to use a rectangular 9 volt battery (like smoke detectors use). There is no way they can put out any where near that much current.

Greg

Budman231 said:

Need to verify with the experts... Is this correct ?

I'm trying to use a 9 volt battery to power a 1 watt Lux Emitter. At 350ma of current and 2.95 forward volts...

I used a resistance calculator. It says I have to use a 2.2 Watt (3) resistor @ 18ohms !!

Does that sound right.? The 3 watt resistors are huge compared to a 1/2 watt even !

Is there any alternative ?

Thanks,
Bob

Click to expand...

Hi Bob, see 5th reply of this threat :
https://www.candlepowerforums.com/threads/105380

greg_in_canada said:

I hope you aren't trying to use a rectangular 9 volt battery (like smoke detectors use). There is no way they can put out any where near that much current.

Greg

Click to expand...

Hi Greg,

Appreciate the reply. It is a Common 9V battery and when I measure the current, a failrly new battery show about ~3.7 amps and ~8.9volts. The Lux I only needs .350 amps to run normally. Am I missing something.. ?

I am fairly new at this stuff so it's entirely possible.

Thanks...

paulr said:

Rather than resistors in series wasting all that power, put two led's in series so the power turns into twice as much light. Calculate the series resistance by adding together both vf's and subtracting from 9v. Better yet, use a Downboy regulator (or similar) instead of resistors.

Click to expand...

Thanks PaulR. The idea about using 2 LEDs was the ticket. It got a bit more expensive (xtra $7 or so) than I was planning but using the 2nd Lux I brought the current to a managable level. I was able to use a 1/4 watt 10ohm resistor and use the extra power to (as Emeril would say) ... "Kick it up a Notch !!" It looks great and very bright too ...

BTW. Made an alternating blinker out out 4 Lux I White Lambs. Kind of looks like the WigWags on a cruiser... Now I can attach that to my pants in the back and watch the cars go way over to the other side of the road when I'm running at night.

Bob

You might find this link useful:

http://metku.net/?sect=view&n=1&path=mods/ledcalc/index_eng