# Maximum reverse voltage applied to a resistor-LED configuration

Consider the simple circuit below:

simulate this circuit – Schematic created using CircuitLab

Will the applying of a reversed 36 V to this circuit damage the LED? Knowing that the maximum current that could flow into the LED is 36 V/2 kΩ = 18 mA (the LED can withstand 30 mA forward current).

The maximum reverse voltage is 5 V. Does that mean that the LED will actually blow up if I applied, say, a reversed 5.1 V on the input?

And in case the circuit is that sensitive to reverse voltage, what is the simplest way to protect this circuit from a reverse voltage?

EDIT: I actually tested the LED and it doesn't seem to be damaged by applying 30 V in reverse (applied for several minutes). Also I connected an ammeter to measure the reverse current (Fluke 87V with μA accuracy) and it shows absolutely no current flow. So what's the deal?

• 5.1 may still be fine, but 36 will sure kill it.
– user76844
Commented Sep 10, 2016 at 11:38
• Regarding the edit, I'd say you got lucky, but don't count on it (since the alternative isn't that hard) Commented Sep 10, 2016 at 12:20
• Leave the 30V few hours. :P Commented Aug 11, 2017 at 14:23
• I have done that at 28V. It worked without problems UNTIL I connected a BIG coil to the same power supply. This obviously spiked the voltage and killed the LED. Commented Jan 15, 2018 at 9:42
• The reverse current, or "dark current" is going to be in the micro amps. So it might be hard to measure it with a DMM. Especially if the DMM has an input impedance less than 10 M ohms. The voltage breakdown happens when the reverse bias exceeds the insulation voltage rating of the doped region separating the "N" and "P" semiconductor material. Commented Apr 14, 2021 at 12:07

"Knowing that the maximum current that could flow into the LED is 36V/2kOhm = 18mA"

Yes, but that statement implies that at 18mA the whole 36V would be dropped over the resistor, and actual volatage over diode would be 0 (zero) volts. :) (I had to include this for newbies, that could be misguided by statement, even as it probably meant just that in no case current can go over 18mA.) For diode it does not matter what voltage you input to the whole circuit, all the diode feels is the voltage between its two legs.

Trouble is voltage drop over resistor is porportional to the current, so if diode blocks the current completly ... no voltage drop on the resistor, and the whole supply voltage lands on diode itself.

In reality actual voltages over diode and the resistor is the result of the equilibrium. Diode might pass some current in reverse ... if it is not destroyed in shortcircuit mode (usualy it fails open to open circuit), so not so much voltage gets droped over resistor.

BUT. But it really depends on the diode characteristics (which you seldom find in spects for range above stated max reevesre voltage, and many times not at all as graph for revesrse voltage (only reverse current at max rev.V).

I have 5mm amber LED with Vf of about 2.2V which I tested and blocks current completly up to 41V (did not use higher voltage source) (voltage drop over 2K resistor was 0 mV).

Red 3mm LED with Vf about 2V, let some current in reverse even before 5V and wih higher voltage it got worse fast. But after putting LED back in "proper" way, it works as usual again.

Most currious is 3mm BLUE LED (rated 5mA, 3V) which gave enough light to be clearly visible in lighted surrounding conditions even at 50 micro amperes (yes, 0.05 mA) originaly. (Now after much missuse (bellow) things are different). This LED lets even more current in reverse than it lets forward at the same voltage. I reversebiased it to first to 22V and later to 40V for few secconds, where it let 18mA through (remember 5mA rated) in reverse! When i forward biased it again it DID NOT LIGT. Dead? No, after first abuse, it took near 16V of forward voltage to let any current through (I monitored voltage on 2K resistor, to detect current) an light it, as it light, forward voltage returned to near original value, and subsequently started at similar voltage as usual. After second abuse (40V reverse) it took even higher voltage (over 20V) to start again forward biased, but Vf permanently dropped to about 2V, and now it takes about 2.5mA at Vd of 2.45V (Supply voltage to circuit 8.9V) to light it the same as it lighted at only 0.05 mA originally.

BTW, I noticed simmilar fast deterioration effect at some chaep red laser diodes run at only close to 20mA (it's rated current).

None of the tested diodes did light reverse biased (not even slightly), at any reverse current (even 3x rated current), and none was destroyed by high reverse bisas. Except last case blue diode, other exibited no harm I coould notice form high reversal.

• Thank you for sharing your experiment. I understand now that in the case of reverse biasing a resistor-LED network, the equilibrium will determine the voltage on the LED and damage will occur if this voltage exceeded the absolute maximum rating. Now if the datasheet contained the current-voltage curve for negative voltages, we could theoretically calculate the operating point of the LED (voltage,current) at a specific reverse voltage applied to the resistor-LED network.
– fhlb
Commented Nov 20, 2020 at 11:10

The LED will definitely pass current in the reverse direction — nearly as much as in the forward direction. (Think of it as a 5-volt Zener diode.) This much current (and the heat that is produced) will definitely damage it, although probably not catastrophically.

The simplest solution is to put an ordinary diode in parallel with the LED, pointing the other way. This allows the reverse current to bypass the LED, limiting the reverse voltage across it to about 0.65 V, which is within its rating.

You can also use a second LED instead of a silicon diode, which will give you more light and reduce flicker, rather than letting the reverse current go to waste. The forward voltage of one LED will be less than the reverse rating of the other.

– fhlb
Commented Sep 10, 2016 at 12:15
• Reverse direction diode parallel to LED is a solution for only some cases. In other cases, when voltage reversed and of same magnitude as normally applied forward, the voltage across the ordinary diode will be only ~0.7V, compared to 1.4 to 2.5V for an LED, so the series current-setting resistor will allow substantially more current through the ordinary diode than originally intended for the LED. If, eg., driving an LED from an op amp, or from logic gates, that could be significant. It could make more sense to place an ordinary diode in series with the LED. Commented Feb 22, 2023 at 6:19
• @gwideman: With a 36-V supply, the ordinary diode will pass about 6% more current than the LED. I would not call this "substantially more". Commented Feb 22, 2023 at 14:26
• @DaveTweed Of course you are right for 36V example, which the OP posed. My comment was taking a more general and very common case of some other lower voltage applied to a single LED and resistor, or a voltage considerably higher but applied to a string of several series LED plus resistor. (Given that most readers are unlikely to have this exact example using 36V.) Commented Feb 28, 2023 at 2:44

Here is what the datasheet says in the ABSOLUTE MAXIMUM section about reverse voltage:

This is completely clear and unambiguous.

If you are still confused, look up "absolute maximum".

– fhlb
Commented Sep 10, 2016 at 12:14

I once saw an LED that is directly connected to mains 220V with only a 2W resistor in series. I saw it in a cheap UPS system and the LED indicated that mains is present. So that LED was actually blocking 220V. That LED is still functioning normally for 5 years now.

Many related posts here at stackexchange mentioned the datasheet being strict, with the LED having a much higher reverse breakdown voltage but with no reference

• I have done that some years ago. The LED worked for about 4 years. Maybe 5. But it was getting fainter in time. Commented Jan 15, 2018 at 9:40

The datasheet of LTL-307EE provides another spec besides the maximum reverse voltage of 5V. It says that IR_max = 100uA when VR = 5V. In that sense, 5V reverse voltage won't damage the part for sure.

Then how to address the concern over even higher reverse voltage cases which are not specified by the datasheets? I think this website provides some useful design guidelines.

Will the applying of a reversed 36 V to this circuit damage the LED?

The datasheet doesn't guarantee damage. It's the same deal as with C or C++ code that has undefined behavior. Software crashes may happen then, but aren't guaranteed. And, according to Mr Murphy, the LED damage and software crashes will happen either during a high-stakes demonstration, or when the system is deployed to an inaccessible location that costs \$10k to get to... Things always tend to "work fine" in the lab.

The LED may not be damaged - who knows. Or it may only get damaged if you power it in reverse when it's hot, or it has aged 1000 hours, etc.

The datasheet only states that if you apply more 5V reverse voltage, then you're on your own: if the LED gets damaged, you can't complain to the manufacturer. If you know enough about LED semiconductor processing and physics, and you can reverse-engineer the LED enough, you might be able to convince yourself that it's OK - or not. But, again, that's entirely on you, and you better be really, really sure.

The LED must not be subjected to reverse voltages of more than -5V if you want the LED to keep working. It's that simple.

Knowing that the maximum current that could flow into the LED is 36 V/2 kΩ = 18 mA

When the LED is reverse biased, no current flows across the resistor, so the resistor voltage is zero, and the entire reverse voltage is across the LED. See the V(D1.nA) plot below.

And in case the circuit is that sensitive to reverse voltage, what is the simplest way to protect this circuit from a reverse voltage?

1. By adding an anti-parallel silicon rectifier diode across the LED. See circuit in the middle below.

2. By adding a diode in series with the LED, and adding a bleeder resistor across the LED. See circuit on the right below.

The resistor is necessary since D5 conducts a tiny current when reverse biased, and this current could charge up the LED junction to a reverse voltage above -5V. This happens because the LED is connected to a high impedance node. By adding a bleeder resistor, the LED always "sees" impedance of 1Mohm or less across it, and no reverse charge can accumulate on it.

We can simulate what happens by driving the LEDs with 36Vrms square wave (72Vpp).

simulate this circuit – Schematic created using CircuitLab

We can plot the voltage across the LED D1 and D2 in the circuit above:

Note that the voltage across the unprotected LED D1 swings way below the maximum allowed -5V! The maximum negative voltage across the protected LED D2 is about -1V - well within the absolute maximum.

I tested a standard red and high-brightness blue LED. Test conditions were 60 V at 2 mA with a 22 kΩ resistor in series and the power supply was limited to 5 mA.

The red LED illuminated in the forward direction and did not conduct at all in the reverse direction. No damage seemed to be taken by the red LED.

The blue LED illuminated in the forward direction and conducted 2mA in the reverse direction.

The test was repeated with a 1 MΩ resistor in series, the current was around 55 μA.

The red LED illuminated very faintly in forward mode. The blue LED that was subjected to the high reverse voltage did not illuminate but was drawing 55 μA.
A brand new blue LED illuminated at 55 μA and was quite visible and bright. The reverse voltage did damage the first blue LED but at a much higher voltage than the manufacturer's specs and with only 2 mA of current.

No damage was observed to the red LED at all.

The question remains as to what happens when an LED is forward conducting but then is suddenly reversed?

I could not get another blue led to show damage at 55 μA of reverse current using the 1 MΩ resistor. It seems there are current levels a blue LED is prepared to tolerate when reverse biased.

While just about every LED datasheet states a maximum reverse voltage of 5 V for any color, red LEDs seem to be able to tolerate much higher voltages without breakdown. My power supply only goes to 60 V so with the particular LED I tested if it would be safe to use with a 60 V AC input.

The blue LED which also has a max. 5 V reverse voltage stated on the datasheet behaves differently and would not be safe at over 5 V. They will not blow up as such but will internally short circuit the silicon and cease to function.

Putting a 1N4007 diode in parallel back to back with your LED should protect any LED in an AC circuit. Datasheets are only a guide. Good engineering practice would dictate that you run tests for the particular part you are thinking of using. So to be clear on this question, the diode won't blow up but may go short-circuit and cease to function, in which case the current limiting resistor will do its job and limit any current to a safe level (usually no more than 10 mA to 20 mA).

• could you edit your answer and make sure it answers the original question? "The maximum reverse voltage is 5 V. Does that mean that the LED will actually blow up if I applied, say, a reversed 5.1 V on the input?" and " I actually tested the LED and it doesn't seem to be damaged by applying 30 V in reverse (applied for several minutes). Also I connected an ammeter to measure the reverse current (Fluke 87V with μA accuracy) and it shows absolutely no current flow. So what's the deal?" Commented Sep 1, 2022 at 16:26

LEDs are available with built-in series resistance. Honestly, if you don't know the part number of the LED, this discussion is pointless.

Vr is the guaranteed voltage that the LED can take and not blow up. It doesn't mean it will blow up at Vr. A batch with the same number may tolerate more voltage. Another batch may not. But Vr is guaranteed.

• “...Honestly, if you don't know the part number of the LED, this discussion is pointless.” I’m confused by this comment. Not only does the OP provide the part number, but links to its datasheet. Other than that, I agree with what you wrote. (Also, note that this post is over 3 years old). Commented Jan 28, 2020 at 3:40