Increasing the Output Amperage of a LED Boost Driver

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HyperSpeed

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I have accumulated quite the supply of flashlight drivers for LEDs. One which strikes me as very interesting is in a setup that currently drives 6 total 3V LEDs in series from 3S lithium.

OK so here is where things get muddy for me: I can lower the current-sense resistance of the driver to tell it to output more power. At some arbitrary amount near 0.135 ohms on the sense, the driver basically stops working. I assume I am exceeding ability of some component and potentially over-heating or instantly frying (LEDs are completely within spec power). It is OK, as I have many of these drivers and wanted to experiment with the first one.

What my mind is so far attempting to deduce is what component(s) on the board appears to be creating the power conversion from (input amps) to higher output volts--as the boost philosophy goes. It appears board components with position addresses beginning with 'Q' are N type MOSFETs. There are 4 on one driver.

What I realized is that I have a few newer PC motherboards headed for junk piles, and they appear to be gold mines for such scrap components as MOSFETs.

Is someone around who is able to confirm if in fact the MOSFETs are the component to upgrade, to increase amp output and thus total power ability? General ideas or guidelines of how you would go after this mod would be highly appreciated. I am able to reflow solder components, or hunt down whatever it may be, or upload photos of whatever you would like to see.

Thanks for any advice towards tackling this project! :wink:
 
Maybe I need to reword this a little.

In a "boost" driver circuit design, the voltage on the output is greater than the voltage on the input. What determines how great the difference between the two can be? In other words, is the driver's ability to boost voltage determined greatly by a single limiting factor, such as the rating of the MOSFETs?
 
Well, I know nothing of the drivers you have but you need a mosfet that can handle the voltage that you want to switch and then make sure the driver can drive said mosfet. Ap notes may help.
 
If the design is compact, it probably doesn't use any discrete FETs. At any rate, a PC board won't have any, power supply probably, but low current high voltage.

In a boost circuit, the FET's go between an inductor and ground, seemingly trying to short out the battery, which perhaps is a clue. If there is a special purpose IC, the datasheet could include the circuit. Possibly someone is familiar with the specific popular circuits/ICs, I know the generalities but would need to analyze the whole actual thing (or start over).
 
Is it all in one chip, or are there discrete parts. It is probably going into current limit. Trying to increase the current past that might blow the FET. Just use 2 chips in parallel!
 
Is it all in one chip, or are there discrete parts. It is probably going into current limit. Trying to increase the current past that might blow the FET. Just use 2 chips in parallel!
I bet you guys never thought I would return to this thread! I've had these things sitting on the shelf, I just pulled them back out to start building Ranger headlights and I did acquire some very high-wattage compact boost drivers that will fit in flashlight hosts (they can reach >150W each and I have 4 of them). The special headlight set using those will probably be a one off for my Dad's machine (I had to machine big solid copper heat sinks for all 4 drivers, then reflow the drivers onto the sinks. These other stock boost drivers though, they are fairly potent. I have got to them to crank just with sense resist mods and a healthy lipo pack feeding them instead of cylindrical cells. However, I would like to wring these drivers for all they're worth (which is about $15-20 ea.) I have 12 remaining, so I would definitely snatch parts from 1 in order to boost another boost driver--as non of the drivers will likely ever get used if I don't.

I took all the photos I could, then chose the best ones as far as detail, to crop down and host for everyone to see.

Here's what I noticed. 4 of the "FETs" are the same, Q1, Q3, Q4, and Q5. One looks like the others but the flat part at the front (I assume for heat dissipation) doesn't look the same as they others, it's not flat, but it has a slight U shape stamped from the very front. The PCB also says D3 (not D1 as I put in the photo, that was a mistake) there instead of a Q. In case any of that matters at least.

Here's the photos. What do I need to do exactly, parallel a certain FET, or multiples? Would I need to change inductance at all? I had to move the inductor so the PCB could be more easily seen. The center of the PCB where the inductor is attached at one lead, goes through the board to the positive spring.

Also the output leads aren't connected to the LED strings. It's normally 6 LEDs in series, 2 strings in parallel (12 LEDs total). But I will connect some new LEDs once I substantially increase the power output. I feel the one driver probably died when the cells dropped voltage, so amps probably went up on the input side. This time, I will power from a PSU, so voltage stays constant, and I will test the power increased over the LED string itself. Thanks for your help!

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Details....as in part numbers on the semi-conductors are the bare minimum to get this conversation started.
Without some idea about the actual circuit, nobody can help you beyond SWAGs.
 
Details....as in part numbers on the semi-conductors are the bare minimum to get this conversation started.
Without some idea about the actual circuit, nobody can help you beyond SWAGs.
I was really hoping you weren't actually going to say I needed those, the foreigners removed the SMD model codes on those with likely sandpaper.

I have multimeters, is there any way I can deduce what performance value they are in any way? Like putting a 1ohm resistor in series with a test load across one, and check voltage parameters, etc? Or would I need a scope? I looked up diodes that another guy said he kept burning up on something similar, the flyback diode. Looks to be a schottky diode. Anything we could even assume, at the least? Or... What if I temp probe each component as it warms up, then create a list of temp @ time after turn on, could that tell me which component would be the likely culprit, if it's in fact getting very hot?
 
I think you are using what's commonly called a "buck converter", which ups the voltage without losing that much amperage. They are generally very efficient devices. See this article: https://en.wikipedia.org/wiki/Buck_converter
Buck converter will drop the voltage (DC step down converter) A boost converter (DC step up converter) will increase the voltage. Both have a range where they wont work. i.e a buck will drop a 5v source from 4.5v down to 0v. A boost converter will increase a 5v source to 6v up to 16v.

If the LED string is 6x 3v LEDs (2.7v to 3.4v is typical), then it is more than likely a boost converter to go from the 3.7v 18650 LiPo (or 7.4v if 2 cells are used) to ~18v If it is 2 strings in parallel, then it should be fine. If the 2 strings are in series, then the boost converter wont source enough voltage to supply them both.
 
That circuit is a boost concerter, it stores energy in the big coil, and it generates a higher voltage, which is rectified bu the parts marked d for diode. Q's are mosfets, as you said.
This is about a 5w unit, judging by the size of the inductor. The ways to get more power thru it is higher frequency, which might not work depending on how the circuit is operating; increasing the size of the caps might help a small amount, but increasing input voltage, or the size of the coil , or the operating frequency is about all you can do.
Look over some driver chips in the package you have, and see if the datasheet recommended circuit matches what you have. People these days generally use the factory circuit.
 
R7,R8,R11,R12 are all in parallel. Only 3 are populated. The values are 0.68||0.68||1.5 which give 0.26 ohms.
https://www.allaboutcircuits.com/tools/parallel-resistance-calculator/

The typical voltage reference for the regulator chips is 1.25V. So, the current is likely set to I=V/R = 1.25/0.26 = 4.8 amps.

To try 6A, for example, you'll need R = V/I = 1.25/6 = 0.21 ohms.
Close enough to this would be to add a 1 ohm resistor on the empty R12 pads. P=V^2 / R = (1.25)^2 / 1 =1.56 watts. So, a 2W resistor at minimum.

It looks like a 2512 size smt package, but measure it and compare to the standard sizes.
Something like this might work:
https://www.digikey.com/en/products/detail/te-connectivity-passive-product/RLP73K3A1R0FTDF/4032093
Or higher wattage:
https://www.digikey.com/en/products/detail/stackpole-electronics-inc/CSRT2512FT1R00-UP/14307403

Order a few sizes from 0.4 to 1.5 ohms.

The unknown is the max current, power, and heat sinking for the rest of the components.
 
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