0
\$\begingroup\$

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

enter image description here

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?
\$\endgroup\$
7
  • \$\begingroup\$ What's the fuse rated for? What's the circuit on the other side of the varistor? \$\endgroup\$
    – Hearth
    Oct 15, 2018 at 23:39
  • \$\begingroup\$ The varistor is rated for 390V, the fuse is rated for 220/250 V, 0.2A. The circuit is currently 5 AC/DC converters (Hi Links, to 5V), connected to nothing (so far). \$\endgroup\$
    – Michel Keijzers
    Oct 15, 2018 at 23:41
  • 1
    \$\begingroup\$ What does the varistor measure on a multimeter? \$\endgroup\$
    – Hearth
    Oct 15, 2018 at 23:45
  • \$\begingroup\$ What's the expected current draw of your power supplies? Are you sure it's less than 200mA? Because of it isn't, your fuse will blow from the load, not the varistor. \$\endgroup\$
    – DSWG
    Oct 15, 2018 at 23:47
  • \$\begingroup\$ @Felthry I took it out already, a bit hard to test it since I had to cut off the legs. \$\endgroup\$
    – Michel Keijzers
    Oct 15, 2018 at 23:48

1 Answer 1

Reset to default
1
\$\begingroup\$

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.

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).

\$\endgroup\$
3
  • \$\begingroup\$ Thanks, I used bipolar TVS for my (RS485/DMX512) signals (for 15V), so I will order some 390V TVS diodes for the AC signals than, thanks. \$\endgroup\$
    – Michel Keijzers
    Oct 16, 2018 at 10:51
  • \$\begingroup\$ I added a new question: electronics.stackexchange.com/questions/401418/… ... in principle: would a P6KE400CA be a good choice? \$\endgroup\$
    – Michel Keijzers
    Oct 16, 2018 at 11:08
  • \$\begingroup\$ I found P6KE440CA TVS diodes, which have a reverse breakdown of 374 V, which should be good I guess. \$\endgroup\$
    – Michel Keijzers
    Oct 16, 2018 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.