Ancient Luxeon III LED question

[elusor]

Greetings. Please, treat me as a complete idiot when it comes to LED and battery mods. I'm learning so much, yet, so far to become somewhat of a amateur LED light tinkerer.

I have an ancient Blackburn X3 headlight system that the battery pack finally crapped out. Well, as an idiot trying to figure things out, I use my rotor tools, split the case apart, and examine the inner working parts. I read previous posts about the X3/X6/X8 mods, but not much on the battery pack connection to the existing circuit board/driver itself. So, out of curiosity, I took the same 5 NiMH, hooked it up in series, took the 4-pins battery pack connector, did the exact hookup as the original layout, and the result is no power. So, I believe the driver for this system is dead.

So, out of curiosity, I grabbed 4 of the NiMH battery, shoved them in the 4-battery housing in series, stripped the wire cable going into the LED (white/red/black), where red is the main power, black is ground, and white is :confused: I'm presuming it's designed for a power step down of some type to toggle between High/Med/Low/Flash. Ran the battery pack directly to the LED with power source pos to red, power source negative to black. LED lights up. A couple of question:

1. 4 NiMH in series measured at 5.4V on the meter, and total of 2 forward Amps. Would the Luxeon III LED be okay in the long run just running the battery straight through like this? The LED is rated at 700mA, 6V? I can tell the light is a bit dimmer at high than before. Should I use the 5 battery housing pack instead of 4 and solder a 700mA resistor to the battery pack?

2. Since most headlights have toggle switch at the top, is the white wire can be used somehow as a toggle switch like before with a new circuit board, or rigging it up with some cheap mod for different light setting or am I screwed, and just use a on/off switch for the LED?

3. I got a 7.4V, 5.2AH lithium ion pack from a toy. Should I use a regulator, resistor, capacitor combo for the LED or should I attempt to purchase a multi-power sources driver for 6V at 1A for the lithium ion pack, as well as the seal lead acid 12V 14AH battery for the long cycling trek with my family ahead?

* I will purchase the Cree/SSP LED later to replace the one in the Blackburn housing like I saw earlier. That will be for the future projects once I'm confident with this one.

Any info is greatly appreciated! Like I said, I'm a moronic newb just trying to start out his tinkering in the LED and power pack areas.

Leo

 

[idleprocess]

1. 4 NiMH in series measured at 5.4V on the meter, and total of 2 forward Amps. Would the Luxeon III LED be okay in the long run just running the battery straight through like this? The LED is rated at 700mA, 6V? I can tell the light is a bit dimmer at high than before. Should I use the 5 battery housing pack instead of 4 and solder a 700mA resistor to the battery pack?

That will likely fry the LED sooner rather than later.

Most Luxeon I's & III's had a forward voltage of around 3.6V and a maximum drive current of 350mA (Luxeon I) or 1000mA (Luxeon III). If your battery pack @ 5.4V (likely hot off the charger; nominal voltage should be closer to 4.8V) can short 2A into a multimeter then it can easily dump 2A into that LED since LED's effetively do not have resistance.

Since direct-drive of LED's isn't an adviseable practice (unless your voltage close to the LED's forward voltage), you will need to limit the current somehow. A resistor is the simplest method if you plan to use a constant voltage circuit; since you're mentioning a variety of potential batteries, some sort of DC-DC "buck" circuit that can efficiently convert voltages in excess of the LED's forward voltage down to a more manageable level and restrict current may be the best method.

 

[Lynx_Arc]

Without really knowing the circuit in the headlamp I would say that as idleprocess said the LED itself is not going to like being driven beyond about 4v or so off nimh batteries. The circuit may include some sort of buck regulator so that the extra voltage is converted to current and regulated to drive the LED at safe levels. If the circuit is bad you will have to repair or replace it and use batteries that are applicable to the new circuitry. And also there is no such thing as a 700ma resistor nor is there 2A batteries. resistors are rated in Ohms and batteries are rated in ampere hours or Ahr.

 

[asdalton]

Resistor calculations for use with an LED and battery are simple.

R = (Vb - Vf)/Imax

where,
Vb = voltage of battery (or battery pack)
Vf = forward voltage of LED at typical drive current (usually 3.6 volts for white LEDs)
Imax = maximum current (in amps) that the LED should receive

Also check power dissipation in the resistor (P = R*Imax²) to make sure that you are not exceeding its maximum thermal rating.

 

[xul]

R = (Vb - Vf)/Imax

where,
Vb = voltage of battery (or battery pack)
Vf = forward voltage of LED at typical drive current (usually 3.6 volts for white LEDs)
Imax = maximum current (in amps) that the LED should receive

I respectfully disagree. Vf is dependent on I so the formula is more complicated. An LED is a sort of current dependent resistor something like an incand. bulb.

But, you could use piece-wise linear analysis for this circuit.

 

[degarb]

Many of us are young enough that 50 years from now, we will say, "I remember the first high powered LED called the Lux I. I gave out a whopping 40 lumens at roughly one watt." Sadly, our grand kids will pull out the latest Sears' Dorcy and say, "Yep, still selling them."