# Reverse current kills the LED, why?



## rizky_p (Jul 15, 2008)

Ok this always bothers me, sorry for such a newbie question. 
Guys why is it when connecting +/- in reverse kills the LED(Cree/SSC), arent they basicaly a diode? diode will prevent reverse current right? and why a 5mm LED simply doesnt lit up and not destroyed when connected in reverse?

Thanks.


----------



## Changchung (Jul 15, 2008)

Hi, it is a diode, but in the small Leds or normal you put only 20ma, but with the big ones you put more current, 700ma or more, that kill the Led, if you wnat to try the Led before connecting use just a little of current, 20ma


----------



## lctorana (Jul 15, 2008)

Think of a normal rectifying diode, say, a 1N4004.

_That takes a continuous forward current of 1A, and exceeding this will burn the diode out._

_In the reverse direction, exceeding 400V by very much will force the diode into reverse breakdown ("avalanche"), where the diode can conduct rather heavily. And at (say) 500V, it doesn't take much current to build up enough wattage to burn the diode out._

Now let's get back to LEDs. The reverse breakdown voltage of a LED is very low indeed. And since the supply impedance to LEDs is inevitably very low, a heavy enough current can damage the LED.


----------



## PhotonWrangler (Jul 15, 2008)

What lctorasna said. A diode's maximum reverse voltage is typically spec'd as "PIV" (Peak Inverse Voltage). In a power diode such as the 1N4004 is substantial because it's designed to throw away half of a large sinewave. LEDs, while technically diodes, aren't designed for rectification. Their structures are optimized for maximum light extraction, which is kind of the opposite of maximum PIV.


----------



## 2xTrinity (Jul 16, 2008)

Practically what I've found is that while the PIV is lower for an LED than it is say a normal diode, it will still generally be higher than the normal forward voltage of the LED. This means if you have a direct drive setup, and put the batteries in backwards, you should survive okay.

However, the problem comes in if you have a regulator, a few scenarios there can cause death:

1) You wire the LED backwards to the regulator -- the board tries to push a constant current, and in doing so, raises the voltage above the PIV of the emitter and cooks it
2) You insert batteries backwards to a correctly wired light -- the LED itself probably won't burn out, but the driver might.

If you're worried, you could put a normal diode (in series with a resistor) in inverse-parallel with the LED. That is, the extra diode would then bypass the LED, and shunt any reverse current through your resistor.


----------



## rizky_p (Jul 19, 2008)

thanks.

cheers
Rizki P


----------



## lctorana (Jul 19, 2008)

2xTrinity said:


> Practically what I've found is that while the PIV is lower for an LED than it is say a normal diode, it will still generally be higher than the normal forward voltage of the LED.


I wouldn't want to rely on that as a golden rule, though.


----------

