Not sure if it’s right thread to ask. but is there any way to know which resistor is responsible for output current in below driver -
this is a from a 13w motorcycle headlight, designed for osram black flat leds (looks like it). output current is 1.5amp. but i want to increase it to 2.5-3amp and replace led to whiteflat.
Yes, that’s why I said “they”. I was refering to Oslon. In the the way Cree released an all parallel wired 3 volt xhp50.2, Oslon needs to release a parallel wired version running at 3 volts.
Granted, it’s not a very helpful post because it’s completely out of our hands. Does Oslon make a parallel version of that led package? I have no idea what leds they make.
I looked at the circuit and my guess were the same resistors. Look at the values. Those are the only ones with a really low value, 0.36 ohm. That is typical for a sense resistor. Plus the package size is about right.
The big ones labeled 2000 (200 ohm) are too high a value plus are probably related to the power input section (middle of the board). They are also right next to the main input wires. Since it’s from a motorcycle I assume it’s a Buck driver so you need the power input section to take the 11v-15v input and smooth it out, get rid of the spikes and ripple. The generator/alternator on a car or bike is creating AC current on 3 output wires (some small bikes might use 2 wires) which then goes to a rectifier bridge turning it into a chopped up DC voltage then to a voltage regulator. Plus voltage can vary with engine rpm. I’m guessing this led driver needs a cleaner, smoother input. I only see resistors and diodes so not much is happening here, just smoothing out as far as I can tell.
Then the input voltage gets sent to the driver side you need with High on one side and Low on the other. The 2 driver sides looks almost identical in design and components. Same inductors, same IC chips.
I guess input voltage still varies up and down and the drivers just compensate so they put out steady voltage and current to the leds.
This is my guess and I’m just learning these things. Hopefully Schoki or another engineer here can tell us for sure.
I don’t know if there is a way to calculate the results of a new sense resistor. You might just have to experiment. R9 and R10 look to be in parallel, same for R4 and R5. This means the pair are creating 0.180 ohms. If you replace one of the R360 for an R250 the total resistance goes down to 0.15 ohm. Try a small change like that with the new led and see what you get.
Keep in mind that driver may not be designed to handle much more current than it does. It’s possible you could damage it. So don’t make big changes.
AFAIK the Oslon was developed for “beam shaping”, I don’t think come in parallel. But the IQ-X embeds the Oslon so a parallel OSLON couldn’t be retrofitted anyhow. I looked into this long and hard because 3v drivers are so much more common. Also the Oslon is the highest lumen density of all LEDs, I do believe.
The manufacturer employed SK26A which is a 2A60V diode for this buck driver. I guess that the drive current is around 1A to 1.5A. You just replace the sense resisters to boost the current to 3A may not be possible. The ability of the inductor and the driving IC are also to take into consideration.
driver ic (A6213) is capable upto 2.5/5a depending on ic version used ) no idea about inductor.
Buck Switch Current Limit Threshold ( min - typ - max )
A6213 - 3.0min - 4.0typ - 5.0 A max
A6213-1 - 1.9min - 2.2typ - 2.5 A max (i guess i should go for 2.5a max)
i have only R180 resistor to experiment right now. will see how it goes. btw i was wondering if output current goes up/down linearly if resistor value is halved/doubled respectively ? or only way to find out is by experimenting ?
You definitely don’t want to take a big step like going half or double. I have only messed with sense resistors in linear and boost drivers, but not Buck. In just about all those instances slightly lowering the resistance saw a noticable increase in output.
This one driver had a 0.04 ohm sense total and put out 4.6A. Changing the value to 0.03 bumped up the current to 7A. So small resistor change equals big output change. It’s probably one of the reasons this driver uses two resistors in place of one bigger one. You can’t always get the exact value resistor that you need, so the manufacturer can use two different values combined to get just the right value.
Using one .360 and one .180 gets you .12 total. That’s a pretty big jump from .18
Maybe get a few .300 and .250 to go along with the .180 you have.
A .360 and .300 will get you .16
Two .300 will get you .15 (same as .360 and .250)
A .300 and .250 will get you .14
Two .250 will get you .13
A .360 and .180 will get you .12
A .300 and .180 will get you .11
A .250 and .180 will get you .10
Yeah, that should work. Buy some R300 and R250 in the correct package size and you can get whatever value you need.
Considering the driver comes stock with R360s, swapping them with R300s nets you a 20% current increase. Using R250s means 44% increase. Its an inversely proportional thing. With R220s the increase is 63.63%, and +80% with R200s. You may want to also swap the smaller schottky diodes in such a case, and even add some extra output capacitance. Ask the experts LoL.
So if he wants to go from 1.5A to 2.5A he needs about 65% increase. When you say “with R220s” you mean two in parallel which equals 0.11?
A .300 and a .180 would also work.
Hmm, just replacing one of the .360 with a .180 is pretty close at .12. That might work. Hopefully it won’t burn up. Lol
well, it does seem like it’s inversely proportional to that resistor. i found this pdf about the driver ic which is used. and it indicates same thing i guess same should apply for my driver too ? does this apply for all buck drivers ?
Wow that MP3431 datasheet shows 97% efficiency @1A which is around the current I need. Just need a driver with more evenly-spaced or configurable modes, and a ~3V cutoff voltage.
EDIT: Appears to be PWM not constant current :person_facepalming: