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Shuyin_Cho
04-11-2010, 10:09 AM
Hi
I need some help, have been looking all over and cant find what im looking for. basically i have a lux III red and want it to be as bright as possible for my first build saber. i have a 2.2ohm 5w resistor and 2 x 700mA bucks, which will give me a brighter blade with 4aaa battery? also how would i wire the 2 bucks to get 1400mA, do i litterally wire them together then wire up the led, batt and switch as normal?
Please help :confused:

Do-Clo
04-11-2010, 12:05 PM
The connecting of two buck pucks together is covered in some post on buck pucks but in my opinion you would need a 1.5 ohm resistor with the 4 aaa pack for max brightness

Shuyin_Cho
04-11-2010, 12:46 PM
thanks, on the wiring guide it shows the 2.2ohm but i can try the 1.5 :-)
will this significantley affect the life of the led?

Kal El Rah
04-11-2010, 01:09 PM
thanks, on the wiring guide it shows the 2.2ohm but i can try the 1.5 :-)
will this significantley affect the life of the led?

Considering that most if not all LEDs go for about 100,000 hrs at full power and voltage, I don't think 1 ohm is gonna make any difference. Most of us will have passed away by the time the LED decides to bite the dust.:rolleyes:

Shadar Al'Niende
04-11-2010, 02:58 PM
And if not, LED's are relatively cheap ;)

cardcollector
04-11-2010, 03:14 PM
And if not, LED's are relatively cheap ;)

It's not the LED....

It's TAKING APART THE DARN SABER!!!:mad::mad::mad::mad:























Ok, so I had a bad LED burnout experience....:rolleyes:

Jedi-Loreen
04-11-2010, 03:22 PM
Sabers aren't hard to take apart to replace LEDs when you use MHS blade holders and quick connects on your LEDs. ;)

:p

Kal El Rah
04-11-2010, 08:15 PM
What J-Lo said!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!:mrgreen:

cardcollector
04-11-2010, 09:11 PM
Yeah, unless you're using the 1.25" bladeholder and you locktighted it to the hilt and locktighted the LED to the heatsink and you didn't have anydethreader and had to get it done by the next day or a birthday kids would be very sad because his lightsaber didn't light up and the parents would be upset because they payed a lot of money for a saber that didn't light up and you stayed up past midnight rewiring the speaker you broke off because you were wrestling with the bladeholder end and didn't think to watch the other electronic componets because you were running on coffee................



















:D;) (yes that really did happen...)

Skottsaber
04-12-2010, 12:52 AM
Yeowch! How'd that turn out for you?

FenderBender
04-12-2010, 05:12 AM
AAA alkalines won't put out 1500MA, so the resistor isn't going to do you any good. A PWM driver like a puck (or two) will convert some excess voltage to more MA giving you what you need.

Shuyin_Cho
04-15-2010, 12:25 PM
Thanks Fender. I have 2 700 pucks so i'll use those. This is my first diy saber so no doubt it will end up like frankensteins monsters will all the things i try out on it :-p would i be able to use non rechargeable lithium aaa without it burning out? its not really a big thing i can buy more leds just wanted to experiment with different things while im learning.

Jedi-Loreen
04-15-2010, 01:23 PM
Without burning out the LED? Yes, you can use those batteries. Buck pucks can take an input voltage of up to 32V, but they will only give the LED the voltage that it needs.

Shuyin_Cho
04-15-2010, 10:01 PM
Cool, thanks for the advise Loreen.

Skottsaber
04-16-2010, 07:58 AM
The only difference between Lithium and Alkaline non-rechargables is what stores the power. They are still 1.5v.

Shuyin_Cho
04-23-2010, 04:16 PM
Yeah they will be 1.5 (note to self CHECK :-p) it I run any higher voltage will the pucks still act the same in that it will only send the voltage needed to the led? In which case the extra voltage just gets waited and causes extra heat? Is that right?

Jedi-Loreen
04-23-2010, 05:13 PM
Buck pucks are more efficient than resistors, they don't bleed off excess voltage as heat. Your batteries will last longer with a buck puck than with a resistor. Pucks can take up to 32V of input voltage, but will only put out what the LED needs.