Conventional wisdom about LEDs says their maximum reverse voltage \$V_{R(max)}\$ is quite limited, usually in the 5V-8V range.

So, for experimentation purposes, I wanted to bring an LED into controlled breakdown, using my current-limited power supply.

Of course I expected the actual breakdown voltage to be somewhat higher than the reported guaranteed \$V_{R(max)}\$, but I could never have expected the outcome I found. I tried with different kinds of "el cheapo" no-brand Chinese indicator LEDs (3mm and 5mm, red, green, blue, yellow and white) and I couldn't bring them in the breakdown region, even at 32V (where my power supply reached its max)!

Therefore I wanted to double check my assumptions and I systematically browsed many datasheets (about 40) of current devices (standard 3mm and 5mm LEDs, for both indicator and lighting applications) from different manufacturers (e.g. Vishay, Nichia, Kingbright, Fairchild, Cree). Almost all of them reported a \$V_{R(max)}=5V\$, with some Vishay devices rated at 6V.

I was extremely puzzled. OK, manufacturers tend to be conservative, but a >25V margin seemed a bit too high. After all, guaranteeing a \$V_{R(max)}=25V\$ (or something like that) could make LEDs good candidates for some useful applications or allow circuit simplifications (e.g., no need to protect LEDs from low-voltage reverse spikes). Anyway, that would be another bullet in the list the marketing people could boast about!

Of course my test was limited to a dozen LEDs of unknown manufacturer, but I suppose they can't be better than those from reputable sources. Or did I experience a sort of reversed Murphy's law, where I found the only box of LEDs on the planet with such a feature?!?

Question(s): Is my finding something that is known in the industry? Why do they keep specifying LEDs with so low a \$V_{R(max)}\$ when the actual devices seems to be much better? Do I miss something?


(to clarify some points that possibly prompted comments/answers that didn't actually give me the explanations I'd like to get)

Things I already know

  • Stresses beyond absolute max ratings reported in the datasheet might damage the device, and usually will damage it if the stresses are well beyond those limits.

  • When you exceed those max ratings you cannot demand anything from the manufacturer. You are on your own in unknown territory. You can neither sue him nor complain.

  • No sane designer would use a part in his design outside the specs given in the datasheet. Good designers will make sure the part will stay well below the stated max ratings. As I stated at the beginning, I was experimenting, purposely entering unknown land to verify my expectations and my knowledge about reverse breakdown.

My assumptions (possibly wrong; and if they are wrong I'd like to know why)

  • The main limiting factor for any diode max reverse voltage rating is its breakdown voltage. In other words, you can safely reverse bias a diode as strongly as you want until breakdown (either Zener or avalanche) kicks in.

  • Breakdown is not destructive in itself. The sudden increase in reverse current causes an enormous increase in dissipated power, especially at high reverse voltages, therefore the PN junction will be destroyed, unless you limit the current somehow.

  • LEDs breakdown mechanism is not different from that of other PN junction diodes, like regular silicon rectifiers or Zeners.

  • Since LEDs are not designed (as opposed to Zeners) to work in breakdown, the BD voltage is not a well specified parameter, so the manufacturing spread could be quite big. Therefore the manufacturers choose a suitable safety margin and declare that as the max reverse voltage.

  • Although some safety margin is needed, it can't be huge. IIRC, BD voltage depends on doping levels and the geometry of the metallurgical junction and those parameters also influence the diode characteristics when forward biased. If the "useful specs" of the LED have to be reasonably consistent, so the doping and the geometry must be; hence also BD voltage values can't be too wildly spread out.

What puzzled me and made me think there are more issues beyond protecting an LED from entering breakdown

  • So big a difference between rated max reverse voltage and actual BD voltage (at least +400%) should mean something and should have a rationale behind it. Given the assumptions above, I can't believe the same model of LED can have a BD voltage spread that big, i.e. I can't believe the same process (even across different batches) can yield one part that enters breakdown at, say, 10V and another one that enters it at 30V (I stand to be corrected).
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    \$\begingroup\$ are you sure that doesn't affect life or output efficacy? \$\endgroup\$
    – dandavis
    Commented Apr 3, 2018 at 2:13
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    \$\begingroup\$ you can survive getting shot, sometimes. the specs reflect reliable long-term operation. \$\endgroup\$
    – dandavis
    Commented Apr 3, 2018 at 2:23
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    \$\begingroup\$ @dandavis To justify my misconception, even Wikipedia seems to reinforce it here: If the reverse voltage grows large enough to exceed the breakdown voltage, a large current flows and the LED may be damaged. If the reverse current is sufficiently limited to avoid damage, the reverse-conducting LED is a useful noise diode. \$\endgroup\$ Commented Apr 3, 2018 at 2:36
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    \$\begingroup\$ Over what temperature range did you repeat your reverse voltage measurements? Also, I wouldn't draw any conclusions by testing devices from just one manufacturer. \$\endgroup\$
    – AlmostDone
    Commented Apr 3, 2018 at 2:43
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    \$\begingroup\$ I would add that whatever the case, in a production system, you should never operate components outside their specifications. If a manufacturer specifies a maximum reverse voltage (which most (all?) of them do), you shouldn't be going above that, because characteristics beyond that point aren't guaranteed by the manufacturer. Anything that you measure at that point could easily change drastically between batches. \$\endgroup\$
    – BeB00
    Commented Apr 3, 2018 at 3:36

5 Answers 5


Yes, this is widely known. Anyone who has tested it knows that. The die manufacturers certainly know it.

They don't specify LEDs for more than 5V reverse voltage because it would not measurably increase sales (ie. very few need that capability) and would require them to actually consider each LED type and what voltage it might withstand (maybe 12V for some, maybe 80V for others). There may also be long term reliability issues that would require quantification or possibly a change in LED design to mitigate.

The 5V rating comes from the reverse voltage experienced by an LED driven in a matrix from a 5V supply with push-pull drivers, which is one of the few times you deliberately reverse bias an LED (back to back LEDs in AC-input optocouplers see the forward voltage of the other LED worst-case, or about -1.2V).

There are many parameters that are unspecified (typical data or no data at all) or only loosely specified because the bulk of the market does not demand it. For example, reverse beta, Vbe breakdown on BJTs, temperature coefficient of Vf on indicator LEDs.

As far as what the actual capability of ordinary LEDs is, there is evidence of reverse bias voltage causing gradual damage to the LED due to hot carriers. For example, DOI 10.1109/LED.2009.2029129 indicates damage to green LEDs with -40V applied, so it would be unwise to blindly design something that depended on the high reverse voltage breakdown.

enter image description here

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    \$\begingroup\$ Thanks! That's very relevant info! I begin to see some light in the issue! (pun intended :-). This also confirms that high Vr degrades LED of types other than Blue GaN, as it seemed reading the articles linked to by DmitryGrigoryev and TonyStewart. \$\endgroup\$ Commented Apr 3, 2018 at 15:32

If you stand under a tree in a lightning storm and you survived, does that mean anything significant? This is somewhat like reverse biasing an LED > -5V.

enter image description here

Graph , courtesy of This shows the sensitivity of LEDs in both reverse and forward bias exposed to ESD. Note below, that it is far more sensitive to the left when Vr goes below -5V

enter image description here

(I could write a book on the subject of Partial Discharge(PD) and Breakdown Voltage (BDV) but I'll keep this edit shorter ;)

WHen a PN junction is reverse biased a charge cloud ( like a cumulonimbus cloud) creates a high E-field charge density where defects are mobile charges (contaminant particles) that are accelerated to form a path that will either detonate the particles (by PD) and "wound" the device ( even MVA transformer insulation) or create a streamer path prior to BDV catastrophic event. ( e.g. like lightning but silent)

Except in a reverse biased LED the E-field I am guessing is on the order of magnitude of 5V/um (like 5kV/mm for almost a spark in air. Then molecular impurities with a slightly lower Er constant will have a greater E-field and charge across it than its neighbour. The charge being built up by the > |-10uA| flow of current where White LEDs are spec'd at -5V in small chips.


A wounded 5 MVA distribution transformer I investigated at a transformer factory in Scarborough had a $m liability problem yet it had a perfect power study performance field test, BUT had dissolved Hydrogen gas proven by frequent dissolved gas (DGA) analysis). This H2 was generated by each PD event in the oil , just like a DIAC Relaxation oscillator, and then reached the well-known (to those in that industry) threshold of explosive levels (4% is the Lower explosive threshold so it was promptly taken out of service , whereafter I performed exhaustive testing to find Root Cause and fix the contamination problem from normal 23kV potentials expected in this dielectric but caused abnormal E fields in particles > 16V/um causing discharge and detonation of oil molecules around it thus breaking down the long CxHy hydrocarbon chains releasing H2.

A similar but different contaminant ( mixed with normal distribution of Nitride, Gallium Phosphide and Arsenic) is accelerated by abnormal E-fields in a reverse biased PN junction and adversely affects LED's life expectancy.

This charge shows the relationship to defects and leakage current but a wounded junction is dense unlike a homogeneous contaminants so the BDV is unpredictable yet known where the stress level begins for many PN junctions (Vbe and LED's, although different in construction exhibit this common failure mechanism with different degrees of accelerated sensitivity.

So to summarize, if a PN junction has a higher tolerance to reverse bias from testing, it does not mean it still is not wounded, just that it has a lower density of particle contaminants in parts per million. The charge acceleration is not linear with contaminant density but rather logarithmic. It is the impact kinetic energy that detonates the micro or nano-sized damage.

end edit

When reverse biased the current is rated usually 1 µA for RY and 10 µA for BGW colors.

Imagine that reverse biasing is extreme micropower and measure it and if there is no ESD clamp that something of the order of 100 µW has more power per square micrometer than forward biased current 100 mW per square mm because the path is MUCH DIFFERENT.

It's not like a Zener diode limited by power in either direction. The band gaps can abruptly fail or softly.

So the stress is invisible and wounds the junctions differently. The result can be seen with a higher junction capacitance or an off-colour or a lower intensity or a wounded to reduce MTBF significantly.

Whether it can withstand it briefly or for a while or not is irrelevant. Experts understand the stress level reduces reliability, or performance.

If you don't understand why absolute maximum ratings exists, don't ignore it or doubt it or when you least expect it... hmm, it's not working.

An Engineering guide I made to client in 2005 before I went on site visit to resolve ESD and solder problems causing 1% field failures fixed later by my process improvement recommendations. enter image description here enter image description here

research article on Reverse Voltage stress in diodes

Trivia test

Why is this a bad idea?


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Thanks for the hints. I don't doubt max ratings, but I'd like to understand why they are the way they are. As I wrote to @dandavis in a comment to my question, I thought the main limitation for that reverse voltage rating was the junction destruction by entering breakdown. From what you say it seems that I had a misconception. But I'm sure I've heard that rationale about low max Vr many times before. Maybe it's a common misconception. \$\endgroup\$ Commented Apr 3, 2018 at 2:33
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    \$\begingroup\$ From what I recall, the breakdown of Base-emitter bipolar junctions would degrade the Noise Figure. \$\endgroup\$ Commented Apr 3, 2018 at 2:50
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    \$\begingroup\$ Having shipped about 1,000,005 mm LEDs in the previous 10 years with no factory defects the only problem is I have seen are from in design and Process failures. For example soldering within 5 mm of the base when the heat velocity is 1 mm per second reaches the junction in five seconds.The Epoxy softens and the gold whisker wirebond can shear under vibration or lead stress. Initially my client was using non-ESD protected LEDs and their process with injected moulded plastic was causing all sorts of EST failures. So after flying halfway down around the world to their factory I fixed the probs \$\endgroup\$ Commented Apr 3, 2018 at 3:15
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    \$\begingroup\$ haha , only from past ESD failures that worked in many cases but off white colour in the y direction not Blue shift and increased capacitance in Huntron style tests from smaller gaps. \$\endgroup\$ Commented Apr 3, 2018 at 6:48
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    \$\begingroup\$ Would you care to add links to some source explaining better what damages (ESD apart) could an LED junction suffer if reverse voltage is too high, but not high enough to make it enter breakdown? \$\endgroup\$ Commented Apr 3, 2018 at 12:53

Exceeding absolute maximums from data sheets does not necessarily mean immediate catastrophic failure. It means that you've gone into a region for which the manufacture no longer sees fit to guarantee that the device will ever perform to spec again, for the remainder of the life of the device.

Does this mean that it won't perform to spec? No, it means the manufacturer no longer guarantees that it will perform to spec.

Also, since your tests were performed on LED's of "unkown manufacture", you have no idea how they are rated.

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    \$\begingroup\$ Right. Programmers call this the realm of nasal demons. \$\endgroup\$ Commented Apr 3, 2018 at 12:29
  • \$\begingroup\$ Thanks to point this out. I know that I cannot expect anything precise from a part outside its specs. My puzzlement arised because the spec in question (Vrmax) was so much less than the possible breakdown voltage (which I wasn't able to reach). What I would expect is that a manufacturer would apply some safety margin to guarantee a certain level of reliability, but not that that margin is so huge. My question was asked to understand why the margin is so huge, since apparently entering breakdown is not the failure mode the manufacturer fear the most. \$\endgroup\$ Commented Apr 3, 2018 at 12:38
  • \$\begingroup\$ To make a comparison: I would not expect an 1N4007 (1000V Vrmax) to have a breakdown voltage 5 or more times higher (maybe I'm wrong), and I would be puzzled as well if I found out it could withstand, say 5000V without entering breakdown (destroyed or not). As I said at the beginning of my question, I was experimenting trying to force an LED into controlled breakdown (current-limiting in order not to cause immediate destruction by too much power dissipation). I expected it to be damaged at some point. \$\endgroup\$ Commented Apr 3, 2018 at 12:46
  • \$\begingroup\$ The fact it was not means that either manufacturing spread in BD voltage is huge (could it be that the reason?) or that there are other, less known, failure modes/reliability issues that enter into play well before getting near the BD voltage. \$\endgroup\$ Commented Apr 3, 2018 at 12:48

Simply put, applying high reverse voltage to new LEDs for a few minutes is not a conclusive test. Reverse current in LEDs increases as they age (1), and I would expect that the breakdown voltage decreases as well. By the end of their lifetime more LEDs will break at lower reverse voltage values.


It would unnecessarily lower production yield.

Once you specify a higher than necessary (for LED usage) breakdown voltage, you would have to reject (or sell as a different grade) any production output that doesn't meet that specification but otherwise works A-OK as a LED. Unless the user needs a LED that can do double duty as a rectifier, this would only drive up cost and/or catalog complexity.

  • \$\begingroup\$ I believe this is the right answer. Other answers are just sugar-coated "I don't know". \$\endgroup\$
    – Damien
    Commented Dec 2, 2019 at 8:05

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