Dead driver replacement Conundrum

Hi Guys, a good friend bought a nice quality 12 LED flashlight,



... . . . . . the driver originally had 2 toroid inductor coils on it,



http://m4dm4x.com/2014/09/07/review-kinfire-k1200s/

^^^ link to the same light,

after a very short while the driver has died,
The 6 batteries are 2S3P, and the 12 Leds are 2S6P,
I assured him I could fix it easily :laughing: :laughing: ...
but i`m stuck as to which driver?????? …the LED V. and battery V. are very close!!
Thanks for any input

Well first of all you need to decide what current you want to run them at.

To get the claimed 8700 lumens from 12 XM-L2s, you'd need to run them at around 2A each, or 12A for the array. There's no way that driver was even close to that.

And with the depth of the toolmarks on the heatsink, you can't have very good contact between the stars and the heatsink, so you wouldn't want to push them very hard. There's also not enough surface area to dissipate the 50-ish watts of heat it would generate at that current, even if there is good thermal path from the head to the body, which I can't really tell if there is.

I'd recommend shooting for 6A or less for the array. Even at that level, the light would get uncomfortable to hold long before the batteries were drained. You'd also be pushing the junction temps pretty high at that point.

If those are genuine Cree parts (I wouldn't bet on it), then the Vf should be below 3V (6V for a 2S array) at that drive level, and any low dropout (LDO) linear driver should be able to push them to full blast until the batteries are pretty nearly dead, and the efficiency would be hard to beat with a buck-boost driver, which is the only type that can handle battery voltage both above and below the array's Vf.

Buck drivers can be built with LDO, and such a one could beat a linear for efficiency (or not, depending on how well it's made), but most of the buck regs I've seen for flashlights have dropout voltages of 1V or more.

It would help to know the actual Vf of the array at the desired drive level. Any way you can measure that?

It would also help to know what modes you are looking for.

If it were my light, I would either see if one of DIWdiver's LDO drivers would fit, or just put some resistors in to manage the current. It is a nice light, but it won't really put out any more light than one with 1/2 as many LEDs due to the thermal path. A "buck" driver will only marginally at best improve the run time, since Vbat is so close to Vf.

If my math skills are working today, (and as a starting point for you)

- assume 1 amp per LED string, so 6 amps total. (just guessing)
- assume a typical LED Vf of 3.3 volts, so 6.6 volts per string
- assume an average Vbat (under load) of 3.8 volts so 7.6 volts for 2 S
- 7.6 - 6.6 = 1 volts to push along 6 amps
V= IR so 1 = 6 x R
R is about .2 - .5 ohms range resistor setup

Power = I x I x R = 6 x 6 x 0.5 = 18 watts rating on the resistor bank

Maybe someone can double check my assumptions and math ?

Harry