When a LED is included in a circuit that applies a reverse voltage that exceeds the reverse breakdown, a reverse current can flow and the LED might be destroyed. But what is it that actually destroys the LED: is it the reverse voltage itself, or is it the reverse current that is made to flow, or is it simply the overall power dissipation caused by the reverse current and voltage exceeding the device rating? Or something else?

So, for instance, if I connect a 12 volt source in reverse to a LED that breaks down at 5 volts, via a resistor, the passage of reverse current will cause a voltage drop across the resistor which in turn can limit the voltage on the device to its reverse breakdown value (and thereby define the current that would flow) - rather similar to what happens in the forward direction. Would this in itself destroy the LED as long as the total power was within the LED rating?

Normally of course, one would place a regular diode in parallel with the LED in the reverse direction to limit the LED reverse voltage to 0.7 volts or so, but there may be situations where this might not be possible or economic to do. I am just trying to understand how much circuit design flexibility I might have to meet different requirements.

And if it is possible to expose a LED to a reverse voltage, what precautions should be taken to avoid damage, and which spec parameters are relevant?

  • \$\begingroup\$ In an attempt to improve ESD protection in RS232 links, I tried a cheap white LED on the TX lines and a clear plastic photo transistor (that looks like an LED) on the RX lines, both were pushed into a 1" piece of plastic tubing. It worked great but I did start seeing a few LED failures that lead me to this discussion. \$\endgroup\$ – chris Sep 16 '17 at 14:36
  • \$\begingroup\$ I tried the following goggle search and the first snippet from some Colorado, USA education institute seemed interesting, Enough reading to make anyone happy. - google.com/search?q=led+reverse+breakdown+mechanisms \$\endgroup\$ – KalleMP Dec 22 '17 at 19:38

ESD appears to cause damage due to hot spots or some other localized damage. I have seen heterojunction LED failures that appear to be partial.

Failure with DC current in reverse is probably related to power dissipation, but it might be unwise to depend on it. Breakdown could be quite high, hence the allowable current could be quite low (maybe less than 1mA).

Safest is to follow the LED data sheet recommendation- usually 5V or so reverse is guaranteed. Many types of LEDs have much higher actual reverse voltage breakdown (perhaps 15V to 70V), but it's unwise to depend on it- the LED maker could change the chip supplier or process or purchasing could go to a different vendor.

The typical situation where LEDs are exposed to reverse voltage is when they are operated in a multiplexed configuration- they will see up to the supply voltage in reverse. It's not really a good idea for efficiency to make the supply voltage much higher than the sum of the series LED forward voltages (often, but not always, just one LED is used). For example, you might use 5V for a single 2-3V LED array, or 12V for an array of series strings with 6-9V per string. Since the individual LEDs can take 5V each (usually guaranteed), you'd be fine in either case.

See this nice instructables gif:

enter image description here

  • \$\begingroup\$ Hmm, If by ESD you might include local heating.. from excess current flow.. then I believe it. \$\endgroup\$ – George Herold Jun 1 '15 at 0:46
  • \$\begingroup\$ @GeorgeHerold Well, I shrewdly avoided discussing pulsed damage other than from ESD, but I think that there are similar issues possible from repetitive overcurrent, maybe damage to metal as well as semiconductor. \$\endgroup\$ – Spehro Pefhany Jun 1 '15 at 1:40
  • \$\begingroup\$ I imagine the high energy involved with reverse breakdown (esp. if it's near the surface) could cause dislocations, introducing new charge traps and encouraging SRH recombination, thus reducing photon emission. \$\endgroup\$ – Zulu Jun 1 '15 at 3:59
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    \$\begingroup\$ Ran across an oldish (2007) Usenet post from Don Klipstein who was of the opinion that even relatively small energy/power (less than a few mW) could cause problems on GaN and InGaN LEDs (most LEDs with shorter wavelength than green). I've played with green LEDs at high reverse voltage/current but would not do this in a product without more research. \$\endgroup\$ – Spehro Pefhany Jun 1 '15 at 4:09
  • \$\begingroup\$ @SpehroPefhany, The most fun I've had with a reversed biased LED is here, teachspin.com/newsletters/TeachSpin_MAY13FINALFOR%20WEB.pdf LED is either an AND113-R or AND114-R (the 113 has a clearer case and is more sensitive.) \$\endgroup\$ – George Herold Jun 1 '15 at 13:12

Most datasheets from famous and respected manufacturers including Vishay and others, show breakdown voltage of 5V and reverse current of 10 to 50uA. That is not true. I just tested white, red and green leds, with a 0-30V power supply and a 1k resistor in series, measuring voltage across the led, and current in the circuit. This is the result: White 9V=0.4uA, 13V=1uA / Red 5.3V=0.3uA, 6.7V=0.5uA, 12.5V=1uA / Green 5.11V=0.3uA, 6.5V=0.5uA, 9.9V=1uA. So, no led presented anything close to half a microAmper at 5V as stated the manufacturers, and only right after 10V they presented ONE microAmpere. That is not enough to damage the component. According to what I read, it would need a minimum of 1mA reversed to increase the depletion area.


  • \$\begingroup\$ I don't think you understand what the datasheet is telling you. The datasheet says that none of the LEDs will ever breakdown at a reverse voltage less than 5V or conduct a reverse current greater than 50uA under those conditions. It sounds like all of your LEDs conform to the datasheet. \$\endgroup\$ – Elliot Alderson Dec 12 '19 at 17:58

Diodes have properties which create what is known as the depletion region. This is the barrier which prevents current from flowing across it with a forward bias until the depletion region is minimized (forward voltage dropout).

Reverse biasing a diode increases the depletion region, acting like a one way door. However if you apply enough voltage to it, the mechanism breaks down and current flows either way, usually after the PN junction is shorted and the diode is effectively destroyed.

Basically what destroys the diode is power dissipation, or whatever causes the diode to be physically altered. The reverse bais case may typically exceed the normal power dissipation of the forward bias case before it damages the device.

However there are certain types of diodes like zener diodes which are made to break down in reverse at a specific voltage, which makes them useful as voltage references and limiters.

For the 12v case on a 5v LED in theory using a limiting resistor to reduce the current (and drop the reverse bias voltage anyways) shouldn't, in theory, destroy the LED. Some are more forgiving than others.

For your last question I'm wondering what scenario you'd have reverse voltage applied to it? Usually there is protection applied before it gets to the LED.


LED's are still diodes, in reverse bias they will avalanche.
(Though I've not done anything like an exhaustive search, I've never found an LED that breaks down in reverse at less than 20V.)
It's my guess, that it's the heat/ power dissipation that would kill an LED in reverese. So as long as you limit the current such that the power is less than ~10 mW LED's can handle reverse bias, IME


In my experience the standard 3mm and 5mm red LEDs can block 12V easily so I have used them as reverse polarity protection in 12V systems where current is about 10 mA DONT use them on a 24 Volt system some LEDs died at approx. 30V

  • \$\begingroup\$ So, is it good if I put an LED in reverse to detect wrong polarity? I mean, I put the LED in reverse in a circuit. I want the LED lights up if I reversed the polarity of the circuit. The circuit is 5V 1A. Will the LED die if I do this? \$\endgroup\$ – Oki Erie Rinaldi Apr 12 '16 at 9:00
  • \$\begingroup\$ At 5V its fine ,you wont have a problem. \$\endgroup\$ – Autistic Apr 12 '16 at 9:01

Your choice of series resistor (required in this case) will protect the LED (a diode by any other function) in an ac circuit or when reverse biased. Choose this resistor based on spec sheet for your particular LED.

  • \$\begingroup\$ Welcome to EE.SE. This isn't quite right. A resistor will limit the current once reverse breakdown occurs but otherwise will allow full reverse voltage to be applied to the LED. Consider the case of a high resistance resistor and series LED across mains supply. It is unlikely to last. I have written more on the topic here. You can be "lucky" though, depending on the batch of LEDs. (I didn't downvote.) \$\endgroup\$ – Transistor Dec 9 '17 at 0:59

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