9
\$\begingroup\$

I wonder this by pure curiousity. If we polarize a LED in avalanche mode by applying a very high reverse voltage (but keep the current low so that the component doesn't fry), is it possible that it also emits light when used this way ?

(The reason I don't "just try it and see" is because of the high voltages involved).

\$\endgroup\$
5
  • 1
    \$\begingroup\$ Try it and see, under proper safety conditions. Remotely activate the high voltage circuit. \$\endgroup\$
    – Passerby
    Apr 6 '17 at 6:19
  • 6
    \$\begingroup\$ Most LEDs have a rather low (around 5V) reverse voltage rating -- getting one to avalanche should be possible with a current-limited lab supply, no fancy high voltage stuff needed. \$\endgroup\$ Apr 6 '17 at 11:44
  • \$\begingroup\$ @ThreePhaseEel yeah, that means that for avalanche breakdown (not Zener-style breakdown that end as soon as the voltage drops below the threshold) you'll need a really strong E-field without breaking down non-avalanchingly. Which means that you'll need extremely short pulses of extremely high voltage – and I agree with Passerby, that's probably nothing you want to touch. \$\endgroup\$ Apr 6 '17 at 11:48
  • 2
    \$\begingroup\$ @ThreePhaseEel No, the LEDs I tested this on didn't break down even at -34V. \$\endgroup\$
    – Bregalad
    Apr 7 '17 at 15:16
  • 1
    \$\begingroup\$ It emits blackbody radiation at substantially above room temperature. And perhaps some smoke. \$\endgroup\$
    – mkeith
    Jan 17 at 5:36
7
\$\begingroup\$

I'll go with a: In general, no, that's not the case.

Light emission in LED-type devices typically happens when electrons and holes recombine, and the energy that gets freed in that process gets converted to a photon with the resulting wavelength. That happens in the transition zone of a doted semiconductor junction, where there's gradient in the band structure.

Let's imagine a diode in reverse bias: In aforementioned transition zone, there's practically no free charge carriers (no holes and electrons), so the device would be a perfect isolator – I say "would be" if not spontaneous creation of such carrier pairs could happen due to thermal effects (and also, things like photon absorption).

Now, under avalanche breakdown conditions, the electric field across that isolating zone is so high that the charge carriers get accelerated very fast – and might "knock" out other charges from the non-conducting bands (to make this feel a bit more scientific: the electric field gives spontaneously created charges an impulse that is enough to transition further charges in k-space to the conduction band).

Now, these charges will just travel to the contacting areas and recombine there - usually nowhere where there's a) a well-defined bandgap to make emission of visible photons likely and b) no optical structures to couple out that light. You just heat up the substrate.

That is not to say there won't be light emissions in all this: purely from a stochastic point of view, some recombination with visible emissions might happen, and also, nothing says that over the temporal process of that avalanche breakdown, there won't be some times where the whole field configuration wouldn't lead to interesting band diagrams where recombination within the optically relevant parts of the LED might take place, at totally different photon energies than the LED was designed for.

\$\endgroup\$
8
\$\begingroup\$

First I thought it won't but surprisingly I found this: THE AVALANCHE-MODE SUPERJUNCTION LIGHT-EMITTING DIODE

So what you wanna do is theoretically supported. But for that, you have to make special diodes. I don't think normal regular diodes will work. Best of luck. Do the experiment by yourself and learn.

\$\endgroup\$
3
\$\begingroup\$

My research is on the same topic and I am the author of the avalanche mode superjunction LED as posted above. You can read these additional following papers if interested.

http://aip.scitation.org/doi/full/10.1063/1.4931056

http://proceedings.spiedigitallibrary.org/proceeding.aspx?articleid=2601523

\$\endgroup\$
2
\$\begingroup\$

As a kid in the 1970s, I bought a large TO-60 IR LED from a surplus seller. It drew an amp, and you could see the dim red glow by eye. Then I immediately killed it by accidentally hooking it up backwards, even using the proper series resistor. It broke down somewhere below 12V, and the edges of the visible electrodes on the chip face gave off a broadband (white) light.

Some kind of avalanche-mode fluorescence? Not colored, or IR, but white glow. It was running at far higher than 1.1V.

\$\endgroup\$
4
  • 5
    \$\begingroup\$ Any reason to assume it wasn't incandescence? \$\endgroup\$ Apr 6 '17 at 12:37
  • 2
    \$\begingroup\$ @rackandboneman Perhaps. But the glowing region was at the die surface, in direct contact with clear epoxy and silicon. To attain white heat (not orange, not yellow,) I assumed it couldn't be hot unless the material first developed an insulating gas layer, leaving visible bubbling and perhaps darkened epoxy. Also, it was quite dim white glow, only visible against color of blue-black Si. The LED soon failed shorted, while I was trying to pulse it and observe w/microscope 40x. Click on Anklon's link above: avalanche-mode superjunctions emit broadband, in reversed silicon diodes. \$\endgroup\$
    – wbeaty
    Apr 7 '17 at 2:15
  • \$\begingroup\$ Bond wires acting as incandescent bulb filament? \$\endgroup\$ Jan 16 at 17:08
  • 2
    \$\begingroup\$ @MicroservicesOnDDD That was in ?1973?. The silicon surface at the edges of the die aluminization were visibly glowing white, and the on/off was instant, like an LED (not anything like a tungsten filament.) This in a gold stud-mount few-watts Si IR LED. I was probably seeing the white "avalanche light" discovered in 2017. Again, see Anklon comment: reverse LED breakdown gives broadband white emission. Link has changed, it's now utwente.nl/en/news/2017/3/73275/… \$\endgroup\$
    – wbeaty
    Jan 17 at 5:02

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.