0
\$\begingroup\$

I understand that a DC fuse's voltage rating is determined by the possibility of a connection-sustaining plasma arc. However, unlike the current rating, which is based on the very well understood principles of resistive heating, the voltage rating is based on the likelihood of plasma creation, and this seems to be much more stochastic. I expect that as a result, the current rating is very precise, but in contrast the voltage rating is very conservative.

I'm interested in how much excess voltage is actually required in typical circumstances to see plasma arcing. I'd also like to know what factors affect the likelihood of plasma creation, and how do they play into the fuse's real-world behavior.

UPDATE: Another way to phrase this is that I'm looking for an understanding of DC fuses' design safety factor and how environmental factors affect the real-world safety margin.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ If you have you not watched BigClive's video on fuses yet, you should: youtube.com/watch?v=kx35WN3uLis Realize that the voltage rating does not need to be accurate, as long as the fuse can at least withstand its rated voltage, there is no issue. I mean, that voltage rating will never be say 200 V +/- 10 %. That 200 V means that it is guaranteed to be at least 200 V. In practice maybe the fuse can withstand 400 V. That's 100% more than 200 V so very inaccurate but that does not matter. \$\endgroup\$ – Bimpelrekkie Mar 30 at 15:16
  • \$\begingroup\$ I have updated the question to clarify that I am looking to understand the upper voltage bounds. \$\endgroup\$ – Kenn Sebesta Mar 30 at 15:30
-1
\$\begingroup\$

The voltage rating (for an AC fuse) is related to the rated breaking current and some assumed frequency.

If the breaking current is higher than rated interrupting rating then the fuse may not open expeditiously (see example below).

If the frequency is much lower than typical mains frequency the voltage may need to be derated or the fuse may not be guaranteed to work at all (DC being an extreme example).

Image from Cooper-Bussmann:

enter image description here

Of course current rating has to be reduced for high altitudes because of the lower convection cooling, that also applies to dielectric strength over 1000m. Some (very few) fuses are hermetic, but the fuse holder may also come into play.

See, for example, IEEE C37.40-1993 which shows a derating of voltage to 0.56 at 20,000 feet (6100m) and of current to 0.90.


Edit:

As a concrete example, compare the Littlefuse 5x10mm fuses on this datasheet:

Voltage rating is 125VAC with 10kA max interrupting rating

Voltage rating is 250VAC with 35A max interrupting rating

If you attempt to interrupt 10kA fault current at 250VAC the fuse will arc from end to end and explode spraying shards of glass everywhere. It has no problem at 125VAC. Same fuse.

\$\endgroup\$
9
  • \$\begingroup\$ This is interesting, but I also feel it's not addressing the question. Can you help me understand how this relates to specific DC voltage ratings vs DC real-world plasma performance? FWIW, temperature decrease due to ambient lapse rate more than compensates for reduced density and thus reduced mass convection capacity, so this needs to be taken into account if derating solely because of cooling issues. \$\endgroup\$ – Kenn Sebesta Mar 30 at 15:58
  • \$\begingroup\$ I think you're getting into fuse design issues, which are outside of the scope of my answer and might be a better question for physics SE. I can state from empirical experience that breaking current is a very important factor, so it's probably meaningless to discuss voltage rating without breaking current. \$\endgroup\$ – Spehro Pefhany Mar 30 at 16:02
  • \$\begingroup\$ I feel like Physics.SE is perhaps too theoretical. I'd like to know the practical effects, born out of experience. By way of analogy, Physics.SE is where I'd go to ask for the Maxwell equations of objects inside a microwave cavity, but here is where I'd go to ask what practical metal shapes tend to arc. For current, I think it's fair to assume that the current in question is the DC fuse's rated current. So a concrete question might be "If I have a 300A/80VDC fuse and I run it at 100V, under what conditions is it likely to arc when the fuse breaks?" \$\endgroup\$ – Kenn Sebesta Mar 30 at 16:14
  • \$\begingroup\$ That is not at all fair to assume. The voltage rating is at the fuses maximum interrupting current which is always far higher than the rated carry current. Might be 10,000A for a 1A fuse. Or it might be 25A. See my edit above. Voltage rating and interrupting rating go hand-in-hand. \$\endgroup\$ – Spehro Pefhany Mar 30 at 16:21
  • \$\begingroup\$ I am as of yet unconvinced that it is not a fair practical assumption. The edits you added are for AC, which, in the sense that it encompasses an infinite frequency range, is a different beast from DC. I am convinced by what you say about AC, but the question is specific to DC. tme.com/Document/e9b9699950c3171aea7f9f2ddcb91cca/… is an example of a high power DC fuse, and the docs do not derate current capacity as a function of voltage. \$\endgroup\$ – Kenn Sebesta Mar 30 at 18:30

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.