A beginner gets hooked. Can the current sensing resistor mod be applied to most led flashlight drivers?

Modestly enough, these two little headlamps at HomeDepot for 9.95 sparked my interest in leds and on to the web’s rabbit warrens of led applications.

Who couldn’t, after reading here for a few days, resist the urge to start taking things apart?
I just received a Trustfire tr-3t6, on its merits accumulated in various forums, since it was available for 28 dollars free ship. I don’t know what board I got or can attest to the lamps’ provenance, but its build quality and heft and threading and board build are just astonishing for 28 dollars. I’d like to resistor mod the board. There are three resistors as described in parallel on the board exactly as presented in this forum in images. One R100 and two R200 current sensing resistors. Then I’d like to go warmer with another choice of xm-l2. I understand that a warmer xm-l is less efficient than the T6. But I really do not care for the look of 6500K plus.

But I’m getting ahead of myself. First to cut my teeth on something even cheaper and just about disposable.
http://reviews.homedepot.com/1999/203409159/high-power-led-headlamps-2-pack-reviews/reviews.htm?sort=rating

Front side of board.

Rear side.

AAA Alkaline Batteries.

Replace the Alkaline with AAA Eneloop.
Little dab of thermal paste under the star.
Which are the current sensing resistors?
What value to mount atop one or more?
Would a mod appear to work on this circuit or is it purely trial and error?

I realize that this may not be a worthwhile modification by most of your lights, so to speak :slight_smile: , but it’d be cheap and easy and a good beginners trial. I actually use the light often for close in work and while working into the late evening in the yard now that winter is setting in and it gets dark so early.

Thanks for any input

None, none of those are current sensing resistors.

As far as I can see the circuit above uses a resistor in series with the LEDs to limit the current.

R5 is the resistor for the bright LED. To double the current, half the resistance of R5.

As a quick and nasty bodge you could de-solder R1 and solder it on top of R5. This would almost double the current to the LED.

Can the LED take it? Is there enough thermal sinking? I have no idea. Also you would lose the small LED in the middle.

R5 is just the resistor before the transistor(gate or base), to lower it would not make it brighter or am I wrong?

You are right.

I was half talking pish.

R5 controls the current flowing into the base of the transistor, not the limiting the current through the LED. oops.

While my reasoning was wrong, the outcome would be the same. The current flowing into the base controls the current flow through the collector and the emitter. Doubling the Base current, should double Ice.

Without looking at the datasheet for that transistor I don't think we can really accurately determine that. While yes, a higher current on the base does allow a higher current on the emitter side of the transistor I'm not sure that is fully applicable here. As I understand it most transistors have a beta value of ~100, this means if you supply 1 ma to the base then the emitter can put out 100 ma. The question is whether that applies to this driver. It could very well be that with this driver the limiting factor as far as the current goes is the LED (it will only take in so much current at a given voltage).

I would say the most sure fire way of doing this is to direct drive the LED off of the battery, this should be fine since the AAA can not supply very much current without massive amounts of voltage sag, something that I would guess is already negatively affecting the brightness of this circuit.

Looks like that XB-D LED is pwm driven by MCU.

Q3 (Y2 marked) transistor drive LED. its PNP transistor. SS8550. rated for 1A
There is no current sensing resistor used. LED is direct driven. but LED drive current depend on max PWM duty, Transistor drop, battery voltage and internal resistance.

To improve LED current:
My idea is add AMC7135s (3 for 1A etc…) to the LED (-) line and connect collector out of q3 to Vdd pins of 7135

I know this is rudimentary stuff, but just to be certain……

Is this right? Do I “add” each AMC7135 Vdd to each led neg, or disconnect the - led first before connecting the 7135?

you should disconnect LED neg first.
connect led neg to 7135’s out.
connect led pos to +B

Thanks again for that. +B = POS Battery directly on the board?

Don’t know why I put three leds in that drawing. I had the 3t6 on my mind lately and had been gathering ideas about it.

LED pos () to battery pos () / +B marked in board
LED neg (-) to 7135’s output pin

Or just stick R1 on top of R5.