2
\$\begingroup\$

These LED's were taken from an old light bulb, a dimmable 800 Lumen Philips EnduraLED A19 Lamp 12.5W Model 12E26A60-1, a light bulb interesting in that a "white" LED usually has phosphors integrated into its epoxy coating, whereas for this light bulb, it contains "Royal Blue" LED's with the outer plastic "shell" of the light bulb containing the phosphors. More pictures of this light bulb are further below, if you are interested. You will see two of the ceramic PCB's in the picture, and one is missing (accidentally destroyed).

enter image description here

This LED is on a ceramic substrate also having a black square on it, and three of these LED's are on a ceramic PCB (there are three of these PCB's -- two are shown in the picture above).

I want to know what the black square is, that is on the ceramic substrate along with the LED.

The picture below was taken with the LED barely lit.

Purplish LED with black component on integral mini circuit board

The following picture is a little more at an angle, and with a bit more current flowing, so closer to the eventual Royal Blue color at normal power. The black square is seen at a different angle.

Bluish LED with black component on integral mini circuit board


Edit 1:

The black square is very shiny on top, like Marcasite.

The light bulb had 9 of these LED's in series, 23.4 volts at 100mA. So about 2.6V per LED at 100mA. The bulb ran hot to the touch, which is one reason that I phased it out. The heat sink states that the bulb was made in China.

enter image description here


LED-01 -- Removed from ceramic PCB, the LED has a ceramic substrate, which I broke, intending to try to measure the black chip separately:

LED-01 Liberated and Zoomed, snapped

LED-02 -- This LED has been kept whole, the epoxy lens popped off. It looks like the black chip is in parallel with the LED die:

LED-02 Liberated and Zoomed, whole, not snapped


According to comment by Tim Williams, proceeded with testing. With current limit set to 10mA, reverse voltage went to -8.1V and once the black square was removed, the voltage then reached -11.0V without conduction (or destruction). So the black square is acting like an 8.2V Zener in the reverse direction. Don't know about the forward direction.

\$\endgroup\$
5
  • 3
    \$\begingroup\$ Reverse protection, most likely. electronics.stackexchange.com/questions/352441/… \$\endgroup\$ Commented Jun 11, 2022 at 0:09
  • \$\begingroup\$ @TimWilliams -- So how would I test that? (Hopefully without destroying the LED). \$\endgroup\$ Commented Jun 11, 2022 at 0:26
  • 1
    \$\begingroup\$ Some protection device, like TVS, zenner, etc. \$\endgroup\$
    – user263983
    Commented Jun 11, 2022 at 1:04
  • 1
    \$\begingroup\$ from which device is that LED? ... what is a commercial LED? ... the item in the photo is a LED \$\endgroup\$
    – jsotola
    Commented Jun 11, 2022 at 1:19
  • 1
    \$\begingroup\$ @MicroservicesOnDDD Test it by applying a small current (say 1mA) and measuring the voltage. Probably somewhere between 5 and 20V; I guess they aren't usually rated? In my experience, GaInN dies fail suddenly somewhat over 20-30V, so it has to be less than that to offer any protection. Note that most DMMs don't supply such high voltages on diode/zener test, you'll just measure open circuit with one. \$\endgroup\$ Commented Jun 11, 2022 at 2:41

1 Answer 1

3
\$\begingroup\$

Thanks for adding the photos and measurements!

That is most likely an ESD protection (TVS) diode, probably anti-series with a regular diode so it doesn't steal forward current from the LED. Apparently it's a flip-chip, so the contacts (and diode structures) are on the bottom side, and what you see is the bottom of the die, just bulk silicon; which indeed looks dark metallic, like some minerals.

The purpose is several:

  • ESD. During handling and assembly, ESD may be encountered, and having a diode internal to the component itself greatly improves survival. This may also apply directly in use, when the LED array can be handled by the user.
  • AC powered chains. Probably not as common for these (higher power) LEDs, but cheap strings (especially Christmas lights, etc.?) can be direct AC powered. In this case, when mains voltage is reversed, the voltage may not distribute evenly among the chain, and thus some become overvoltaged and fail. The TVS handily prevents this.

As for motivation, as far as I know, GaInN junctions can't handle any avalanche (reverse breakdown current). At least not LED types. Perhaps it can be made to handle it, but it seems at present, this isn't enough of a priority that anyone is doing it (even for power transistors in similar materials). My tests have shown blue LEDs break down around 20-30V, resulting in sudden failure, even at low currents (<1mA). So some protection is welcome, especially on these higher cost types like you have.

\$\endgroup\$
2
  • \$\begingroup\$ Is that why you suggested only -1mA test current -- because "GaInN junctions can't handle any avalanche"? \$\endgroup\$ Commented Jun 11, 2022 at 22:06
  • 2
    \$\begingroup\$ 1mA is a reasonable test current -- it's unlikely to damage anything that can take it, and it's unlikely to overheat things (unless it's several hundred volts). Perhaps ~uA or even ~nA would permit testing GaN and relatives, but in any case it's quite a low threshold. \$\endgroup\$ Commented Jun 11, 2022 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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