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I recently bought a very cheap assortment of glass fuses rated for 250V (AC, presumably). Just to be on the safe side I decided to test that the fuses actually trip at the currents I want them to trip at.

To do that I used a constant-current DC power supply connected directly across the one of the fuses that was rated for a 0.2A current. I was able to gradually increase the current to above 2A before the fuse blew.

What I expected to see instead was the fuse blowing at around 0.2A. How come my expectation doesn't match reality? Are the particular fuses I got just out of spec? If what I saw in the experiment is to be expected, how would I derive the actual trip current of a fuse within the context of a larger circuit based on its rating?

Thanks!

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  • \$\begingroup\$ Maybe its a slow-burn fuse. \$\endgroup\$
    – Eugene Sh.
    Commented Jun 10, 2021 at 17:39
  • \$\begingroup\$ @EugeneSh. I've sustained the current for several minutes at above 2A without it blowing. They're also advertised as fast-blow fuses. \$\endgroup\$ Commented Jun 10, 2021 at 17:42
  • \$\begingroup\$ Well, we can't tell anything without seeing the specs. \$\endgroup\$
    – Eugene Sh.
    Commented Jun 10, 2021 at 17:43
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    \$\begingroup\$ You should watch this video: youtube.com/watch?v=WG11rVcMOnY where Dave from the EEVBlog shows how unpredictable fuses are. A fuse rated 0.2 A means that you should use that to protect circuits that need up to 0.2 A. It does not mean that the fuse will blow immediately above 0.2 A. If you put 10 A through that fuse it will blow almost immediately. \$\endgroup\$ Commented Jun 10, 2021 at 17:44
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    \$\begingroup\$ Related: electronics.stackexchange.com/questions/148156/… \$\endgroup\$
    – MarkU
    Commented Jun 11, 2021 at 0:28

3 Answers 3

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Like @EugeneSh. says, without knowing the spec for your fuses, there's no way to tell if it's out of spec or not. I will say that 10x overcurrent for 100s of seconds is a lot for a "fast blow" fuse but if you're buying a cheap assortment of fuses you're kind of rolling the dice.

Here's a datasheet for a fuse from a reputable manufacturer: https://www.littelfuse.com/~/media/electronics/datasheets/fuses/littelfuse_fuse_235_datasheet.pdf.pdf

You can see that their fuses should be able to withstand a ~10% overcurrent indefinitely, a 50% overcurrent for about 10 seconds, etc.

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Stop buying components out of Amazon.

Most are fake, either fuses, chips, passives, batteries etc..

If you supply your components through Amazon, aliexpress or the likes, be ready to waste a significant amount of time trying to understand why your circuit does not work properly.

Use reputable suppliers like digikey, mouser and so forth. Yes it's more expensive, and there is a good reason: You actually get what you ordered.

How much do you value your time, and your hair?

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  • \$\begingroup\$ This. Over and over again. The difference in cost isn't even that high. (if it is; you can be 100% sure that you are getting waste-products from the assembly line - instead of going in the trash, they go on Amazon and the likes) \$\endgroup\$
    – Arcatus
    Commented Jun 11, 2021 at 9:03
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Rated current will have a long holding time max (hrs) and wide tolerance for min, so they are rated like \$0.02 A^2t @ 0.2A\$ at 10xI

E.g. at 10 x 0.2A rating , =2A, t=0.02/A^2 = 0.5s fuse time.

Melting I2t 10.0 In typ. [A2s] for 0.2A @ 10x rating.

Thus accuracy of time improves with overcurrent load and not reliable is loaded just at fuse rating. So you choose a rating slightly over demand current averaged over a period of time like 150%.

What this means is fuses get warm with about 1/4W loss just below rated current, yet not enough to fuse it. Depending on the ambient and filament temp, exceeding this current above rating causes the temperature and due PTC also resistance, which in turn accelerates power loss in a \$I^2t\$ quadratic function thus reducing the fuse time. This is called thermal runaway but is slow at rated current in hours or less then accelerates with this quadratic characteristic.

Because the mass is small, it can only protect devices from damage due to thermal rise damage in a similar fusing quadratic characteristic and not say for semiconductors with wirebonds much smaller than the fuse unless of course rated higher than the fuse. In those cases, component protection demands other means for abrupt current cutoff like an open switch or a “crowbar” shorting the fuse with some current limit.

For example gastubes are often used as a over-voltage protection (OVP) crowbar and when triggered, drop in voltage to something like 60V across the grid source and need a fuse between gas tube and circuit to prevent melting a hole the board. ( seen that before) But that can protect the circuit from the following grid current due to a voltage breakdown on the power supply, triggered by lightning.

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    \$\begingroup\$ Thank you, Tony. I think there's a lot of good information in your answer that's very relevant to my original question, but I'm having trouble following some of the notation/abbreviations/lingo you've used in your answer. In particular, the statement Melting I2t 10.0 In typ. [A2s] for 0.2A @ 10x rating. looks very opaque/ambiguous to me. Would you perhaps be able to expand on that or make the notation a bit clearer? Thanks! \$\endgroup\$ Commented Jun 10, 2021 at 22:16
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    \$\begingroup\$ Ok added more details. \$\endgroup\$
    – D.A.S.
    Commented Jun 10, 2021 at 23:35

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