Diode failure due to wire self-inductance?

Thread Starter


Joined Jan 2, 2014
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.