I remember playing with signal LEDs and observing what happens when I misuse them, and one particular LED behaved very strangely, as if it were stateful.

Using a drained 9V battery (that can only deliver about 100-200 mA), the (probably green) signal LED and a tester, I remember observing the following cycle:

  • I apply the current forward, the LED shines dimly, current is rather constant (5-15 mA?)
  • I apply the current backwards, the LED obviously does not emit any light and the meter showed small, but raising current (beginning under 1mA)
  • The current keeps raising and suddenly spikes (50-100 mA?), as if the LED is almost short circuited
  • I reverse the polarity again. The same relatively high current flows in other direction, no light is emitted. But the current slowly decreases.
  • Suddenly the current drops to about 5-10 mA and the LED starts emitting light again (first dim, then a bit brighter).
  • The more time the LED is turned on "normally", the more time of reverse 9V needed to trigger the breakthough. The more current flowed backward, the more forward-flowing time needed to "repair" the LED.

It didn't look like as if it were just temperature. If I waited a bit the "state" preserved.

How can this stateful behaviour be explained? What happens when current flows backwards though LED? Why could it be "healed" by forward current?

  • 1
    \$\begingroup\$ You've almost certainly exceeded the reverse breakdown voltage with a 9V battery. What were you were trying to achieve, exactly? \$\endgroup\$
    – Finbarr
    Commented Nov 21, 2017 at 14:04
  • \$\begingroup\$ @Finbarr, Just playing with it. Testing if will work almost without issues, shine wrong colour, become degraded (e.g. conducting current in both directions even at low voltage like a wire, but still emitting light), "magic smoke" (and shine like an incandencement bulb the last time for a brief moment) or just scatter around with a bang (if discharge a big 200V capacitor into it, regardless of polarity). The same reason as it experiment with physically destroying chip of a cheap pocket calculator to see how functions disappear one by one, buttons stop working, display segments go off... \$\endgroup\$
    – Vi0
    Commented Nov 21, 2017 at 14:44

1 Answer 1


The reverse current heats the crystal and reduces the breakdown voltage. Then there is a breakdown.

The area of the semiconductor transition is significant in relation to its thickness. The breakdown takes place in a much smaller area. In a place that is more prone to breakdown because of the lack of manufacturing.

The breakdown location behaves like a resistor and not as a semiconductor.

Direct current flows mainly through this "resistor". It works like a shunt so there is no light. Next, the "resistor" warms up and blows off.The breakthrough area is separated from the surviving part of the crystal. The LED starts to shine.

But in fact, LED is irreversibly damaged.


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