Why does my varistor cause my fuse to blew up?

Question

I have made the following circuit (copied from a very similar picture on internet):

At one side of AC/DC I put a normal fuse.

The MOV is a 14D391K 14K391 390V varistor diameter 14mm AC 250V DC 320V (similar to this datasheet.

However, as soon as I put 220 AC on the circuit, the fuse blew. I would expect that it would blow when it gets a power spike of over 390 V (AC).

I don't know if the varistor still works, but I removed it and the circuit works as it should be.

What did I do wrong?

Update

I see that the voltage rating is 390 VDC, and the clamping rating is 650 V

1. Does this mean that this varistor is not meant to be for AC ?
2. How can the clamping rating be higher than the voltage rating?
3. In case this varistor is to be used for AC, why does my fuse blows up when using this varistor?

Answer 1

The MOV is probably toast. They fail shorted. You can measure it with an ohmmeter.

The rating (normal operation) is 250VAC (assumed to be RMS voltage of a sine wave), which is \$\sqrt{2}\cdot 250 \$, or about 354V peak.

The 390V refers to the (nominal) peak voltage when the MOV is just starting to turn on (1mA). It's not clamping much, but it could get hot if that condition persisted, especially with DC. It could be as much as 429V or as little as 351V.

When it is clamping a heavy transient (50A) the voltage can be as much as 650V.

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So if you anticipate surges no greater than 50A you can use it to protect a circuit that is capable of withstanding 650V peak.

Perhaps worth noting that MOVs "wear out" and a series of heavy transients can lead them to failing shorted. Of course if the surge is large enough they can fail open because the lead wires have been blown off etc.

Silicon devices such as bipolar TVS devices can provide sharper clamping without the wear-out mechanism (they will still fail- often shorted- if overloaded, of course).