QX9920 Driver Board - Stock Sense Resistor = 1K Ohm?

I have a flashlight w/ a driver board that has a driver IC marked “LEDA”. As far as I can tell, it’s a QX9920? However, it looks like Pin 6 (CS) is connected to VSS through R5 (marked 102(1K Ohm)) and in series with another resistor marked R6 (R510).

According to the datasheet, I expected Rcs to be around 0.1ohm to 0.25ohm depending on desired current output. Am I looking at the wrong resistor? Is it really 1000ohm on this board? I measured 60mV on Pin 6 so i should be getting around 0.6A output?

Welcome to the forum turbo_slug. You are correct. That appears to be a QX9920.

Some drivers, like the HX-1175b have a fairly high value resistor (100 ohms on the HX-1175b) that is in series with the current sense resistors ("CSR's"). I've been meaning to play experiement with bypassing the series resistor, but haven't had a chance yet. I think it may be just a buffer to absorb spikes and such. If that is the case, I'm not sure I want one on my drivers.

Your current CSR of .51 ohms should yield about 2 amps (.51/.25) of current, if everything else is working correctly and you have adequate voltage overhead.

I’m pretty sure it’s not putting out anywhere near 2amps. Stock led appears to be as dim as a 1watt. Even with a Cree xm-l led, it’s not that bright. Let me measure the current through the led.

This flashlight has a very good body and battery system for supporting a cree xm-l but the stock driver isnt putting out enough current. My goal is to modify the stock driver to increase the current to around 1.5 to 2amps.

Sounds like you have a single XM-L in this light. What is the cell arrangement (2S, etc)?

You should be able to safely swap in a 100 ohm resistor for that 1000 ohm resistor and see how current is effected.

OK I just looked up the specs for this flashlight and it says it’s a 1watt flashlight but the heatsink looks beefy enough to handle at least 5-8watts.

I also just measured the current through the stock LED and it’s 0.370Amps which confirms the manufacturer 1watt spec is correct.

Battery arrangement is 3 18650 cells in series (it’s a tool battery pack).

Also, is anyone able to identify the transistor labeled WFAOH? I can’t seem to find any info for it.

Interesting. I wonder if the current if being further limited by PWM. You could short the Pin powering the MCU to the PWM Pin to simulate 100% "On" and see if that impacts your current.

Do you have a picture of the other side of your driver?

If it’s being limited by PWM, I’d expect another IC connected to pin 3 EN? Pin 3(EN) is connected to pin 4 (VDD) through R4 (1000ohm) so it doesn’t look like there’s any additional PWM going on?

I’ll have to take some photos of the othe side of the board. There are no IC’s on the other side. Just the inductor, resistors and few diodes that look like zeners.

I see, no MCU. I don't need to see a pic of the other side. I would think that the 1000 ohm resistor to the EN pin might reduce current if the PWM signal is "consumed" by the QX9920. I would think it would have to be "consumed" as I know that the QX9920 works fine even without a pull-down resistor on the PWM feed. If you look at the Data Sheet for the QX9920, it shows no resistor between VDD and EN in a similar set up to yours.

Regarding the component labeled WFAOH, I'm guessing you know that is your N-Channel FET. At least it's wired like it is one. I don't know anything about that particular chip though.


EDIT: Added clarification.

OK so I did some experimenting and piggybacked some resistors on top of R5 until I got the resistance down to 111ohm. Output current is still around 380mA.

What I did notice though is that if I bridge the ve led pin through the resistance of my finger to CS (pin6), output current increases to 0.5amp 1amp depending on how hard I press

Thanks for that report. Kind of confirms that the resister may just be some kind of buffer.

R5 is the one that interests me. It looks like more is going on with the EN Pin than just R5 though. I assume the through hole above the R4 resistor brings VDD to the Buck Converter's Pin 4. But below the resistor, it looks like there are a couple triodes that I'm guessing drain from the voltage that would go to Pin 3 EN. It's hard to tell with the shadows and all.

You may want to take a pic of each side of the driver with some back lighting to make the traces more visible.

OK, see photos below.




At a glance, R4 and that dual triode circuit appears to be a fancy voltage divider to reduce voltage to the Pin 3 EN. Seems like a lot of trouble they went through for that. If you measure voltage, you should find that voltage at Pin 3 should be much lower than Pin 4. This will definitely reduce current. You could just slice the trace between R4 and triode network and then then solder a lower resistance resistor on R4 or just bypass with a solder blog.

I haven't tried to look at the data sheet to confirm how those triodes work, but I'm confident that is what they are doing. Besides, the QX-9920 data sheet doesn't call for anything special between Pin 3 and 4.

I have a feeling the manufacturer put the cct in there as a way to turn off the led when the battery voltage drops below a certain voltage.

The battery pack does not have any low voltage cutoff circuitry.

^

Possibly. Could be a way to even output at various battery voltage levels. At lower voltages, the triodes might quit working and, therefore, not pull down voltage at Pin 3. Not sure without really studying it.

If it was my light. I would do what I recommend above. If it works the way I wanted, I would then pull all the components around the triodes as they may using some power. It appears that R1 and VD1 on the other side of the board also are part of that triode circuit. the R2 and VD2 also tie into the triode net work, but they are also feeding your Buck Convert.

By the way. If you think you will be modding more buck drivers, I would order some QX-9920's. They are pretty cheap and they can fail more easily than most of the other components in flashlight drivers. I just buy them from the lowest priced vendor on AliExpress.

Have you seen this English version data sheet? R5 and VD2 appear to be the equivalent of R1 and DZ in the data sheet diagram.

an R510 resistor is 51 ohms, not 0.51 ohms.

I suspect that the 1K/51 ohm resistor combo is used as a voltage divider to kludge up a low-voltage cutoff circuit. If the battery voltage falls too low, the pin will be low and shut down the light.

R represents the decimal point. marking 510 is 51 ohms. R510 is 0.51 ohms.

Hmmm… I just checked a R510 here and it is 51 ohms… ahhh, the wonders of cheap ass China components…

Old thread I know but found This similar driver from FastTech that also uses the QX9920(marked LEDA1514) with a sot 23-3 (marked A1FK) connected to EN through a 102(1k) resistor. According to the data sheets for the qx9920 it’s capable of 2.5A and pwm control via EN. I’m guessing that it’s possible to eliminate the 3-pin chip and 1k in favor of an ldo with an attiny and voltage divider Is anyone else interested in having a go at this as a project? As it is right now the FT driver can be sanded to 15mm and is single sided. The goal would be to come up with a 15mm two sided 2.5A max buck driver for Oshpark DIY with Attiny 13A mcu control. Here’s another circuit diagram similar to the one above.

I’m gonna high jack this thread for a bit since for now it just adds to what we know about the QX9920. When it comes time to design a board we can start a new thread. For a test rig I’ll need an attiny mcu and an LDO voltage regulator. Fortunately my parts bin had this still in it. It dates back to my first year here from This thread it was used to drive 3 xml’s from 3 cells in series. It’s basically a linear driver with the 8 x 7135’s mounted on a separate copper heat sink and a 5V regulator instead of a Zener/resistor. Just to be sure it still works I hooked it up to an XHP 50 and two 14500’s. Low, medium, high, no problem, so I desoldered the mcu board from the heat sinked chips since I won’t be using them. I’ve removed the 1k resistor and 3 pin chip from the buck driver and my next session will be to connect it to the mcu board(LDO out to QX Vdd and mcu PWM to QX EN. Both boards get B+/B~~, led+/~~ still come from the buck board.

Ok. I’m watching :slight_smile: