I'm not EE, so I would just guessing. I've put a question on DX forum.
Can you compare your driver with my pictures posted there. Is it more like stock DX pictures or is more like my pictures. There are some differencies between the two.
"This one is a bit tricky. The microcontroller turns on/off the QX5241 chips, most likely via the DIM pin. The DIM pin is pin #2, which is the bottom middle pin when viewing the chip with the "5241a" markings right side up. You could try isolating it and forcing it high, but then it will always be on as soon as you plug in the battery. The switch won't work at all, and it won't go into low when the battery voltage gets too low. Pin #4 (upper right) is a 5V LDO output, so you should be able to connect DIM to it in order to force it high, after you've isolated it. In other words, cut the trace going to pin 2, then connect pin 2 and 4 together.
If you want to only remove the modes, but preserve the function of the on/off button, then you will have to reprogram the microcontroller. I've never done that before. I don't even know which microcontroller this is. "
Can you get/use external switch? This would be the easiest way. Reprograming MC would be hard to do.
Hi! Yes i Can use an external switch! When you said “cut the trace going to pin 2, then connect pin 2 and 4 together” can you show mw in the driver where are the pin 2 and pin 4? and what i have to do to conect them?
"The DIM pin is pin #2, which is the bottom middle pin when viewing the chip with the "5241a" markings right side up
Pin #4 (upper right) is a 5V LDO output,"
You have to desolder (or cut off from the circuit) pin #2 bent it up a little bit and solder it to pin #4 with some tiny wire between them. See related PDF how the chip is constructed.
BTW, be aware your switch should stand current of 2-3A.