3
\$\begingroup\$

I'm trying to reverse engineer a broken control box for some dual intensity motorcycle LED spotlights, and I note on the circuit board there is an AX 3007-50 12V input, 5V output PWM chip.

I assumed the 5V PWD output was to drive the low beam of the LED spotlight, and in a bid to simplify the new circuit I wanted to replace with a 5V linear regulator.

In my tests, this hasn't worked. I've been trying to understand why, and I'm thinking that the LED needs 12v to run, the 5V output from the PWM is 12V, but just cut up so it's only on for ~40% of the time. My constant 5v output is not enough to drive the LED, so it simply won't turn on.

Would this line of thinking be correct? I know the LED works with 12V, should I be looking to replace my linear actuator for a PWM setup like I think it used before?

Some pictures of the LED, and the PWM chip on the control board:

enter image description here enter image description here

\$\endgroup\$
  • \$\begingroup\$ No coils on the other side? \$\endgroup\$ – Sunnyskyguy EE75 Apr 1 '18 at 21:44
  • \$\begingroup\$ Looks like your LED is a 5050 size 3A 2.85V with a buck regulator using a coil and large Schottky diode in the middle probably operating between 0.5A dim and 2~2.5A bright. But it must have a coil to store the 12V current and stepdown to higher current. \$\endgroup\$ – Sunnyskyguy EE75 Apr 1 '18 at 21:53
  • \$\begingroup\$ It might have, it's certainly doing something as I have three wires coming into the LED unit, but there are four visible internally. Unfortunately, I don't have an allen key small enough to remove and look behind. \$\endgroup\$ – Ben Everard Apr 1 '18 at 22:05
  • \$\begingroup\$ Sounds like maybe the 5v PWM drives a FET, not the LED. It would indeed be strange to see an LED accept both 5v and 12v, as would using PWM with capacitors. A clue could be GND != LED-, which makes me wonder if there's not 2 arrays of LEDs in the COB, one in reverse polarity, like a bicolor LED. With a 12v PWM signal and no caps and some small resistors, it seems likely. \$\endgroup\$ – dandavis Apr 2 '18 at 1:05
  • \$\begingroup\$ @dandavis it's not a pwm ic, it's a switching regulator, with an internal 2 amp fet. \$\endgroup\$ – Passerby Apr 2 '18 at 2:07
1
\$\begingroup\$

The circuit is more complex than just a dc-to-dc switching regulator.. You need to sketch out the circuit to find out how it works. The circuit board is simple enough it's a single-sided board that you can trace.

That said, the LED is a simple 3.3 Volt or so multiple amp package. With the right voltage and current limiting device, AKA a resistor, it should turn on if you solder directly to it. If it's not turning on then it's due to your power supply not giving enough current. That said, a switching regulator is more efficient than a linear regulator. If you're dropping 7 volts at 2 amps that's 14 watts of power it's wasting in heat. Your regulator may be shutting off in thermal protection mode.

\$\endgroup\$
  • \$\begingroup\$ The controller is indeed two-sided (only pictured the side with the PWM on). I've been trying to trace the layout but in places it jumps from layer to layer (and underneath a larger 12v relay!). Thanks for the information, from my research on DC-DC conversion it seemed it was more efficient, I wasn't aware that'd be as much of an issue but it seems to make sense, I'll factor this into my next tests. \$\endgroup\$ – Ben Everard Apr 1 '18 at 22:19
  • \$\begingroup\$ I wasn't able to test this before the manufacturer responded to my request (see my new answer), and in doing so have confirmed that a 5V input won't work... so I guess I now have a spare 5v regulator! Thanks for your input. \$\endgroup\$ – Ben Everard Apr 3 '18 at 17:56
0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ if that's the circuit behind the lamp a DC signal can be used for dimming. \$\endgroup\$ – Jasen Apr 2 '18 at 0:37
  • \$\begingroup\$ Yes as the LPF filtered PWM results in a DC control signal \$\endgroup\$ – Sunnyskyguy EE75 Apr 2 '18 at 1:04
  • \$\begingroup\$ It seems there is a circuit behind the LED, unfortunately, when I remove the front-most PCB I don't have enough room to see underneath. The manufacturer has actually told me what I need to provide to operate the dual intensity functionality (see me other answer), does this marry with your answer? I'm a bit out of my depth so not sure if this is what you were getting at. \$\endgroup\$ – Ben Everard Apr 3 '18 at 18:06
  • \$\begingroup\$ I meant is there a coil on the bottom of green board. This is what i expect. Then the dual dim function is regulated by the pin called FB (feedback) with a slightly different Resistor ratio to dim the light and reduce current. \$\endgroup\$ – Sunnyskyguy EE75 Apr 3 '18 at 19:32
0
\$\begingroup\$

As well as asking this question, I had also asked the manufacturer for their input also. I wasn't expecting an answer but they did get back to me with information on what I need to provide the LED unit to operate the dual intensity functionality.

  • Red: 12 Volt Postive
  • Black: 12 Volt Negative
  • Yellow: Duty Cycle of 10% to 90%, 500Hz, 0-5 Volts 10% duty cycle will be 90% light output, 20% duty cycle will be 80% light output and so on. (100% duty cycle is off, ignoring the input will provide 100% light output )
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.