I saw Djozz you mention recoil hunting, so i thought you used lights on rifles.
(edited)
Jerommel,
Magnum centerfires are rifles. Magnum means badass rifles :laughing:
Lets stick to my primary task on this thread. Maybe you could help?
Do you know what resistor should I buy and where should I place it in a circuit of this Djozz FET driver? Pretty please?
Edit: Simon drivers has mode memory and is OFF time based which I don’t prefer. Old grandpa hunters likes simplicity and any kind of complicated UI is out of a question.
Oh, i see.
We here in EU only have cutlery for weapons, so i’m not familiar with gun / rifle lingo.
You mean it will change modes when you use it again within seconds, i guess.
No, i wouldn’t know what value resistor you would need.
Also, i assume you want to use the 1mm² white flat?
You should know it’s not worth driving it on 5 Amperes.
It maxes out on 4.5 Amperes and the difference between 4.0 and 4.5 Amperes is quite negligible.
I see 2 possibilities to best solve your question:
1] Don’t use fully charged cells, and forget about resistors and thin wires and crappy springs and what not.
You’ll have much more maximum output runtime when you start with 3.8 Volts cells, the discharge curve of the battery is much less steep under 3.8 Volts.
2] Get a Nanjg driver with 8x 7135 and the four stars to choose groups, and stack 3 or 4 more 7135 on top of the original 7135’s.
It’s not that hard if you have a fine soldering tip and some flux grease.
The trick is to
a) remove the middle leg
b) bend down the 2 legs
c) stick them on top of the original 7135’s with a tiny drop of super glue
(the bent down legs will help positioning them, but you can also first glue them in place and then bend down the legs, but then the glue may break loose))
d) solder bridge them to the 2 outer legs of the originals
e) bend down the thermal tab (which is also the middle leg)
f) solder thermal tab to the original’s thermal tab
The second possibility gives you safely regulated current for a considerable time.
Use fairly high drain cells (30T or VTC6 or similar) and optimize the electrical path to prolong regulated current drive.
You eliminate many unknown / inconsistent variables with 11 or 12x 7135.
You will never over drive the LED.
(I assume you don’t use a remote switch, with long thin wires.)
Jerommel,
Thanks on a help but please tell me something I don’t know! :person_facepalming: 
I have asked you simple question: “Do you know what resistor should I buy and where should I place it in a circuit of this Djozz FET driver?” If you know ok if you don’t know that is also fine… That was the question that demanded answer. It is ok if you don’t know that :innocent:
I sold over 1000 modded lights in mine life. I have a master degree on soldering AMC 7135 and 7138 chips to linear drivers and you really don’t need to cut middle leg. I solder all 3 of them even without bending.
I know what are my potential solutions but I refuse them since I have hundreds of AK 47 C1 drivers and if there is possibility for current restriction than I will try to mod them.
It is the most stupid thing to stack so many amc chips on small linear drivers cause it heats a lot and failure will obviously happen sooner or later. FET driver is much much better than any high current 17mm linear driver with hundreds of stacked 7135 chips cause it does not heat that much at all…
I also know how FET drivers work since I made hundreds of them this diy Djozz type and Texas ace one (without stencil; just soldering iron and calm hand).
If you want to run FET with Osram WF just use low current 18650 cell at any charge state and don’t bypass springs and that is it. Or add thick steel contact springs for example.
But this is not solution for the problem of how to restrict current in FET driver.
Potential solution in this below mention style but I would like to take that to higher level and make restriction on driver only… EasyB!
So any advice regarding my question should led to solution(or possible solution) how we can simply restrict current draw in our FET drivers? Simple resistor mod or something else? It is probably doable but we need electronic engineer for this.
Maybe youŕe not the only reader here.
Demand…
I gave you the answer to that particular question:
I don’t know what value you need (but i could calculate it if i look up the specs / test results), and it also depends on the battery you use, the health of the battery, how well the contact between battery and springs is, the condition of the switch you use, the ambient temperature, differences between individual LEDs, LED wires.
All uncertain variables.
Oh and of course the charge of the battery…
So they stand upright then, and have no decent thermal path to the driver.
Okay, i didn’t realize you’ve got hundreds of those 3x 7135 drivers.
I can understand you’re not planning to stack those up to 12x 7135.
It depends on how much work you put into it.
at 4.1 Volts (freshly charged with 0.1 Volts battery sag) it’s less than 1 Volt x 4.5 Amperes, is a maximum of around 4 Watts of heat the driver has to dissipate. With the thermal tabs soldered together and the driver sits tight with a alu / brass retainer ring in the pill or head, it will survive.
That’s true.
The maximum 4 Watts will be dissipated elsewhere in the electrical path, including inside the battery itself.
But it’s a series chain, so it’s the same as adding a resistor to the driver, but the resistor is approximately the same whether it’s hot or cold.
ok.
“So they stand upright then, and have no decent thermal path to the driver.” Hmmm 
Yes they stand above or “upright” soldered directly on their twin AMC 7135 brother soldered at 4 points: 3 legs and one back soldered so it is the most superior way of stacking 7135 chips with the best thermal path unlike your 2 legged one 
But lets stay on topic…
Edit:
“But it’s a series chain, so it’s the same as adding a resistor to the driver, but the resistor is approximately the same whether it’s hot or cold.”
When you sell thing to potential buyer you really don’t know which cell he plans to use so you must protect yourself.
Sorry, i don’t understand what you mean by ‘back soldered’ regarding the larger tab.
I can’t visualize what you mean.
Do you solder a piece of copper between the back of the standing one and the larger tab of the surface mounted one?
That would indeed be better than how i do it.
If i had to use the AK47 with 12x 7135 on it, i would give it a thin sheet copper wall around it, also soldering the PCB outer perimeter.
But enough about 7135’s
And sorry number 2 for forgetting who you were.
I’ve been away from here and the flashlight hobby for a while.
But now i remember.
And now that i know you decided to use the drivers you have in stock, i understand.
And yes, i understand your point about buyers.
But frankly that sounds to me like 1 more reason to opt for current regulation…
Single 1x18650 cell configurations(I use 1x18650 configurations only) will have even less runtime on high mode with regulated 5A driver than the one without regulation (FET Driver). Only difference is they are truly in regulation while FET not.
With FET driver (once when modded or just by using lets say Cree XP G2 S4 2B) will pull from 5>3.5A 4.2>3.3V of INR 30Q battery on high mode.
With mentioned drop 5>3.5A 4.2>3.3V LED will not loose much visually on performance and it should have even better runtime than regulated one since at certain point of battery discharge will have less current draw which again like I said visually will not have any drastically visual experience for user… Especially with Osram wf led emitters. I really don’t visually see any difference between 5A draw and 3.5A draw on mine white flat. Only when I put on lux meter I see the difference.
Not to mention that FET driver does not generates heat at all. Build like a tank(especially modded and potted Djozz style Nanjg AK 47 which has thicker PCB than any 17mm driver I ever tried or seen) . Virtually indestructible setup.
The maximum current is dependant on many factors, so a single resistance value can’t be calculated that covers all those factors.
However, if you know the maximum current your driver pulls as it stands, and the voltage across the LED at that current, it should be possible to calculate a value that reduces the current to 5A for this single instance.
As for where the resistor goes, in theory only, you either need to limit the current coming out of the battery(so somewhere between the battery out and FET in) or out of the FET (so somewhere between FET out and LED in) or somewhere in the negative path of the battery, LED or FET, though i’m not sure of the heat implications with regards to the different placements.
Changing the resistance path of the MCU would have no effect on LED current.
I would try using a fet with a higher RDSon. We have always looked for fet’s with a low resistance. If you trying to limit current to 5 amps search thru Digikey for a fet with higher resistance. Some experimenting would have to be done to get the right fet resistance but it could be done for a particular setup.
Thanks Marc E and moderator007…
So possible solution is adding a resistor in a circuit or different FET with a higher RDSon.
FET with a higher RDSon sound as an easiest solution once when we find proper FET.
Test subject should be high current cell (INR 30Q) and Osram White Flat emitters.
About resistor mod. I uploaded picture so I would like to hear your opinion guys…
Once when I find suitable resistor( I still don’t have it so help what to choose would be appreciated) where should I put it?
May not be the best way, but I would try adding one end to the diode pad and the other end to the 7135 center pad close to were it says Q3. Cut the trace going to ground before it gets to the pad with Q3 right beside. Make sure you cut it good and wide, dremel cut off wheel works good. This way you can solder on the resistor and solder on the led positive wire without anything being fragile.
So directly solder resistor from diode pad to Q3. We cut the trace that goes to Q3?
Look at this:
What about FET side? That would not work?
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.
.
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.
.
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.
.
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?