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I have this LED light that was over the aquarium at my work desk. There's a circuit board (not pictured) that controls when the lights go on and off but, after a power outage a few weeks ago, it seems like something got damaged because it doesn't cycle the lights.

I thought I'd just rewire the lights directly to the power supply (12V at 1amp) and use a plug in timer to control the on/off cycle. The problem is that, when I had wired them up like that, they'd all power on and then about 10 seconds later, they'd all go off. Then a second or two after that, they'd all go back on again, and then off a couple seconds later.

Was it due to heat? Maybe the circuit board had them controlled to run at a duty cycle lower than 100%? Or maybe the circuit board, which has a backlit LCD display, consumed a significant amount of voltage and now, with it removed, the lights were getting too much?

Or maybe I had wired them up incorrectly. I had the W and B circuits powered in parallel but when I rewired them in series, only a few of the lights came on at very, very low brightness.

What is happening to the lights?

Picture of Light with SMD detail Circuit Board

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    \$\begingroup\$ Most likely it is the heat. Your board apparently has developed a micro-crack in the major supply trace,or the ballast resistors (R1/R2) have a bad solder joint. The coefficient of thermal expansion for aluminium is 35% greater than that of copper, so this should happen sooner or later if traces are done in straight way. You can try an experiment by heating your board with heat gun, and then apply the voltage. \$\endgroup\$ Nov 21, 2018 at 21:21
  • \$\begingroup\$ Yeah, try to gently bend the board in various directions, see if that makes it turn on or off. Emphasis on "gently". \$\endgroup\$
    – bobflux
    Nov 21, 2018 at 22:06
  • \$\begingroup\$ @Ale..chenski how does a power outage cause a bad solder joint? No, it is not a crack in the board either. And no it is not likely heat, these are not high power LEDs. How did you come to the conclusion that R1/R2 are "ballast resistors"? The power comes in on the other side of the PCB. 10 and 100 ohm quarter watt ballast resistors? The W and B LEDs are driven with CC drivers. The LEDs are likely wired 3 series LEDs in parallel 3S3P and 3S4P so there would need to be more than 2 current limiting resistors. \$\endgroup\$ Nov 29, 2018 at 6:02

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UPDATE

The controller board has two Constant Current drivers, but both are not driving with the same current. R12 and R14 are not the same value as R15 and R18. These very likely are the current adjust resistors. These two drivers are are most likely for the W and B terminals labeled LEDA and LEDB on the PCB. They appear to be buck step down drivers (inductor on output), so if the power supply is 12v the string's forward voltages must be less than 12V.

I was thinking the Moowell chip is likely the drivers for the RGB LEDs. But the RGB LEDs look to be WS2812 with its own PWM shift register driver. there should be a microcontroller with two wires for the serial data.

A pic of the flip side of both boards would help. Where do the outputs of the Moowell chip go to?

Check the VCC voltage, and then voltage on the track between the 7805 (U5) and C3, it should be 5V. The 7805 or the VCC power supply (Red and Yellow wires?) are the most likely components to cause a catastrophic failure. My money is on the power supply failed and not the controller board.

END OF UPDATE


The following is all based on guesstimates.

The on and off must be a protection circuit (e.g. over current) kicking in on the power supply. Then it goes into recovery and starts working again. Then the protection kicks in again and the cycle repeats.

If you still have the on/off controller post a pic of the PCB and list the part numbers of any ICs.

The on/off controller likely had a couple of constant current drivers. One each for the white,D1-D18, (W+, W-) and one for blue, D19-D39 (B+, B-).

You also have the older 6 pin WS2812 RGB LEDs with the builtin shift register PWM driver. But I do not see the 5VDC supply or the serial data line for programming them. Maybe that is four black wires on the left hand side?

...I'd rewire the lights directly to the power supply (12V at 1amp)

This was the incorrect thing to do.

Both white and blue LEDs have a forward voltage of about 3V. If powered with a 12V supply they would be wired in sets of 3 or 4 LEDs with a current limiting resistor. There are only 2 resistors 10Ω and 100Ω 1/4 W so they could not be current limiting and the W and B terminals must have been driven with a constant current (CC) source.

My guess is there is two strings of nine white LEDs and one string of 12 blue LEDs and the two LED drivers were 36V or 40V drivers. The string could be broken down into smaller series/parallel strings with 3 LEDs per string and a 12V driver, but doubtful as many LED driver ICs are 40V max. That is IF D19-D30 are blue LEDs and W and B stands for white and blue.


What to do?

Put a 5Ω, 5W power resistor in series with the supply to the W and B terminals. Measure the voltage between W+ & W- and between B+ & B-.

I am expecting about 36V across the B terminals and 27V across the W terminals. Or about 9V across either or both.

Follow the PCB tracks and see if the LEDs are serial string of 9 and 12 or sets or parallel sets of 3 serial LEDs.


If strings of 9 and 12

The strings can be driven with a 40V CC driver.

For the white you could try a Mean Well LDB-300L 40V boost driver ($5.85 at TRC). If there are two strings of 9 white LEDs in parallel they would each get about 150 mA.

It would be better to test the LEDs with a 48V supply and use various resistors to gauge the current needed before buying new drivers.


If parallel sets of 3 LEDs

Use your 12V supply with various resistors to find the current through for single string of 3 LEDs.

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  • \$\begingroup\$ I updated the question with a picture of the circuit board. I haven't gotten around to playing with adding resistors yet (due to the holiday). \$\endgroup\$
    – Rykara
    Nov 28, 2018 at 2:59
  • \$\begingroup\$ I updated my answer. \$\endgroup\$ Nov 29, 2018 at 5:44

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