Reverse biased LED

25.11.2009 16:10

Here's an interesting question: If you force current through a LED in the reverse direction (i.e. by causing a breakdown in the junction), will it emit light, and if not, why?

Fear the glowing spheres

Let's start with an experiment. I took an old low-intensity 3 mm yellow LED (I'm guessing GaAsP chemistry) - I want to limit this discussion only to simple P-N junction devices and ignore for the moment complicated new LEDs with quantum wells and such.

I applied a high reverse bias through a large resistor (1 MΩ, to limit power dissipation) and visually checked for any light. The diode started conducting a significant current at 185 V (which, by the way, is surprisingly high as all datasheets I've seen rate such LEDs at maximum 5 V reverse bias). At that reverse voltage I let 0.1 mA of current through it (which gave total power of something around the rated 20 mW). I couldn't see any light coming from the LED.

As a control I then reversed the polarity and let 0.1 mA flow in the forward direction. In that case I could clearly see the LED lit up in a darkened room. So, the LED wasn't destroyed in the experiment and 0.1 mA was enough to produce a visible effect.

So, the experiment confirms that a LED will not light up when the current flows in the reverse direction. But what is the theory behind it?

Visible photons emitted by the LED are generated by electron-hole recombinations in the semiconductor. The material has just the right energy gap that an electron in the conducting band can give away it's excess energy to a visible photon when it fills a hole in the valence band. In forward operation most of these recombinations happen after charge carriers travel through the depletion region and are diffusing into N or P material as minority carriers.

When the junction is in breakdown, a similar thing happens (from the high reverse voltage I measured I'm guessing the avalanche mode of breakdown). However this time carriers don't diffuse through the junction but are generated there through impact ionization. But because the voltage is reversed, electrons enter the N material as majority carriers (same goes for holes in the P material). So gone are the recombinations near the depletion region and no significant number of photons is produced.

Interesting though, carriers from an avalanche breakdown have significantly more energy than thermal ones in forward operation. And they do in fact emit light after they pass through the junction (i.e. through bremsstrahlung and hot carrier recombinations). However the light's wavelength is no longer defined by the material's band gap and its spectrum is completely different to that of forward operation. So the LED might have as well lit up in my experiment, but with intensity and wavelengths that were invisible to an unaided eye.

Posted by Tomaž | Categories: Ideas


I'll not protect LED from reverse bias from now on! I've seen some destroyed by large forward current. Probably junction overheating?

I would not trust the well-being of your LEDs to this one experiment.

If the manufacturer says 5 V is the maximum allowable reverse voltage then that's what you must design for.

Posted by Tomaž

LEDs are, electronically speaking, just like any other semiconductor diode. They *usually* go into constant-voltage breakdown somewhere between 5V and 6V. They will be quite comfortable in this mode, as long as the maximum rated power dissipation is not exceeded. After all, Zener diodes spend their working lives in reverse breakdown ..... but they are made with a clearly-defined PIV. Rectifier diodes are made to have a high PIV and low forward voltage drop. LEDs are made so the energy gap corresponds to a photon of (usually visible) light. With diodes, one size does not fit all!

Zener diodes can even be used as rectifiers in an emergency, as long as the peak supply voltage (possibly in series with the charged reservoir capacitor, depending upon circuit topology) doesn't exceed the Zeners' PIV. And the humble 1N4007 can be used as a "photodiode" for X-rays.

I find that my cellphone camera can pick up a much wider spectrum than my eye. I routinely use it too "see" if IR LEDs are hooked up properly. Maybe a camera could detect the reverse-biased frequencies?

Posted by Spencer

This is interesting. I was trying to figure out why just about every reference out there says you shouldn't run LED's off 120VAC with just a resistor when you can buy this combination as an indicator light from Blue Sea Systems and they work just fine.

That made me curious so like our blogger, I hooked up a green LED with a 33 kohm resistor to a variable DC supply. I found the following when reverse biased - at 100 VDC - diode voltage was -88V, current .38 ma. At 200V - diode voltage was -98 V and current 3.2 ma.

And this is the weird part. I turned down the lights to check for photodiode behavior with a flashlight and there it was - Above about 140 volts, the diode was clearly emitting yellow-orange light. Its intensity increased with voltage up to 200 volts (-98 volts on the diode) and the diode was unharmed at this level. It still emits green light when forward biased.

So, it was very clear that at least this type of LED does emit visible light when reverse biased and isn't harmed (at least in the short term) by this with current levels around 3 ma.

Thanks for creating and maintaining this blog - JB

Posted by JB

I ran this very same experiment a while ago, to see what happens to LEDs in avalanche mode. While I did not notice any visible photons emitted from the junction, the LED's behaviour matches any other diode's:
As long as it doesn't reach thermal breakdown due to overcurrent or (severe) overvoltage, you will observe current increasing slowly with voltage, then a sudden spike of current once the avalanche region is reached.

However, I did not use any current limiting resistor, as my power supply can limit current just fine. 1MΩ is probably a bit much anyway, and very likely the reason why you needed 185V. Running the numbers: Reverse current 0.1mA, serial resistance 1MΩ, voltage 185V, giving: Vtot = Vled + R · I => Vled = 185V - 1MΩ · 0.1mA = 85V. Still seems a bit excessive. 185V is well above what's safe for human consumption, though the resistor and power supply would limit any unsafe currents.

Anyway, are you sure about LEDs producing X-ray bremsstrahlung? I know about diode X-ray detectors, but emitters?

JB: I usually see this effect when I (accidentally) pass excessive current through a LED.

Posted by onitake

I built a Joule Thief ( that ran a 3.2v white LED from a "dead" AA cell (if Vbatt > ~0.5v). I was watching youtube videos of other people's Joule Thieves, and one guy said "...and the weird part is, you can put the LED in either way and it still works." and that caught my attention.

I tried it with the 3.2v white LED in reverse bias, and it DID work! (emit light) I put it on the scope and the Joule Thief was raising its output voltage to the reverse breakdown voltage of the LED at around 50v. (The Joule Thief raises the voltage with an inductive effect, but the output power impulse lasts on the order of microseconds or less - little heat is produced) The "white" LED I used is a bright blue LED with a phosphorescent coating that, when forward biased, absorbs blue and re-emits the energy as a wideband signal in the visible spectrum (there is still a huge blue spike in the spectrum).

I showed all my colleagues and nobody else had expected that LEDs would emit in reverse bias, either. I am confused by the physics of how the photoelectric effect is working here.

Please somebody else try this and report back.

Posted by Cam

Cam, interesting observation. JB in a comment above also mentions that he saw a green LED glow yellow-orange in reverse bias. So it might be that some diodes do in fact emit light.

My reasoning in the original post applies only to simple P-N junctions. As far as I understand, newer high intensity LEDs have somewhat more complicated designs, so the situation there might be different. I haven't looked into semiconductor physics for many years now.

Another possibility is that your joule thief circuit (briefly) creates negative voltages and puts a forward bias on your LED. Such inductive circuits like to produce large voltage swings in both directions.

I might revisit this experiment when I have a chance. Do you have a link to the video you mentioned?

Posted by Tomaž

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