Hi Folks,

I have a cruddy bike torch which came with a Luxeon LED in it. It's not very bright, so I went by the luxeon star website in search for a better "globe".

I found one that sounds much better (lots more lumens), so I brought it. But then in a panic I emailed the folks there to confirm it would be a straight swap, and they said no because of the forward voltage.

They are both rated at the same current draw (700ma) but one is rated at 3.9Vf and the other at 9.45Vf

My question is, what does this mean, does it need more volts ? My battery delivers 11v @ 2200mAH ... so could it be as simple as swapping a resistor to allow more power to drive the brighter LED ?

Tim.

 

That is pretty much it. I wonder at the 11 volts out of the battery (12 or 13.2 is more common), but the resistor calculation is simple. You subtract the forward drop of the LED from the battery voltage, and get the resistor value by the formula R = E/I.

For the 3.9 volt LED, there is 7.1 volts left pushing current, and the formula gives 10 ohms (actually a bit more, but 10 is a standard value). The higher drop gives 2.2 ohms, another standard value.

There is a resistor power issue, too. The power dissipated in a resistor is I^2 * R. I^2 is .49. The 10 ohm resistor will have to dissipate 5 watts of heat, the 2.2 will have 2 watts dissipation. And it's always safer to get more dissipation than you need, so use 10 watts for the one, and 5 for the other.

 

Thanks. Its getting clearer (mind the pun).

Now I just have to pull out the multi-meter and figure out this circuit. Its a bit tricky because the torch has a button that changes between 3 modes, steady low beam, steady high beam and a flashing mode. So I presume theres a different set of resistors for each mode, and the chip (which I cant read) is probably something like a 555 timer for the flashing mode.

For reference, here's a pic of the circut. But thats getting off topic for this thread I expect.

Many thanks again for the response.

Tim.

 

exie, when i spotted the toroidal inductor under the pcb, i guess the whole circuit is driven by a microcontroller (the 8 pin ic), and the current (and voltage too?) is regulated in switching mode to preserve energy.
if that's the case, then you should either accept the condition as is (maybe it'll glow brighter, maybe darker), or try to adjust the regulated current by pin-pointing the right resistor used in the current (and / or voltage) sensor feedback.