# Diode failure due to wire self-inductance?

#### ericar

Joined Jan 2, 2014
1
See the attached jpg file that is a photo of a circuit with some notes.

I have a string of LED xmas lights that has failed with bridge diodes D1 and D2 shorted out.

The switch SW1 is actually the location of the strings female AC plug, where additional strings can be plugged in. In my case, I had four of these strings connected serially. Post mortem analysis found fuse F1 blown in all four strings, diodes D1 and D2 were shorted out only in string #3, the other three strings (string #1, string #2, and string #4) were all okay after replacing the blown fuses.

My hypothesis is that a short-circuit occurred in string #4 in its female plug that caused diodes D1 and D2 in string #3 to fail due to the self-inductance of the very long 22 AWG wire. The reason I think the short-circuit occurred in string #4 instead of in string #3 is because the fuse F1 was also blown in string #4.

In the attached circuit diagram, I have simplified the situation by showing only one string.

An unexpected short-circuit (shown as SW1) will cause fuse F1 to blow. After the short-circuit has occurred, but just prior to the fuse blowing open, the short-circuit current will be about (169V/0.8 = 211A). The long 22 AWG wire has self-inductance of about (25 [feet] * 2 * 0.414 [uH/ft] = 20.7 uH).

When the fuse blows with 211A flowing in the circuit (which will cause the circuit to suddenly open up) will there be a large voltage spike (V) across diodes D1 and D2 because of the wire inductance? If so, how do I calculate this voltage. Could this large voltage damage diodes D1 and D2 and cause them to fail?

P.S. I do not think that string #3 diodes failed first and then fuses blew because (if that were the case) the (down-stream) fuse in string #4 would NOT have blown.

Thanks.

Eric

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