While searching for fuse holders in the web, I found an infinity of pages like this, where the rating of the fuse holder is e.g.: "10A/250V AC, 15A/125V AC" (it is an example, values do not matter). I can't figure out what is the meaning these ratings: For the current rating, I can understand that the fuse holder can support at most e.g. 10A, since the contact between the fuse and the fuse holder has a small resistance that generates heat when crossed by big currents. But in this case, this rating should depend only on the current and not on the potential, so why a potential of 125V in the example above allows the larger current 15A? Moreover, the spec "AC" for the fuse holder is rather strange: what has a fuse holder to do with AC or DC ?
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1\$\begingroup\$ Fuses have to be bigger and better to do DC .Arcs are much worse on DC because there are no zero crossings like on AC for the ARC to go out and hopefuly not do too much harm. This difference becomes very significant at mains voltages . In fact it was one of the arguements for AC power Vs DC power when according to history Tesla and Edison had a spat. \$\endgroup\$– AutisticCommented Nov 21, 2015 at 22:18
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\$\begingroup\$ Thank you - So, this is related to the fuse spec. It would be interesting also to know if one can deduce some DC rating from the AC rating (a rule of thumbs, say) \$\endgroup\$– MikeTeXCommented Nov 21, 2015 at 23:02
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1\$\begingroup\$ "Rules of thumb" are not appropriate for fuse ratings. Fuses must always be applied according to information provided by the manufacturer. There may be exceptions to that, but they would require testing by the product manufacturer with independent testing lab verification. An AC rating on a fuse holder is most likely required because the holder was tested only with AC rated fuses. \$\endgroup\$– user80875Commented Nov 22, 2015 at 0:50
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2 Answers
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Fuses first: -
Fuses do need/have a voltage rating - you wouldn't use a fuse that is only qualified for 125V on a 230V AC system. A fuse that is specified for only 125V AC may fail to "break" adequately at 250V AC. These are safety devices after all.
You'll probably also find that some fuses have a "high-rupture-current" rating too. This defines the sudden massive surge of current that may cause a lesser fuse to form a plasma inside the glass/ceramic tube and therefore still conduct and be unsafe.
Fuseholders: -
They have to handle the current and they have to not go unsafe on the voltage that may be placed across them.
answered Nov 21, 2015 at 17:45
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\$\begingroup\$ My question was not about fuses, but since you have spoken about them, I'm surprised by what you said: I always believed that the voltage rating of a fuse was only relative to the rupture voltage of the fuse, which is an essential characteristic that comes into play only after the fuse has burnt. Regarding your answer about fuse holders, I am not entirely satisfied with it since it does not explain why there are two ratings 10A/250V and 15A/125V (if the fuse is safe with a voltage of 250V, what has this to do with the current rating?) \$\endgroup\$– MikeTeXCommented Nov 21, 2015 at 18:02
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\$\begingroup\$ @MikeTeX fuses and fuseholders have to pass certain safety tests and it is unlikely (given the small number of applications at 250V and 15A) that this scenario justified expensive testing. \$\endgroup\$– Andy akaCommented Nov 21, 2015 at 18:09
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\$\begingroup\$ Thank you for answering me. Regarding your first answer, are you saying that the fuse holder has, like fuses, a rupture voltage? if so, why does it depend on the current? Regarding your comment above, I don't understand how this relates to the cost of the testing. \$\endgroup\$– MikeTeXCommented Nov 21, 2015 at 18:26
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\$\begingroup\$ The thickness and area of the metal part of the fuse holder, the area of contact with the fuse and the spring force that maintains the contact has some effect on the heat generated by the current. The heat generated determines the current at which the fuse element melts. That means that each fuse rating and fuse holder must be tested together as a matching pair. It is not a question of the fuse holder failing, but causing the fuse to not melt at the proper current. \$\endgroup\$– user80875Commented Nov 21, 2015 at 19:22
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1\$\begingroup\$ The holder is suitable for fuses rated 10A and less rated 250V AC or less and fuses rated 15A and less rated 125V AC or less. In addition, fuses are usually designed so that they won't fit in holders that are not suitable. You need to make sure the fuses fit properly. I have seen fuses that appear to fit, but can be seen to not fit properly because there is a similar holder that they are designed for. Also make sure both fuse an holder have the proper test agency mark for the part of the world where they are used. \$\endgroup\$– user80875Commented Nov 21, 2015 at 20:50
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Since a fuseholder's contacts are resistive, and ohmic, increasing the voltage across them will cause the current through them to increase as well.
That increase in current will increase the \$ I^2 R\$ losses, causing the contacts to heat up and, since the contacts are in intimate electrical and thermal contact with the fuse's metallic contacts, much of that heat will be conducted into the fuse element itself, which will cause its temperature to rise, but not solely due to just the current through the fusible element.
answered Nov 21, 2015 at 19:19
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1\$\begingroup\$ I completely disagree with you. You are not supposed to apply any voltage between the pin of a fuse (this voltage appears only after the fuse have burnt, and should not exceed the rupture voltage of the fuse). Of course, since the fuse and the fuse holder have a very small resistance, such a voltage drop exist even when the current flows, but it is negligible and will never attain 250V ! \$\endgroup\$– MikeTeXCommented Nov 21, 2015 at 20:40
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\$\begingroup\$ @Mike Tex: You don't know what you're talking about, since what's being discussed is the power dissipated by the resistance of the fuseholder contacts (and, by the way, the contact resistances between the fuse contacts and the fuseholder contacts) at the two voltages in question and the temperature rise that'll cause being conducted into the fuse, which will lower its capacity. \$\endgroup\$ Commented Nov 21, 2015 at 20:56
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\$\begingroup\$ What I am saying is the following: Let A and B be the pins of the fuse holder that contains the fuse, and R be the total resistance due to the fuse and the contacts between the fuse holder and the fuse. You are not supposed to apply any voltage between A and B (this voltage appears only if the fuse has burnt). When some current flows, there does exist a voltage drop between A and B, but it is negligible and will never attain 250V (otherwise, R being so small, that would imply a current of thousands of Amperes flowing through the fuse) \$\endgroup\$– MikeTeXCommented Nov 21, 2015 at 21:08
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\$\begingroup\$ You still don't understand that you don't know what you're talking about since if there's current through the fuse, that same charge will flow through the fuseholder as well, heating up its contacts and transferring that heat into the fuse, lowering its capacity. \$\endgroup\$ Commented Nov 21, 2015 at 23:03
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\$\begingroup\$ If your aim is to explain that the contact between the fuse and the fuse holder generates heat, this is what I wrote in the question of the thread above, so you are not adding anything new. The fact that this lowers the capacity of the fuse is hence so evident that I didn't feel the need to write this in the question. Anyway, this does not help much to answer to the questions in the thread. \$\endgroup\$– MikeTeXCommented Nov 22, 2015 at 5:09