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Fuses and circuit breakers are often specified at a certain current they will 'blow'.

Increasing the current will also increase the power.

So if a fuse is rated for 12V DC and 20 A, this would be equal to 240 watts. If a different voltage is supplied, will this change the current at which the fuse will break? Does the fuse technically 'blow' at 240 watts?

If 6V DC was applied to this example fuse, 240 watts in this condition would be 40 A when the fuse would 'blow'. Am I correct? Or does the fuse always 'blow' at 20 A, regardless of the voltage?

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    \$\begingroup\$ What causes car crash mortalities? The speed or the deceleration? \$\endgroup\$ – Stian Yttervik Sep 21 at 14:40
  • \$\begingroup\$ Also look into fast-blow vs slow-blow if you're jumping in and learning things. Different fuses have been engineered for the link to melt after a different amount heating. Fuses protecting motors can typically sustain quite a lot more current than their listed rated for a very short time, whereas a similarly rated fuse for an electronic piece of equipment might blow near instantly when it's over the limit. \$\endgroup\$ – Sean Boddy Sep 21 at 16:35
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    \$\begingroup\$ Nobody has yet pointed out clearly that the rated current of a fuse is the maximum current that it will happily pass forever. Over that and it will blow, the greater the overcurrent the quicker it blows, depending on slow- or quick-blow design, etc. \$\endgroup\$ – Michael Harvey Sep 21 at 21:54
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    \$\begingroup\$ The likelihood of a fuse blowing is based mainly on how inconvenient it will be. \$\endgroup\$ – Hot Licks Sep 22 at 1:26
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    \$\begingroup\$ @StianYttervik Car mortalities are caused by energy absorbed by victim. Energy is proportional to speed squared. Deacceleration is proportional to speed squared for a given distance. When you drop an egg the chicken and the egg arrive at the impact point at close enough to the same time. | The old adage says "Speed kills" - I've not seen it posited that "deceleration kills' - even if it does :-) \$\endgroup\$ – Russell McMahon Sep 22 at 12:46

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It's the watts dissipated in the fuse itself not the watts in the system. Therefore since the fuse has resistance (R) it's the current, which provides that power I^2*R.

The voltage has nothing to do with it : at 6V, 12V or 240V, the fuse still blows at 20A. However you cannot use a low voltage fuse in high voltage applications : it will still blow at (strictly, slightly above) its rated current, but may sustain an arc that a HV fuse would extinguish.

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    \$\begingroup\$ @MikeBrockington How would the fuse know what the supply voltage is? No matter if the supply is 10 or 1000 volts, the voltage drop will be the same. \$\endgroup\$ – pipe Sep 21 at 16:48
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    \$\begingroup\$ @pipe it doesn't of course. The voltage drop across the fuse depends on the current through it, not the supply voltage. \$\endgroup\$ – Brian Drummond Sep 21 at 17:08
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    \$\begingroup\$ @pipe I think Mike Brockington is right, strictly speaking. The fuse's resistance, together with the load, forms a voltage divider. If the fuse's resistance increases due to heat, and the load's resistance remains unchanged, the voltage across the fuse will also increase slightly, heating the fuse further and thus accelerating the blowing process. In practice, this effect is hopefully minimal. Especially because a fuse is supposed to heat up before blowing. \$\endgroup\$ – Kevin Keane Sep 22 at 4:06
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    \$\begingroup\$ @MikeBrockington is correct. The voltage is relevant. I think that most people ignore the voltage because they assume that it remains fairly constant, which is true in the vast majority of the cases. It is incorrect to say that the fuse will blow all the same at 6V, 12V, or 240V. If you apply 240V to a circuit designed for 6V, the fuse will probably blow because the current in the circuit would be multiplied by 40 (240V/6V), thus the fuse would dissipate 40^2=1600 times more power. \$\endgroup\$ – luiscolorado Sep 22 at 22:23
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    \$\begingroup\$ Well, if you want to get pedantic, ultimately it's heat that causes it to blow. The temperature is a function of several things, including not only the current, but also how long it is applied and the thermal characteristics of the fuse. The fuse does not heat up instantly; it takes some time to warm up, therefore it is possible to significantly exceed the current limit without blowing the fuse if the pulse is short enough. This is also related to the energy dissipated into the fuse - volts (across fuse) times amps times time, also written as I^2 T after factoring in the fuse resistance. \$\endgroup\$ – alex.forencich Sep 23 at 8:33
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So if a fuse is rated for 12V DC and 20 Amps, this would be equal to 240 Watts. If a different voltage is supplied, will this change the Amps at which the fuse will break? Does the fuse technically 'blow' at 240 Watts?

All the fuse knows (before it blows) is the current passing through it. This might be: -

  • 20 amps from a 1 volt supply feeding a 0.05 ohm load or,
  • 20 amps from a 100 volts supply feeding a 5 ohm load.

The fuse knows nothing about load power. It is \$I^2 R_{FUSE}\$ dissipation in the fuse that causes it to heat and eventually blow (due to a combination of internal power dissipation and time).

Make sure the voltage rating is also sufficient or the fuse may not disconnect correctly. Also make sure that the fuse is capable of handling the large rupture current that could flow in some circuits; example: you can get fuses that are only 100 mA rated but, they have a rupture current rating of hundreds of amps.

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  • \$\begingroup\$ +1. And what about i²t rating? Does it play a role here? \$\endgroup\$ – Rohat Kılıç Sep 21 at 6:17
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    \$\begingroup\$ I've amended to make that bit clearer @Rohat \$\endgroup\$ – Andy aka Sep 21 at 7:09
  • \$\begingroup\$ Nice answer. Folks can also search on "current limiting fuses" to see a neat fuse we have used for a long time in grid. It forces a premature zero crossing (before the natural current zero would otherwise occur). Greatly limits how much energy can get downstream of the fuse. See Fig 6 here. \$\endgroup\$ – relayman357 23 hours ago
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The rating in current is the characteristic that defines when the fuse will blow up. The rating in voltage is the characteristic that defines how much the voltage can be without producing an arc after or while blowing up the fuse. Multiplying both values has no meaning.

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    \$\begingroup\$ Very important point, well made. And a very good illustration of why the question 'does a fuse blow due to excess power or excess current?' is equally meaningless. Might as well compare apples to elephants. \$\endgroup\$ – Simon Tillson Sep 23 at 10:13
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The correct answer is heat.

When current passes through a fuse, the fuse gets heated up due to the non zero resistance. More current means more heating. If the current and duration is enough to raise the temperature of the fuse above its melting point, the fuse will melt (blow).

It means you can push a higher than rated current for a very brief period of time without blowing the fuse.

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    \$\begingroup\$ And we have different types of fuses like slow-blow fuses vs. fast acting fuses, so we can kinda choose how "brief" the period of overcurrent is. There are even TT and FF ones. --- I can imagine some fuses being placed near a cooler. Would that ever be significant? \$\endgroup\$ – Thomas Weller Sep 22 at 8:54
  • \$\begingroup\$ Presumably, you could even hit it with a stream of liquid nitrogen (or whatever) to keep it from ever blowing. \$\endgroup\$ – fectin Sep 22 at 20:27
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    \$\begingroup\$ I knew some folks who participated in a particular autonomous aircraft competition, and as part of the competition requirements the electric motor that drove the propeller had to be powered via a very specific fuse. So, they attached a large heatsink to each terminal of said fuse and mounted it sticking out of the side of the plane so they could exceed the current rating without blowing the fuse. \$\endgroup\$ – alex.forencich Sep 23 at 2:07
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    \$\begingroup\$ @alex.forencich - that's a nice hack. Thanks for sharing this fun fact. :) \$\endgroup\$ – Whiskeyjack Sep 23 at 8:25
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As mentioned in the other answers, the fuse blows due to too much current flowing.

After the fuse has blown the circuit becomes open so a voltage develops across the fuse (usually the supply voltage like mains voltage or the battery voltage). The fuse must be able to withstand that voltage and keep the circuit open. That means that the voltage rating on the fuse must be higher than the voltages used in the circuit you're protecting.

Suppose you have a circuit that runs on 240 V and uses 0.5 A. You (wrongly) protect this circuit with a 1 A, 50 V fuse. When the fuse is intact (not blown) there is no issue, no more than 0.5 A flows through the fuse so it does not blow.

Then a fault develops in the circuit making more current flow and blowing the fuse. The fuse then opens the circuit and the 240 V develops across the fuse. 240 V across a fuse rated for 50 V! So the fuse might break or arc-over and no longer protect the circuit. This is why the voltage rating is also important but it only becomes important after the fuse has blown.

BigClive made a very interesting video about fuses, find it here.

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  • \$\begingroup\$ Wouldn't it be more correct to say "the fuse blows due to a too high current" as current is already defined at the rate of flow of electric charges? \$\endgroup\$ – geauser Sep 22 at 10:37
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It's technically something like "timeConstant/wattsOverLimit". Fuses are thermal, meaning they trip once they reach a certain temperature.

For reference, Watts = Current^2*R

The voltage doesn't matter, unless it changes the current and thus the watts.

The voltage does matter, because if you use a fuse at a higher-than-rated voltage, the fuse might fail.

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    \$\begingroup\$ Second paragraph: It is power, not watts. Watt is the unit. \$\endgroup\$ – Peter Mortensen Sep 21 at 15:09
  • \$\begingroup\$ I said watts, because I assumed he might not know that power and watts are the same thing. \$\endgroup\$ – Colin Marcus Sep 21 at 17:15
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    \$\begingroup\$ @ColinMarcus You should write either Power = Current^2 * Resistance or Watts = Ampers^2 * Ohms, not to mix different kinds of terms. Power and Watts are not the same thing at all. \$\endgroup\$ – Arvo Sep 22 at 10:25
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The current. The fuse has no idea how much voltage is involved.

There's only a small fraction of a volt drop across the fuse. The fuse has no terminals connected to common, neutral, ground or any other voltage ref. The entire fuse floats at supply voltage.

Until the fuse blows; then it has working voltage across it as long as the switch is on. That is the only reason fuses have voltage ratings.

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So first we have fuses and circuit breakers

And all kinds of fuses have defined characteristics like A,B,C ,..., fast, slow, lazy,...

Standards normally define a characteristic curve for fuses or circuit breakers. Usually these characteristics are very non-linear. This characteristic defines tripping points by multiples of nominal current. So while a minimum violation of the nominal current may require a fuse to trigger after minutes/hours, a violation by 3 or 5 might require the fuse to trigger in no time.

And while it's true, most fuses/circuit breakers work with a thermal principle, they are (always referencing the normal, common types) just monitoring the integral of I² over time.

https://en.wikipedia.org/wiki/Circuit_breaker#/media/File:Standard_Trip_Characteristic_of_a_Thermomagnetic_Circuit_Breaker.svg

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The fuse blows when it get too much energy. That is watts x time. If the energy in IR^2 x time is higher than it can dissipate, then it heats up and eventually blows. Time is critical.

A 1 amp fuse can withstand a 100 amp pulse, provided it is short. See fuse tables. Similarly a 1 amp diode can can take many more amps provided it is short.

Energy absorbed is the criterion.

See here Fuse characteristics

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Your questions: "So if a fuse is rated for 12V DC and 20 A, this would be equal to 240 watts. If a different voltage is supplied, will this change the current at which the fuse will break? Does the fuse technically 'blow' at 240 watts? If 6V DC was applied to this example fuse, 240 watts in this condition would be 40 A when the fuse would 'blow'. Am I correct? Or does the fuse always 'blow' at 20 A, regardless of the voltage?"

The fuse does blow due to power (heat), but the voltage rating of a fuse is not the voltage drop across the fuse in operation, so isn't used to calculate the power required to blow the fuse. A fuse is a non-linear device and it is designed to take advantage of being in series with a load. In normal operation, it dissipates very little power. But, as the current reaches the fuse's rated current, the power goes up, the heat goes up and the Resistance goes up, which in turn increases the V*I=Power=Heat... and poof goes the fuse's conductor opening the circuit as it is designed to do. To answer the question about a fuse blowing at rated current regardless of applied voltage... yes, but it's because it is a non-linear resistance in series with the load.

It is fundamentally wrong to say 'it has nothing to do with the voltage' though. Ohm's Law tells us that without voltage, in this case the voltage drop across the fuse, there is no current (V/R=I). Power is defined as V*I=P.

From Wikipedia: "The electric power in watts produced by an electric current I consisting of a charge of Q coulombs every t seconds passing through an electric potential (voltage) difference of V is

P=Work done per unit time=(VQ)/t=VI

Q is electric charge in coulombs t is time in seconds I is electric current in amperes V is electric potential or voltage in volts"

I^2R is the same as (V^2)/R is the same as VI

I believe the confusion with voltage 'not mattering' in this discussion, is that, the R of a fuse is NOT constant and non-linear... on purpose. Similar to an incandescent lamp filament, (for a 100W standard household filament, the R when cold is low (5-10 Ohms) and higher when hot (100 Ohms)). The fuse typically has a very low R when cold, but as the Power goes up, Vfuse*I=P), so does the heat.

The fuse's conductor has a very non-linear resistance to temperature coefficient, meaning, that as the power (V*I) dissipated goes up, it reaches a point at which R rises rapidly, the VI applied melts the conductor and blows the fuse. The power distribution in this series circuit shows the power dissipated by the fuse is very low when its R is low (in safe operation) and the power redistributes from the load to the fuse when the current reaches a critical point (trip or 'blow' point) in the R to Temp curve.

Voltage is very much in play. But, it's not the supply voltage or rated voltage of the fuse, it's the Voltage drop across the fuse along the R/Temp curve. Perhaps I'm being pedantic, but a fuse's usefulness really depends on it's non-linear R combined with the fundamental Power transfer law in a series circuit and power is a function of Voltage and Current.

Other answers are correct in that the Voltage rating of the fuse is important to prevent arcing when blown. The bottom line here is that for fuse protection, use rated current to protect the circuit against overcurrent conditions and use the rated voltage to ensure that the fuse doesn't blow and continue to conduct via arcing.

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