nice - I got the exact same driver in a 3x XM-L2 light
but not any heatsink at all in mine 
so I didnāt play with the driver
It is very little... nanoamps of leakage once charged, but the pulldown resistor will also consume some. Unless a huge resistor is used on the gate, I think that the voltage drop there is negligible.
Ah, thank you RMM.
I could be wrong though, as it is only a slightly-educated guess. The only way to know for sure is to get out some testing equipment, control the variables, and figure out what is really going on.
Thank you all for your responses.
I have compared the output before and after to my 3x XM-L Fandyfire, and it still compares pretty much equally, with a slightly dimmer centre spot, but more overall flood.
As you rightly say though RMM, it is not an accurate measurement for sure.
If we (hypothetically) accept that both the high and low modes are unchanged, what were the resistors doing before I bypassed them?
I thought perhaps I should post pics of what I did, just to be sure I've done it right...
That length of wire probably has significantly more than zero ohms; remember, these resistors you're trying to bypass are already very low, so whatever you use to bypass will have to be extremely low to beat the resistors. Pluck off one of the resistors, or scrape the solder mask to make your own 'pads' to solder to, and then use a flat piece of scrap copper sheet instead of a curly wire earthworm thing. Or at least lay the wire down flat between the pads, make it as short as possible.
I just bought this light and am waiting for it to come in.
I too have read that this light seems underpowered and I agree it is not due to the LEDs, but the driver
Looking at the specs from DX, the current is 2800mA which is 400mA per LED which is great for long term use. This is only about 50% of the 700mA rated.
http://flashlightwiki.com/Cree
While bridging those resistors seems like a good idea, your experiment leads me to believe that the PWM is ultimately the limiting factor for the current. Your short is in the mOhms while those resistors are in the 3.25 total. Not much more room to move that.
Changing the 1k resistor wont do much either as it is a mosfet, a voltage device and since there is no voltage divider, the FET, according to the datasheet, is being driven in saturation (fully ON) mode.
The chip doesnt look like it has any feedback and is more or less open loop PWM, which is a good reason why the current is so low as to protect the LEDās. The only way to increase the light is to either bypass the PWM and add current limiting resistors. Or the second way is to replace the IC that will monitor the current provided. This conclusion still has to be confirmed, probably when I get the flashlight I can scope it.
I have a couple questions which may help:
- What is in R9?
- What is in C3?
- Can you image the bottom side?
On a side note, for short term use, you can pump up to 200% of the operating current (really depends on the ambient) for even more light, so theoretically, we can pump in 1400 x 7 = 10A @ 3 ish volts = 30W : )
It's a very plain direct-drive-with-limiting-resistors driver. To check for PWM being less than 100% on high, that's easy, just temporarily jumper B+ onto the FET's gate, and see if the output current goes up or stays the same. If it goes up, then the FET's not being fully switched on as a ham-handed way of limiting the output. If it doesn't go up, then the limit is coming from somewhere else - junk batteries that can't keep the voltage up high enough to push more current through the LEDs, a crap FET (first step really should have been to track down exactly what the FET is, and find the official datasheet on it).
It's a very simple circuit, it can be figured out with very simple diagnostics.
He needs to remove that wire bridge mod first or he could fry led. Fet data sheet was found in post 2 or 3.
Yes, that's why I was careful to include the word temporarily in 'just temporarily jumper B+ onto the FET's gate'.
I would just bridge here with a small wireā¦ Should bypass the pwm for a quick test. You could even keep it like that if you just wanted high all the time too. Thatās how I got past modes on a next memory driver.
No... that will bypass the FET entirely. What needs to be figured out is if the signal fed to the FET is the issue, or the FET itself. Bridging the FET's source & drain just makes it essentially the same as if you replaced the driver with a contact board with the LEDs connected straight to the batteries.
But isnāt that pretty much what it is at 255 pwm? Just to see if the pwm is set right on high. Heās bridged the resistors I assumed (yeah I know) it was to go direct drive. Iāve done this with my 15 led light.
But overriding the MCU by jumping B+ to the gate is very different than bypassing the FET altogether. In a light that has no clicky switch and therefore no way to turn it off with the FET bridged, that's not really an option and it's not a test that will help figure out anything about how to get more current to the LEDs while still having a usable flashlight.
Oh I should have read a little more, I thought it was a tail switch 
Yeah donāt do that torch! I should stick to the misc section 
Hi comfychair, if the resistors are for current limiting purpose then we can bridge them directly and get direct-drive; while if the resistors are for current sensing purpose (hence itās called āsense resistorā?) which is usually seen on buck drivers then we cannot simply bridge them, is my understanding correct?
And is there a way to identify whether the resistors are for limiting the current or sensing it just by looking at the driver itself?
Thanks in advance.
I apologise humbly for subjecting you to the sight of my horrendous soldering ("curly wire earthworm thing" had me in hysterics! :bigsmile:). I shall remove it forthwith.
I do have some copper tape, but if the resistance is so low, would bypassing the resistors with that really make much difference to the output? I feel the discussion is moving away from that idea
I'm not seeing anything marked B+ on the board, do you mean the positive battery terminal?
The comment about crappy batteries could well be pertinent. Although these are fairly new cells from an OEM Dell laptop battery (fewer than 30 cycles), I recall connecting a 2 cell lipo to my 3x XM-L torch and it was noticeably brighter than with my lions.
I could try one on this torch, but I'm not sure how many cells to use. I have 1,2,3 or 4 cell lipos at up to 2200mah/50c discharge. I assume it would be 4 cell but I'm not certain if the 4 lions are in parallel, series, or 2s2p, and I don't want to fry the driver.
This is a direct drive driver, meaning it just connects the battery/batteries to the LEDs. Less-than-100% modes come from pulsing the LED's connection to ground, through the FET, the FET is essentially nothing more than a solid state relay. The LEDs are all connected in parallel, which means if you feed them more than about 4 volts they will pop whether you have that driver sitting inbetween batteries & LEDs or not. With a driver like this, what goes in is what comes out, it doesn't have any method for stepping down (or up) the input voltage, battery voltage has to be something the parallel LEDs can tolerate.
Poke around on the driver with an ohmmeter, you'll find that B+ on the input side goes straight thru the board and reappears where the red LED+ wire is attached. B+ and LED+ are the same thing.
bibihang: It depends on the controller used on the buck driver. Some are fine with bridged resistors, some are not, and there are so many variants and clones of variants that unless you can find a trustworthy datasheet for each particular controller, all you can do is try it and see what happens.
comfy is pretty much hitting the nail on the head, the only thing i would add is that the capacitor and the series resistors act as a lowpass filter for the pwm meaning that the even the peak voltage is reduced somewhat so not 100% of the battery is hitting the leds
looking at the xml datasheet, the resistors really have to go, but a dead short is kinda dangerous if you are going to pull the gate high on the fet. ledtorch do you have access to other resistors?
i would bridge k+ to any part of R5 to pull the fet high while maintaining clicky clicky