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Consider the simple circuit below:

schematic

simulate this circuit – Schematic created using CircuitLab

LED datasheet: link

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

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

And in case the circuit is that sensitive to reverse voltage, what are 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 30V in reverse(applied for several minutes). Also I connected an ammeter to measure the reverse current (Fluke 87V with uA accuracy) and it shows absolutely no current flow. So what's the deal ?

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  • \$\begingroup\$ 5.1 may still be fine, but 36 will sure kill it. \$\endgroup\$ – Gregory Kornblum Sep 10 '16 at 11:38
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    \$\begingroup\$ Regarding the edit, I'd say you got lucky, but don't count on it (since the alternative isn't that hard) \$\endgroup\$ – ilkkachu Sep 10 '16 at 12:20
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    \$\begingroup\$ Leave the 30V few hours. :P \$\endgroup\$ – Jason Han Aug 11 '17 at 14:23
  • \$\begingroup\$ 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. \$\endgroup\$ – Gravity Jan 15 '18 at 9:42
  • \$\begingroup\$ 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. \$\endgroup\$ – David Mikeska Apr 14 at 12:07
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"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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – fhlb Nov 20 '20 at 11:10
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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.

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  • \$\begingroup\$ please see edit \$\endgroup\$ – fhlb Sep 10 '16 at 12:15
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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".

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  • \$\begingroup\$ please see edit \$\endgroup\$ – fhlb Sep 10 '16 at 12:14
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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

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  • \$\begingroup\$ I have done that some years ago. The LED worked for about 4 years. Maybe 5. But it was getting fainter in time. \$\endgroup\$ – Gravity Jan 15 '18 at 9:40
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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.

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  • \$\begingroup\$ this is the leakage current when a reverse 5V is applied directly to the LED. when a resistor is added in series, it will limit the reverse current. If the datasheet supplied the "reverse current as a function of the reverse voltage" curve, the reverse current would be easily calculated and hence the power dissipated in the LED. Then we would figure whether the diode will get damaged knowing that the absolute max power dissipation is 100mW, \$\endgroup\$ – fhlb Feb 21 '20 at 12:15
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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.

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    \$\begingroup\$ “...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). \$\endgroup\$ – Blair Fonville Jan 28 '20 at 3:40

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