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Reverse voltage on LED.

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MikeMl

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High brightness red indicator LED, the kind that nominally operates at forward currents of ~10mA, at a Vf of ~1.8V. The LED is operated from a nominal 12V with a series resistor of ~ 1KΩ.

What happens to the LED if the voltage is reversed for a few tens of ms as it might due to an inductive spike? The current would be limited by the series resistor. Is special clamping with an inverse diode required, or is the intrinsic reverse breakdown voltage of the LED low enough to protect the LED?
 
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Most LEDs have an absolute maximum reverse voltage of 5V but some are 4V. Add a diode in series with it to prevent damage.
 
Most LEDs have an absolute maximum reverse voltage of 5V but some are 4V. Add a diode in series with it to prevent damage.

Wouldn't that be a diode in inverse parallel to shunt the reverse current and clamp it to ~-0.65V?

I do not have a data sheet for the LEDs I have. I have several hundred of them. I tested a half-dozen with my Tektronics curve tracer, and they all seem to have a Zener-like reverse breakdown at -15V +-1V (very sharp knee, wonder if you could use them as a Zener?).

Since I am limiting the current to a few tens of mA, driving them into the region where they conduct in the reverse direction doesn't seem to damage them. The instantaneous power dissipation at -15V at -20mA is 300mW, which seems within the ability of the package to dissipate.
 
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I just picked a datasheet randomly: https://www.electro-tech-online.com/custompdfs/2012/10/APT3216QWF-D.pdf

And the absolute max is 5V.

With diodes, the one thing that you have to remember is that reverse voltage is non-destructive If the current is limited. The Zener diode operates in reverse breakdown.

In an automotive electrical system, the negative spike could be on the order of -200 V. Damaging? Probably not. Degrading? Possibly.
LED's don;t burn out, but their intensity gets lower.
 
Since you have many volts to throw away with your 12V supply but only a single LED then I thought a series diode would be good to stop the reverse voltage. A reverse shunt diode would also work well.
 
With a diode in series with the LED the LED would see half the reverse voltage of the LED. Right?
 
Why?

The reverse voltages should add. e.g. 5V for the LED + 200 PRV for a diode, so 205 PRV.

What happens with 2 reverse biased Zeners in series, The voltages will ADD.
 
With a diode in series with the LED the LED would see half the reverse voltage of the LED. Right?
No.
The diode does not conduct when the voltage is reversed so the LED gets nothing.
 
Quality commercial products (especially in the old days) put a cap in parallel with the LED, right on the LED. I still do this now on power LEDs. The combination of series resistor and parallel cap gives excellent LED life.

I've replaced a huge number of dead LEDs in appliances over the years, and they were always LEDs without caps. I can't remember ever replacing a LED that did have a parallel cap. As far as putting diodes in series or reverse parallel that sounds a bit gimicky and I have not seen that done much in any commercial products.
 
... I can't remember ever replacing a LED that did have a parallel cap. As far as putting diodes in series or reverse parallel that sounds a bit gimicky and I have not seen that done much in any commercial products.

What value of cap? I have seen the inverse-parallel diode used in automotive applications, likely because a diode is cheaper and smaller than a cap?
 
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My application is light aircraft, which has a similar electrical environment to automotive. Here is a definitive reference that shows there is a potential problem, but doesn't offer a solution.
 
shows there is a potential problem, but doesn't offer a solution
It does mention using a high voltage silicon diode in series with the LED.
 
Seems like Fast Recovery has everything to do with the diode turn off. Turning on, not so much. I'm still favoring the inverse parallel Silicon diode connection. The diode capacitance helps protect the LED, the diode turn-on is what protects the LED, and the turn-off time of the diode doesn't matter. Looking for a definitive reference.
 
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Being that bi-color LEDs are available internally connected inverse parallel to each other, I'd say that's fairly definitive.
 
Fast recovery has to do with the ability to handle a second surge. Fast recovery basically means it's able to handle repetitive fast signals.

Inverse parallel in general if you don't include the LED resistor might also be prone with slowness. It is used for reverse spike protection.

Two LEDS back to back are often seen in AC applications. Your effectively clamping at 2.1V, so I Bi-color LED (one color mode) may give you the lowest component count.

Take a look at these products fron Linear technology: https://www.electro-tech-online.com/custompdfs/2012/10/2PB_surgestoppers.pdf
 
Fast recovery has to do with the ability to handle a second surge. Fast recovery basically means it's able to handle repetitive fast signals.
...
If there are two or more fast negative transients in rapid succession, do I care if the protection diode turns off between the transients? If anything, the slower the diode recovery is, the better, because it will still be conducting (will not have "recovered") when the second and subsequent transients come along...
 
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