First of all... yeah, I know, up there I found and read carefully about the topic, but since the last post posted there was complaining about someone elses resurrecting such an old thread, I decided to start a new one. I apologize for it. =P
The thing is, I was checking the wiring guide and I find it somewhat erratic... maybe... I don't know, just an idea.
As you probably all know, a LED acts, when forward biased, pretty much as a closed circuit. Near to zero resistance. Or at least that applies for regular LEDs.
Now, the Wiring guide says that for a Luxeon V led, ranked at 6.8V 700mA, a 1 Ohm resistor should be in series in the LED with a 7.2 voltage source...
If I recall correctly, Ohm's simplified law says: V/R = I. So, supposing a 7.2v unlimited current source, I'll be feeding the LED with 7.2 Amperes!!!!!!!!!!!!! ten times the nominal current.
So, I'm assuming either two things of this:
1. Luxeon LEDs DO NOT act like regular leds, i. e. they DO have an internal considerable resistance. And doing the math, I'm thinking of a 10 ohms resitance. Dynamic resistance in function of temperature for sure, but nevertheless always close to 10 ohms.
or
2. The battery pack of 7.2v referred in the wiring guide can output a maximum of 700mA at any given time. Which would be a weird thing, not to mention that would make the buckpacks and drivers almost completly uneccesary. So I'm going for my first guess.
So if luxeon leds already have their inner resistance, the resistors at the wiring guide aren't really that necessary.
Because anyway, 1 ohm or 3.3 ohm (signaled for 9v) are ridiculous values of resistors. Chances are high that a poorly done soldering from my part ends up giving me those couple of ohms.
So, I wish somebody can tell me about my deductions are right or completly wrong.
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