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I have some LED modules powered with 12 VDC. Each module has 3 LEDs and 3 resistors all in series. After working for one year, in some modules, one or two LEDs turned off but another one is on.

I wonder because all LEDs are series and must conduct current because one or two led are on but one or two are off.

enter image description here enter image description here

Why does this happen?


I choose three 27ohm resistors for sinking 30ma current if LEDs fail in a "short" mode, why when I increase input voltage, the off LEDs become on?

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    \$\begingroup\$ The question is rather why these have a CE mark in the plastic. \$\endgroup\$ – Lundin May 7 '19 at 11:53
  • \$\begingroup\$ 12 volts / (3 * 27) ohms = 148mA, not 30mA. \$\endgroup\$ – srl100 May 8 '19 at 8:39
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    \$\begingroup\$ @Lundin This only looks like a CE mark, but if you'd close the C to a circle, it goes right through the E instead of touching it. It's more like "China Engineering" rather than the CE Mark \$\endgroup\$ – Alexander von Wernherr May 8 '19 at 8:44
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    \$\begingroup\$ @AlexandervonWernherr The original question had a different pic but yeah you are probably right. At any rate, this is how you get yourselves banned from the European market, hopefully accompanied with a lawsuit given that this is a fire hazard. \$\endgroup\$ – Lundin May 8 '19 at 8:47
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    \$\begingroup\$ Can you double check that these modules have been "working for one year" as the date code on the PCB (2018/12) suggests that version of artwork may only be 6 months old? \$\endgroup\$ – srl100 May 8 '19 at 8:50
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I can't speak for the first photo, but for the second one - three resistors & three LED's suggests that it's one LED per resistor, so the LED's in each module are actually in parallel, not series. The modules are in series, but clearly the power & ground connections are pass-through.

Judging by the scorching, and the fact that the most scorched ones have the LED's out, I'd say it looks like the resistors have burnt out on the failed ones. The resistors are clearly under-sized for the power they are required to dissipate.

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  • \$\begingroup\$ Clearly, I'm sure that all components are in series. \$\endgroup\$ – Ali Abrishami May 7 '19 at 13:34
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    \$\begingroup\$ We disagree, but it's easy to be sure with a DVM. I suppose there could be three resistors in series to share the power, but I'd say it's unlikely. The fact that the modules with the most scorched resistors are the ones with LED's makes it obvious, IMO. \$\endgroup\$ – SiHa May 7 '19 at 14:39
  • \$\begingroup\$ i design pcb by myself \$\endgroup\$ – Ali Abrishami May 7 '19 at 16:52
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    \$\begingroup\$ Fine. If you designed the PCB, then I'll stop arguing, and say that the diodes must have failed short-circuit, as described in the other answer. Could you explain why there are three resistors, then? \$\endgroup\$ – SiHa May 7 '19 at 18:26
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    \$\begingroup\$ if you designed the PCB, why didn't you say so in the question? Why is there no schematic in the question? What thought went into using three resistors? Did you make sure heat transport from the LED and resistors were sufficient? \$\endgroup\$ – Marcus Müller May 7 '19 at 18:41
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Diodes like to fail in a "short" mode, and it basically means that your diode's chemical composition changes from being a junction between a P and an N semiconductor to being a "well-mixed through silicon and other elements" kind of conductor.

That happens when you overheat the diode, but not enough to make it explode.

When I look at your picture, I'd say I see multiple badly overheated arrays of resistors. So, you're probably not cooling these modules sufficiently or are operating them at a voltage that is too high.

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  • \$\begingroup\$ all modules power up with 12 volt DC. if I increase the input voltage, turned off leds, become on again. I think threshold voltage (forward voltage) of leds increase. \$\endgroup\$ – Ali Abrishami May 7 '19 at 13:39
  • \$\begingroup\$ DON'T DO THAT! That overheats LEDs. LED'S forward voltage doesn't increase over time - the photonic efficiency decreased, probably due to THERMAL DAMAGE, and you're making that WORSE by increasing the supply voltage. \$\endgroup\$ – Marcus Müller May 7 '19 at 15:14
  • \$\begingroup\$ According to a paper written by OSRAM almost all LED fail open due to heat and the bond wire disconnecting. \$\endgroup\$ – Misunderstood May 8 '19 at 1:05
  • \$\begingroup\$ @Misunderstood definitely not my or OP's experience. Link to paper or didn't happen :) \$\endgroup\$ – Marcus Müller May 8 '19 at 6:15
  • \$\begingroup\$ Could not quickly find the OSRAM paper (I have a copy somewhere) but this link says the same: Failure analysis of LEDs often finds that the LEDs fail due to de-bonding of the lens material from the cup or the surface of the die due to thermal stress resulting from a large thermal expansion mismatch between the lens material and the other materials used in their construction. Source: semlab.com/papers/failure-analysis-of-leds The OSRAM paper said explicitly LEDs almost always fail open. Show me your link that says they fail shorted. \$\endgroup\$ – Misunderstood May 8 '19 at 7:38
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The only conclusion I can determine is the LEDs have a very short life span. They could have been over driven (e.g. driving a 20 mA LED at 30 mA) or they are poor quality LEDs with low quality epitaxy layers.

Where two LEDs are out, the resistor PCB scorching is worse than the modules where one LED is out. Scorching is least noticeable where all 3 LEDs are still illuminated.

This would lead me to believe the Vf is decreasing (current increasing) as they deteriorate.

Accelerated ageing, i.e. LED efficiency loss within a period lower than expected life, is caused by adverse factors like low quality of epitaxy layers as well as, often, an excess junction temperature due to insufficient heat dissipation. Furthermore, penetration of humidity or other contaminants, latent ESD (Electro Static Discharge) damage as well as an instable power supply can result in an accelerated degradation of epitaxy layers.
Source: LED Failure Modes and Methods for Analysis, LED Professional, Aug 01, 2010

Measure the Vf of the LEDs. Is the Vf of the unilluminated LEDs less than the illuminated?

If the Vf is decreasing with age, then the aging of the LEDs would increase due to the over stress of the excess current.

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Your design is wrong:

You've got a desired forward current of 30 mA, and you've got 3·27 Ω = 81 Ω series resistance.

30 mA · 81 Ω = 2.43 V

That means over your three LEDs, a voltage of 12 V - 2.43 V = 9.57 V drops. That makes, per LED, a voltage of 3.19 V.

That's pretty typical for blue LEDs, but not for red ones. Red LEDs are manufactured with band gaps of 2 V to at most 2.5 V.

So, you're driving your LEDs in constant overvoltage. No wonder they fail all the time!

Let's check this: If we have three 2.5 Vf LED, then there's 7.5 V across the three of them, which means that 12 V - 2.5 V = 9.5 V drop over the 81 Ω. That's 117 mA, not 30 mA.

That means the LEDs get 16 times as hot as they should be getting! No wonder they become worse at emitting light, and better at working like a resistor.

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