It could be mounted on the fet but only one end of the resistor would be soldered to the fet and the other end to the negative wire basically in mid air. That is a little to fragile for my liking unless your going to use a large resistor 1206 or larger. Personally, I would prefer that the resistor can be soldered on both ends to the board and wire soldered to it. The resistor is in series with the led, doesn’t matter which end you connect it to the led.
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In your pic where the red cut line is, from there back to the 7135 center pad you can strip the masking to fit a larger resistor if needed. Move the red line closer to the larger ground pad for the most clearance.
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All I have done here is give you a place to solder the resistor to the pcb pads for durability and isolate the other end of the resistor for the led output.
You are going to want to use a larger resistor since it’s going to be dissipating 2W or more.
Did you run the numbers to come up with that figure? 2 watts or more is going to take a huge resistor, might as well go with through hole resistor.
Yeah it’s a lot of heat. In the resistor mods I’ve done with the 2mm WF to limit to 7A it takes a 0.1 ohm resistor added which is 4.9W dissipated. It will be a bit less for the 1mm WF at 5A. I used nichrome wire but I had to get creative since you can’t solder to it.
Thanks moderator007!
Sorry I did not get you at first… My above posted picture shows + soldered to - which means puuffffff! Ignore pic above guys.
So this below should be right solution?
That is only 5mm distance. Can you please suggest me(link) suitable resistor for that?
Thanks 
Edit: Calculations!
Fully charged Samsung INR 30Q and White flat 1mm (CSLNM1.TG) pulls about 7-8A of current with FET driver
So at 4.2 v current draw = 7.5A
So if you can please calculate me right resistor to lower that to no more than 5A? Resistor needs to fitted into 5mm space like in picture above…
Thanks.
If that’s the case it might still be better to use a higher resistance FET then. I have not done any modding with the osram leds or looking up data.
What about using the old East-092 FET T70N03 or T70N02 or DTU40n06? East-92 4amp 17mm driver
These are probably still to low considering a 30Q can produce way more current than batteries of that time.
Experimenting is probably in order unless someone here can figure out exactly what specs in a FET you would need to acheive 5 amps with the Osram led using a 30Q battery.
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Or another idea might be to use a lower internal resistance battery like a NCR18650GA.
moderator007 we posted at same time… Did you seen pic above your last post?
EasyB… I have only 5mm distance… So at 4.2 v current draw = 7.5A
So if you can please calculate me right resistor to lower that to no more than 5A?
Maybe somehow reducing the gate voltage to the FET would be a better way. In either case something is going to have to drop the voltage and dissipate ~2W. You would have to mount the FET better; it would get really hot. Do all your FET drivers have the FET upside-down?
Here’s how you estimate the resistance to add. First estimate the resistance in your circuit now. 30Q has ~25mOhms, tailswitch has ~15mOhms, wires have maybe 5~mOhms, and FET has maybe 5mOhms. So with a fully charged cell the voltage left at 5A would be 4.2V-0.05ohms(5A)=3.95V. The 1mm WF Vf is 3.35V at 5A, so you need to drop 0.6V at 5A, so add a resistance of 0.12ohms.
Your last pic looks good but EasyB has tried a similar resistor mod and is saying the resistor wattage would have to be 2 watts or more. Thats a lot of wasted energy as heat.
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There’s alot of other factures that are coming in to play here but using a online calculator with a battery voltage of 4 volts (battery voltage 4.2v will drop once load is applied 4.0v or less) and a led vf of 3.4v comes out to 120mohm with a 3 watt rating. https://www.amplifiedparts.com/tech-articles/led-current-limiting-resistor
You still have other resistances that are not calculated in that figure. Guessing 100mohm at a 3 watt rating would work close. Might be best to use 3 1206 stacked in parallel at 300mohm to get to 3 watts if using the pads. If your solder it directly to the FET you could use a much larger resistor (2512) and only one needed. https://www.digikey.com/product-detail/en/bourns-inc/CRA2512-FZ-R100ELF/CRA2512-FZ-R100ELFCT-ND/1775059
I would order a few values up and down to see which gives the best results. EasyB might be able to add some more info as what he did and used.
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Edit: EasyB beat me to it. Since the battery is only at 4.2v for a fraction of second you still figure that as the supply voltage?
Depends what you want. If you want to limit the peak current then use fully charged voltage of 4.2 or 4.15V or whatever it is. Or you could use a little smaller resistance and have it be a bit over 5A at first. It’s probably going to require some trial and error anyway.
Isn’t that drawn resistor connecting batt+ and batt- ? Instead of being in series with the leds??
Or was the idea cutting the trace at the red line and use it as an isolated solder pad?
Edit: never mind, that was indeed the idea :person_facepalming:
The driver design is in the first post of this thread, it will explain why all the FETs are upside-down 
Luminarium, given that you have hundreds of these to do i would look at Moderator’s suggestion of finding the right FET first, which would be like having a FET with the resisitor already built in which would save you SO much time and effort if you find the right one. You could then add a heat sink to the back of the FET so it copes better with the heat, much less fiddly than cutting traces and soldering resistors, and still as robust as your previous drivers.
Hi guys,
Yes Djozz. To cut the trace at red line… I really must do something so I appreciate any good idea.
Guys don’t worry about FET turned upside down. This is picture of Djozz diy basic “how to” work. Djozz thanks for the pics! Thank you for this driver!
Mine FET is permanently epoxied with serious 2 component epoxy from bottom side plus it is potted with super 2 component cold weld compound from above so there are no any exposed parts plus this driver does not generates much heat at all.
The copper mcpcb and LED above it generates heat which is dissipated by pill, flashlight body so potted driver as a pill part just perfectly fit into heat dissipation.
Hundreds of drivers done without any issue at first at harder way by de soldering 7135 chips but thanks to Simon that managed to get me Ak 47 C1 without any soldered AMC 7135 chips to make my work easier and faster.
If you order larger quantities and have right ingredients it can be done for 1.8$ 
Finding right FET could do. I would like to try both methods (resistor and different FET) but I will obviously have problem of choosing, ordering and getting right components for the job.
Here is a FET with 0.1 ohms on-resistance.
For the resistor method, you could sink a 2512 size resistor to the MCPCB.
Thanks! :sunglasses: Can you find me FET with legs? Sorry for bothering you 
but it will be almost impossible to improvise something with FET without any legs into this diy driver.
EDIT:
Found resistors source here
Should I order 0.12, 0.13, 0.15 ? Could this 3 types be enough for the job or should I try with lower values like 0.1 or 0.11? Cost like peanuts…
Size could be a problem 6.5mm…
You could do the trace-cut around the Q3-pad instead of before it, a bit more work but it gives you an extra 2 or so mm. Getting close to the edge though (possible short with pill).
Or put the resistor in the negative path by simply glueing and soldering it on top of the FET.
Or order large resistor and solder one end directly on a FET (direction of FET like in picture above) and other end left exposed for soldering minus black wire directly on it? It should hold especially if it will be epoxied?
Opinion?…
See my edit 