comfychair-inspired quick+cheap+lazy single-sided 17DD FET-driver (poor man's nanjg92)

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Jerommel
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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 12× 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…

luminarium iaculator
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Single 1×18650 cell configurations(I use 1×18650 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.

Marc E
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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.

moderator007
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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.

luminarium iaculator
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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?

moderator007
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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.

luminarium iaculator
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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?

moderator007
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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.

EasyB
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You are going to want to use a larger resistor since it’s going to be dissipating 2W or more.

moderator007
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EasyB wrote:
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.
EasyB
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moderator007 wrote:
EasyB wrote:
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.
luminarium iaculator
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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 Beer

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.

moderator007
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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? http://budgetlightforum.com/node/21657?page=3
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.

luminarium iaculator
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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?

EasyB
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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?

EasyB
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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.

moderator007
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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/...
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.
.
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?

EasyB
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moderator007 wrote:

.
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.
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luminarium iaculator wrote:
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 Beer

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.

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 Facepalm

Marc E
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EasyB wrote:
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?
The driver design is in the first post of this thread, it will explain why all the FETs are upside-down Smile

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.

luminarium iaculator
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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! Beer

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$ Thumbs Up

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.

EasyB
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Here is a FET with 0.1 ohms on-resistance.
https://www.mouser.com/ProductDetail/Vishay-Siliconix/Si7190ADP-T1-RE3?q...

For the resistor method, you could sink a 2512 size resistor to the MCPCB.

luminarium iaculator
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EasyB wrote:
Here is a FET with 0.1 ohms on-resistance.
https://www.mouser.com/ProductDetail/Vishay-Siliconix/Si7190ADP-T1-RE3?q...

For the resistor method, you could sink a 2512 size resistor to the MCPCB.

Thanks! Cool Can you find me FET with legs? Sorry for bothering you Sad 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…

djozz
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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.

luminarium iaculator
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djozz wrote:
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.

Or order large resistor and solder one end directly on a FET and other end left exposed for soldering minus black wire directly on it? It should hold especially if it will be epoxied?
Opinion?…

djozz
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See my edit Big Smile

luminarium iaculator
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djozz wrote:
See my edit Big Smile
Thumbs Up
luminarium iaculator
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Axial resistor vs SMD resistor

If axial can be used; one end could be bent and soldered directly to exposed fet side while we could solder minus wire to other exposed side of axial?

But if SMD resistor is better for some reason than I don’t want axial… Opinion?

Jerommel
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luminarium iaculator wrote:
Axial resistor vs SMD resistor

If axial can be used; one end could be bent and soldered directly to exposed fet side while we could solder minus wire to other exposed side of axial?

But if SMD resistor is better for some reason than I don’t want axial… Opinion?


You can cut off the legs and scrape off the paint on the ends and use them as SMD components.
luminarium iaculator
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Which one of those 2 kind can take heat better?

Edit: I bought smd resistors. 2515 package (0.1, 0.11, 0.12, 0.13, 0.15R) plus some ceramic cement resistors ones that looked good to me but guy has only 0.1 & 0.15 versions.

So that is total 270 resistors for 10$ Smile

Now I’ll try to search for higher resistance MOSFET with legs… Thanks.

Another edit: Those ceramic I bought seems to be larger than a whole driver. I saw that on one of user feedback.

However that ceramic one surely looks like it could be more thermally stable than regular smd one… IT would be great to find smaller version of that. There are no leg MOSFET with high resistance on ali.

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