Your first picture shows a place labeled:

R13 - R12 -R8 - R5

On those places you will see resistors with the following writing on top of them:

R820- R820 -2R00 - empty/no resistor

(0,8 ohm - 0,8 ohm - 2 ohm - empty/no resistor)

Those are most likely the sense resistors, all are in parallel (you might want to measure that they are in parallel just to double check).

On R5, the empty spot, you could add say an R900.  If you want higher current add a lower value. An R900 should give you a good increase though. But I don't even know what this driver does stock..

As with all resistor mods that have not been tested. The lower the value, the higher the current. The higher the current the higher chance of destroying it... Some drivers does not mind a resistor mod, some does and can not be pushed that far. So if you follow this advice, do it at your own risk.

You might find this thread relevant despite that its a different driver circuit... 

Yes.. looks like R8,R12,R13 sense resistors.

thse 3 give 0.34 ohms sense resistor.

Do you know stock drive current of this driver ?

if it 2A, 2×0.34 = .68V feedback, then to get 3A, 0.68/3 = 0.22 ohms.

here we go.. I wrote simple diagram

Old-Lumens wrote:

Well, I see from reading threads here, that the driver gives different values from 2× 26650, 3×26650 and even 4C batteries, so I can only say that my plan was to use 2×26650/18650, which should give about 4 amps, 4.5 amps from what I read in the threads, like this thread., which has a lot of info on what people were getting for amp draws. I did not read amp draw with 3×26650, because I wanted to go with two and cut the body down.

 

in his kind of drivers, input current always higher than LED drive current. because it step up voltage to higher than input voltage. other way, input power = output power + convertor loss

to measure LED drive current, you need to disconnect a led and connect amp meter.

actually you dont need to do it now. simply add between 1.0 ohm to 0.68ohm resistor to exiting sense resistors. then check for any improvement. large current increase may not be possible without changing drive FET or add another FET parallel.

Old-Lumens wrote:

Pavithra_uk wrote:

large current increase may not be possible without changing drive FET or add another FET parallel.

Yes, I wondered, if the driver was capable, why didn’t others already do it? It seems most everyone just changed out the driver with a different one.

Thanks for all the information. I may try it. Changing resistors on that SRK driver was about the most nerve wracking thing I ever did.

What are is the code number on an 0805 SMD 0.68 Ohm resistor? Is there a place that shows the resistor values and matching 3 digit code numbers?

use this onlince smd resistor calculator

code:
0.68 ohm = R68 / R680
0.82 = R82 / R820

If stock driver designed to 2A LED current, adding 0.68ohm will increase current to 3A.
So its safe to start from 1 ohm resistor first.

Pavithra_uk wrote:

in his kind of drivers, input current always higher than LED drive current. because it step up voltage to higher than input voltage. other way, input power = output power + convertor loss

to measure LED drive current, you need to disconnect a led and connect amp meter.

actually you dont need to do it now. simply add between 1.0 ohm to 0.68ohm resistor to exiting sense resistors. then check for any improvement. large current increase may not be possible without changing drive FET or add another FET parallel.

Since the FET is only providing a path to ground, why would adding another FET in parallel increase current?

Or, are you just saying that if he WAS able to increase current, the existing FET wouldn’t be able to support that much current going through it (the existing FET), so he’d either need a different FET that could support the higher current or put an additional FET in parallel with the existing FET to split the current between the existing FET and the new, additional FET?

Just trying to understand…

2 FET’s in parallel shares the load between them, so they together could switch higher currents. The rds is also reduced, the same way two resistors in parallel have a lower resistance. So with a lower rds together, a higher current could be achieved with more current handling capacity. If they where the exact same 2 FET’s basically half the rds of one and twice the load capable.

Pavithra_uk wrote:

 . . . large current increase may not be possible without changing drive FET or add another FET parallel.

This is a huge.  Many of us have run across drivers we can't realized the desired level of current when even shorting the resister bank. Heat may be an issue, but at least the parallel FET would be sharing the current load .

Finding a compatible FET may be a challenge.  Many have the identification labeling ground off by the manufacture.  I guess one could find a compatible FET by measuring the physical size of the driver's FET and the voltage differential across the voltage sense resister bank.  Do you have any FET recommendations that would likely work in most drivers.  It seems that most are the same physical size (whatever size that is) and use a sense current somewhere in the range of .2 to .3 volts.

ImA4Wheelr wrote:

Pavithra_uk wrote:

 . . . large current increase may not be possible without changing drive FET or add another FET parallel.

This is a huge.  Many of us have run across drivers we can’t realized the desired level of current when even shorting the resister bank. Heat may be an issue, but at least the parallel FET would be sharing the current load .

Finding a compatible FET may be a challenge.  Many have the identification labeling ground off by the manufacture.  I guess one could find a compatible FET by measuring the physical size of the driver’s FET and the voltage differential across the voltage sense resister bank.  Do you have any FET recommendations that would likely work in most drivers.  It seems that most are the same physical size (whatever size that is) and use a sense current somewhere in the range of .2 to .3 volts.

http://budgetlightforum.com/node/21657?page=3#comment-473532
This one is a N channel FET some drivers could use a P channel FET so you may want to check before replacing.

Thanks Moderator007.  Sounds like a promising candidate.  One of the SRK driver variants out there uses N channel FETs too.  So maybe that FET is similar to what is used in most drivers.

Speaking of FET's.  The ones in the first picture of the OP were poorly seated.  Probably connected good enough though.

Looking at a SRK driver (version with a large capacitor and one inductor coil).  It has two N channel FET's, but I can't tell if they are wired parallel.  My labels are ground off, but GTamazing has one too and his is labeled as follows:

DTU
06N03
DD13U

I think the one I have reads over 6 amps at the tail.

Right now you are at .34ohms resistance.  If that gets you to 2 amps, the voltage differential is about .68.  So adding a R200 will get your resistance down to .126 and your current to the emitters up to 5.4 amps.

Do you know the size of your R200 resistors?  If they are 0805's, they can only handle 1/10th of a watt and would not probably be able to be used in place of the other resistors.

EDIT: If you want to be safer, you can first remove the 2 R820's.  Then you would be at about .18 ohms resistance and 3.75 amps current.  The best way for me to remove a resistor is to use a blob of solder that touches both ends of the resister.  It usually lifts off with the soldering iron.

Old-Lumens wrote:

Well, I’m not having much luck finding .68 ohm or .82 ohm 0805 resistors around here. I even tried ebay us sellers. Nothing unless I want to go to a kit for big bucks. I don’t have anything at home except for a couple R200 here, which won’t help and I don’t even know for sure if it will do what I want. I’m about to just use five of the NANJG drivers and call it a day, with a shortened tube and one of the Sony 50amp 26650 batteries. 15+ amps, that ought to fry eggs in a few seconds, in that light.

I just don’t want to order from China. I’ve got a hundred bucks worth of orders already aiming for 45 days from, several sellers and nothing can be tracked and none of them want to refund, so I will have to go the paypal route soon, so no more orders from China.

I can't tell without measuring, but the 3 resistors share the current.  So each gets 1/3 of the total current going through them.  They should be bigger because 1.36 watts (.68 volts x 2 amps) is going through that bank (if driver is delivering 2 amps stock).

0805 sized resistors measure 2mm by 1.25mm.

Old-Lumens wrote:

Well, I'm not having much luck finding .68 ohm or .82 ohm 0805 resistors around here. I even tried ebay us sellers. Nothing unless I want to go to a kit for big bucks. I don't have anything at home except for a couple R200 here, which won't help and I don't even know for sure if it will do what I want.

If you have a couple of R200 and want a higher value.. Set them in series..(not stacked directly on top of each other/parallel)

2x R200 in series = R400  Build a pyramid on top of another resistor (you might want to remove one of the stock resistors then).

3xR200 in series = R600  Build a house (with a flat roof) with three resistors

I hope you get the picture... 

another idea is put 3 or 4 R200 resistors in series in a piece of strip board and solder tiny wires from it.

That is an electrical junction box cover. I love to use things for other than what they were intended.

I find that commodity items are cheap and sometimes offer a cheap and convenient solution to construction projects.

I would have to wonder how much added resistance all that wiring adds to your circuit. You may not be getting an accurate picture of what the draw would really be. If you are only interested in a before and after comparison then maybe it wouldn’t matter so much.

You could remove one of the R820's and remeasure to see how much current changed.

think if batteries at 3.7v load voltage,

with 3 batteries = 3.7v x 3 × 2A = 22.2W
with 2 batteries = 3.7v x 2 × 3.75A = 27.75W

that mean higher the input voltage, higher efficiency.

Now LED current confirmed 1.5A so
voltage drop in sense resistors 0.34 ohm x 1.5A = 0.51v
on this bench test, you can solder regular resistors and check LED current improvement.
for 2.5A you need 0.51 ohm, for 3A its 0.33 ohms.

2A – 2.5A more practical than 3A.

I would not bank on 1.5 amps to the emitter.  Dchomak had a good point about all the wire and connections in the rig pictured above.  It's quite possible that the driver will deliver the rated 2 amps in the light.  Especially with O-L's skill at lowering resistance in a light.

EDIT: Added "not" above.

Wiring resistance can’t affect LED drive current because this is a boost driver. in here, boost convertor regulated to constant current. not voltage.

here simple test to prove that:
Add 0.5-1 ohm resistor series with LEDs
then check drive current again

Didn't realize that.  Thank you for that info.  In Post 26 above, don't you mean he needs about .2 ohms overall resistance to get 2.5 amps and .17 ohms for 3 amps?  Resistance is already at .34 ohms.

yes he need overall resistance of 0.2 for 2.5A.
exiting resistance is 0.34 so we need 0.51 additional resistor to make 0.2 ohm overall.