Salvaging ebay LED bulb?

[Vikas]

More like a challenge than saving the money; I have got lots of LED bulb so that I can easily throw this away but I enjoy debugging and fixing.

All of the 24 LED are in series. I checked few by applying 3.3V directly and they are working. Looking that circuit board and the components on it, it is straight forward full wave rectifier with 4 diodes and couple of resistors in series. Except the big filtering smoothing capacitors, everything is surface mounted. I am not able to measure the resistors (R1, R2) leading me to believe that those have burned out. I have no idea what are those two unmarked components.


24 LEDs in series drop the voltage by 80V and few in the bridge. The resistors don't have to take too much of the rest of the drop.

I have zero experience with the surface mount soldering or de-soldering. Buying new surface mount resistors would cost way more than the money I paid for this bulb!

I am open to suggestions.

 

[Kestas]

During assembly and manufacture of surface mounted devices, they are first spot-glued into place before wave soldering.

To desolder, I would use a solder sucker. They're inexpensive.

 

[AirgunSavant]

I know those bulbs all too well.
I bought a ton load and I have found very limited life from them.
I just replaced one that lasted about 60 hrs. Also those plastic vented ones I see
insects get in and it fries the bulb. Junk.
I also seem to remember the maker says that after 8 hours of continuous use they will fail rapidly
because they are not made to handle long durations of "on" time so a bad deal for a dusk to dawn light setup...

 

[Papa Bear]

I often use (when in a hurry) 2 small soldering irons at the same time.

Pops the component right off the board. A little solder wick cleans the pads

If it's not a double sided (multi layer) board then you may be able to drill a couple of small holes and install regular resistors in place of the SMDs

Even SMDs have some marking on them to signify value (small numbers eg 103)

 

[spackard]

Maybe use one of those Portasol or Weller butane torches, that has a tool that can
blow hot exhaust gasses (but not flame).

 

[mk378]

This looks to be a capacitor ballasted design. There is going to be a film capacitor in series with the power line before the diode bridge (leads come through near D2 and R1). The impedance of this capacitor at 60 Hz limits the current that can be developed on the DC side. After the diode bridge there would be an electrolytic capacitor near R2, it has the conventional function of reducing ripple.

R1 and R2 are 470 k ohm "bleeder" resistors-- their only function is for safety to discharge each capacitor when the bulb is off. Otherwise someone could get a shock touching the base of the bulb after removing it from the socket. The two 47 ohm resistors (unmarked above D1) are in series with the rectifier. They are very important to absorb voltage spikes. The impedance of the input capacitor is very low for fast voltage spikes, that will lead to damage if there is no resistance in the circuit.

The most likely failure mode here would be an open LED-- they are all in series, so one bad one kills the whole bulb. Or maybe open input capacitor, or a shorted diode in the bridge.

Overall, yes it's a cheap [censored] design that is likely to have a short life. The better bulbs have an actual switchmode power converter driving the LEDs with a well-controlled current.

You could apply power to the bulb and measure the voltage across the LED string, if there is a higher than normal voltage it is a bad LED, find which one and just bypass it.

 

[zzyzzx]

Originally Posted By: Papa Bear
If it's not a double sided (multi layer) board then you may be able to drill a couple of small holes and install regular resistors in place of the SMDs


+1 for the above comment.

 

[EdwardC]

Originally Posted By: mk378
This looks to be a capacitor ballasted design. There is going to be a film capacitor in series with the power line before the diode bridge (leads come through near D2 and R1). The impedance of this capacitor at 60 Hz limits the current that can be developed on the DC side. After the diode bridge there would be an electrolytic capacitor near R2, it has the conventional function of reducing ripple.

R1 and R2 are 470 k ohm "bleeder" resistors-- their only function is for safety to discharge each capacitor when the bulb is off. Otherwise someone could get a shock touching the base of the bulb after removing it from the socket. The two 47 ohm resistors (unmarked above D1) are in series with the rectifier. They are very important to absorb voltage spikes. The impedance of the input capacitor is very low for fast voltage spikes, that will lead to damage if there is no resistance in the circuit.

The most likely failure mode here would be an open LED-- they are all in series, so one bad one kills the whole bulb. Or maybe open input capacitor, or a shorted diode in the bridge.

Overall, yes it's a cheap [censored] design that is likely to have a short life. The better bulbs have an actual switchmode power converter driving the LEDs with a well-controlled current.

You could apply power to the bulb and measure the voltage across the LED string, if there is a higher than normal voltage it is a bad LED, find which one and just bypass it.


I think you're exactly right.

When you say that you can't measure D1 or D2, does it come up as a short or very high (over limits)? An open capacitor should still allow measurement of those components. If it's shorted, you could do some testing and remove the capacitors (since it's an easy through hole) and retest the resistors. While they're out, double check that the capacitors aren't dead. If D1 or D2 are dead, pop those out, put the caps back, and try again.

GL!

 

[cheetah50]

I'm not familiar with this bulb, but resistor burnout in a marginal design is very common. Please keep in mind that although the the Bulb socket voltage is 120VAC(RMS), the output DC voltage (Assuming their is no voltage conversion before hand) is sqrt(2)*120=169.7VDC, assuming a perfect sinewave, minimal ripple,etc.....

 

[Vikas]

Originally Posted By: mk378
This looks to be a capacitor ballasted design. There is going to be a film capacitor in series with the power line before the diode bridge (leads come through near D2 and R1). The impedance of this capacitor at 60 Hz limits the current that can be developed on the DC side. After the diode bridge there would be an electrolytic capacitor near R2, it has the conventional function of reducing ripple.

R1 and R2 are 470 k ohm "bleeder" resistors-- their only function is for safety to discharge each capacitor when the bulb is off. Otherwise someone could get a shock touching the base of the bulb after removing it from the socket. The two 47 ohm resistors (unmarked above D1) are in series with the rectifier. They are very important to absorb voltage spikes. The impedance of the input capacitor is very low for fast voltage spikes, that will lead to damage if there is no resistance in the circuit.

The most likely failure mode here would be an open LED-- they are all in series, so one bad one kills the whole bulb. Or maybe open input capacitor, or a shorted diode in the bridge.

Overall, yes it's a cheap [censored] design that is likely to have a short life. The better bulbs have an actual switchmode power converter driving the LEDs with a well-controlled current.

You could apply power to the bulb and measure the voltage across the LED string, if there is a higher than normal voltage it is a bad LED, find which one and just bypass it.

You seem to know what you are talking about. I would prefer not to do any live measurements. I have computer power supply with 3.3, 5, and 12V. I could check 4 LEDs at time to see if one of them is broken and then bypass it. I did not realize the unmarked ones are in series with the rectifiers. Aren't the capacitors in parallel? I think rectifiers are ok.

There was a time when I had paid megabucks for a single LED and that is why it seems to be so hard to throw away 24 LED board.

 

[Vikas]

Originally Posted By: mk378
This looks to be a capacitor ballasted design. There is going to be a film capacitor in series with the power line before the diode bridge (leads come through near D2 and R1). The impedance of this capacitor at 60 Hz limits the current that can be developed on the DC side. After the diode bridge there would be an electrolytic capacitor near R2, it has the conventional function of reducing ripple.

R1 and R2 are 470 k ohm "bleeder" resistors-- their only function is for safety to discharge each capacitor when the bulb is off. Otherwise someone could get a shock touching the base of the bulb after removing it from the socket. The two 47 ohm resistors (unmarked above D1) are in series with the rectifier. They are very important to absorb voltage spikes. The impedance of the input capacitor is very low for fast voltage spikes, that will lead to damage if there is no resistance in the circuit.

The most likely failure mode here would be an open LED-- they are all in series, so one bad one kills the whole bulb. Or maybe open input capacitor, or a shorted diode in the bridge.

Overall, yes it's a cheap [censored] design that is likely to have a short life. The better bulbs have an actual switchmode power converter driving the LEDs with a well-controlled current.

You could apply power to the bulb and measure the voltage across the LED string, if there is a higher than normal voltage it is a bad LED, find which one and just bypass it.

You seem to know what you are talking about. I would prefer not to do any live measurements. I have computer power supply with 3.3, 5, and 12V. I could check 4 LEDs at time to see if one of them is broken and then bypass it. I did not realize the unmarked ones are in series with the rectifiers. Aren't the capacitors in parallel? I think rectifiers are ok.

There was a time when I had paid megabucks for a single LED and that is why it seems to be so hard to throw away 24 LED board.

By the way, how did you see 470K and 47 ? I see the digits on the components but I could not see "k" or "0" etc.

 

[mk378]

Surface mount resistors are marked with the same coding scheme used on thru-hole resistors, but with numbers instead of colors. The first two digits are the value and the last one is the exponent. So "470" = 47 x 10 ^ 0 = 47 ohms and "474" = 47 * 10 ^ 4 = 470,000 ohms. Not the case here, but when there is a resistor less than 10 ohms, those are marked with an "R" as one of the digits. "4R7" = 4.7 and "R47" = 0.47)

It is fine to test the LEDs one or a small group at a time with low voltage, but be sure to have a resistor in series with your power supply. You don't need to drive them very hard just to see that they can light up instead of being completely open. Applying a constant voltage with unlimited current is likely to burn a LED out.

 

[supton]

Those LED's are likely at much higher than 3.3V in actual usage, if I had to guess.

Simple soldering on a board like this is easy. Use a pencil tip iron. I can usually work fast, heat one side, then the other, after a couple of trys the part will move.

 

[kc8adu]

bad led at 7oclock inner circle.
test with dvm in diode test.good ones light dimly.once you know polarity markings its easy to test them quickly.if it does not light swap probes.if still no light its a DED.

 

[Vikas]

Originally Posted By: kc8adu
bad led at 7oclock inner circle.
test with dvm in diode test.good ones light dimly.once you know polarity markings its easy to test them quickly.if it does not light swap probes.if still no light its a DED.
Wow; you are a genius! That exactly was it. Thank you and mk378 for giving great help. This is what makes BITOG great.



After jumpering around that LED, the light is alive!