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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?

Peter Mortensen
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Jackson Harvey
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    What causes car crash mortalities? The speed or the deceleration? – Stian Sep 21 '20 at 14:40
  • 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. –  Sep 21 '20 at 16:35
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    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. – Michael Harvey Sep 21 '20 at 21:54
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    The likelihood of a fuse blowing is based mainly on how inconvenient it will be. – Hot Licks Sep 22 '20 at 01:26
  • ..@Hot Licks as it will be if it is directly soldered to a PSU PCB, rather than in clips or a barrel. – mckenzm Sep 22 '20 at 11:04
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    @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 :-) – Russell McMahon Sep 22 '20 at 12:46
  • @Russel Well, power and amperage is related by time integral, I thought I'd try to hilight it by a familiar example... – Stian Sep 22 '20 at 15:00
  • Isn't it just a matter of how the fuse is connected in the circuit? Since the fuse is *in series* with the load, the current is common to both fuse and load. If they made fuses that worked in parallel to the load (MOV?) then it would be voltage the variable that defines the behavior of the device. The "V or I is the cause" is a moot question when V and I are across and through the fuse (they are just the two faces of the same coin: power). – Sredni Vashtar Sep 22 '20 at 22:23
  • @StianYttervik I understood your point :-). That was (only somewhat) tongue in cheek. A key point is when the OP asks "Power or Amperage" they will almost certainly be referring to those parameters in the load in both cases. Whereas fuse blowing relates to current and power in the **fuse** (as I know you know). The OP would almost certainly not have had a concept of "power dissipation in the fuse" when they asked the question. ... – Russell McMahon Sep 22 '20 at 23:55
  • ... |ie a suitably voltage rated HRC fuse will blow at the same current in a 12V 50V 110V 230V 1kV 11kV circuit but the load power will vary in proportion to the load voltage (and I know you know that too). – Russell McMahon Sep 22 '20 at 23:55
  • @StianYttervik The negative distance between you and whatever was supposed to be in front of you tends to be a problem too. – Mast Sep 23 '20 at 06:34
  • @MichaelHarvey True. Which means that the exact answer to the question is a hearty "neither-nor"! If we take the fuse's resistance for constant both measures are equivalent; the answer is something like "the net energy dumped into the fuse's wire (electric power minus dissipated heat) exceeds X". That can be a very slow build-up, involving the fuse's casing heating up until heat dissipation becomes too slow, or a massive short spike which dumps the required energy without any significant time to dissipate. There is no fix maximum current/power. – Peter - Reinstate Monica Sep 23 '20 at 11:11
  • @Peter-ReinstateMonica Fuse makers publish graphs showing time to blow at various levels of overcurrent. – Michael Harvey Sep 23 '20 at 11:28
  • Does this answer your question? [Do fuse blow on ampere or effect?](https://electronics.stackexchange.com/questions/21794/do-fuse-blow-on-ampere-or-effect) – Doug Deden Sep 23 '20 at 22:38
  • Until a point of time, we had fuses that liquefied when the wire heated up too much. – Aravind Suresh Thakidayil Sep 24 '20 at 12:05
  • Duplicate/related (or at least I didn't word it as well): https://electronics.stackexchange.com/q/51366/2028 – JYelton Sep 24 '20 at 20:41

11 Answers11

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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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    The voltage does indeed have a slight effect - as the fuse warms up, the resistance will increase, and at that point the voltage drop will also rise in proportion to the supply voltage. – MikeB Sep 21 '20 at 16:15
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    @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. – pipe Sep 21 '20 at 16:48
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    @pipe it doesn't of course. The voltage drop across the fuse depends on the current through it, not the supply voltage. –  Sep 21 '20 at 17:08
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    @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. – Kevin Keane Sep 22 '20 at 04:06
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    @Kevin That's trivially true, arising because the fuse resistance R is temperature depnendent, but still doesn't imply a dependence on the supply voltage. –  Sep 22 '20 at 09:10
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    @pipe Simple application of Kirchoff really - the fuse has a very small resistance, but could not possibly function if it was a superconductor. It therefore always has a some voltage across it, proportional to BOTH the input voltage and the resistance of the rest of the circuit. – MikeB Sep 22 '20 at 15:44
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    @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. – luiscolorado Sep 22 '20 at 22:23
  • @luiscolorado It's not relevant. Fuses rate by [I^2t](https://www.littelfuse.com/~/media/automotive/catalogs/littelfuse_fuseology.pdf) - what matters is the current through them and the time that the current is applied. – J... Sep 22 '20 at 23:35
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    Is this true in the case where the load's resistance is very close to zero, i.e. where the fuse is blowing because of a short? While not the only mode of failure, it's a very common one. – R.. GitHub STOP HELPING ICE Sep 23 '20 at 03:20
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    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. – alex.forencich Sep 23 '20 at 08:33
  • @R..GitHubSTOPHELPINGICE That question really is moot, isn't it? If a fuse blows essentially immediately, then asking whether voltage affects the time it takes becomes meaningless. Of course, in reality there is no such thing as "immediate" and, yes, the laws don't change, so it is still true, even if we are now only talking about milliseconds or nanoseconds or less. – Kevin Keane Sep 23 '20 at 18:24
  • @luiscolorado Say you apply 240V to a 6V circuit. And say the resistance of the fuse is something like 1 ohm. And say the circuit is currently consuming 1A. The voltage accross the fuse is V = IR, which is 1A * 1 ohm which is 1V for a 240V circuit or a 6V circuit – slebetman Sep 23 '20 at 23:51
  • @slebetman It depends on the resistance of the main load. Let's say that a circuit is designed to be fed with 6 V, the fuse's resistance is 1 ohm, and the current is 1A. That means that the resistance of the load would be 5 ohms. The fuse is dissipating 1W in normal conditions. If you put the same circuit on 240V, the current would be I=V/(R1+R2)=240/(1+5)=40A. The fuse is using now I^2*R1=40^2*1=1600 W. More than most microwave ovens. The fuse will probably melt. – luiscolorado Sep 30 '20 at 18:40
  • @J..., As you can see at that site, I^2t applies when the fuse is operating under the nominally rated voltage: "System voltage exceeding the fuse’s rated voltage may result in fuse damage." The voltage is relevant, but it's excluded from most discussions because it is assumed that voltage will not exceed the established limits. – luiscolorado Sep 30 '20 at 18:46
  • @luiscolorado That note is strictly for arcing considerations. If the voltage is too high, the blown fuse could still sustain an arc if it is not designed to expect a higher voltage on the terminals. This has nothing to do with the blow time since a fuse carrying an arc will already have blown. Energy is energy - the voltage doesn't matter. – J... Sep 30 '20 at 18:49
  • @J... As you can see in the article, there are separate I^2t limits for arcing and melting. A higher voltage doesn't necessarily mean arcing. We can't simply say "voltage is irrelevant for fuse behavior", but we can say "voltage is irrelevant for fuse behavior only under certain circumstances." About your statement "energy is energy - the voltage doesn't matter", I'm not sure what you mean. Voltage, current and resistance are related. Power can be calculated as I^2R, but I is V/R. If you vary voltage, the current changes. The current is not independent of the voltage. – luiscolorado Sep 30 '20 at 19:18
  • @luiscolorado In a fuse it is, because the voltage across a fuse will only ever depend on the current through it, and a fuse blows on current. Obviously no component specifications apply when taken outside of the design range - for a fuse used ***within its design voltage limitations*** the current and time are the *only* parameters that matter. – J... Sep 30 '20 at 19:26
  • @luiscolorado It's like saying the weight of an object doesn't affect the accuracy of a scale - if you put 2kg on a scale the indicator will raise by 2kg, whether going from 0kg to 2kg, or 100kg to 102kg. The mass doesn't matter. Until you put 1000kg on a scale rated for 20kg and the whole thing is crushed. Obviously that's not a reasonable evaluation of the linearity of the scale. Same with putting 1kV across a 10V fuse. That's just silly - it doesn't have any impact on the I^2t discussion. – J... Sep 30 '20 at 19:32
  • @J... You are correct, it is silly to apply conditions out of the specs of the components. I'm just responding to the comments of Brian Dummond in the context of the main question. Brian claims incorrectly that the fuse blows "at 6V, 12V or 240V, the fuse still blows at 20A." As discussed, you can't change the voltage without affecting the current, so a voltage high enough would make the fuse melt or arc. – luiscolorado Sep 30 '20 at 19:44
  • @luiscolorado Yes, but you never use a fuse as the only element in the circuit. The fuse is never limiting current - it's designed to introduce as small a v-drop as needed to perform its function, so the voltage *drop* across the fuse will always be the same for a given current, regardless of the operating voltage of that circuit - because the full operating voltage is **never** applied to the fuse directly (unless there's a fault, in which case it will blow). Any fuse that you directly apply rated voltage across will blow, because they're basically negligible resistors over their rated range. – J... Sep 30 '20 at 21:56
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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.

Andy aka
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  • +1. And what about **i²t** rating? Does it play a role here? – Rohat Kılıç Sep 21 '20 at 06:17
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    I've amended to make that bit clearer @Rohat – Andy aka Sep 21 '20 at 07:09
  • 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](https://m.littelfuse.com/~/media/electrical/application-notes/powr-gard-technical-application-guide.pdf). – relayman357 Oct 23 '20 at 01:08
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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.

Paul Ghobril
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    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. – Simon Tillson Sep 23 '20 at 10:13
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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.

Bimpelrekkie
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  • 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? – geauser Sep 22 '20 at 10:37
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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.

Whiskeyjack
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    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? – Thomas Weller Sep 22 '20 at 08:54
  • Presumably, you could even hit it with a stream of liquid nitrogen (or whatever) to keep it from ever blowing. – fectin Sep 22 '20 at 20:27
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    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. – alex.forencich Sep 23 '20 at 02:07
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    @alex.forencich - that's a nice hack. Thanks for sharing this fun fact. :) – Whiskeyjack Sep 23 '20 at 08:25
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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.

Harper - Reinstate Monica
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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.

Colin Marcus
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    Second paragraph: It is power, not watts. [Watt](https://en.wikipedia.org/wiki/Watt) is the unit. – Peter Mortensen Sep 21 '20 at 15:09
  • I said watts, because I assumed he might not know that power and watts are the same thing. – Colin Marcus Sep 21 '20 at 17:15
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    @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. – Arvo Sep 22 '20 at 10:25
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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

JRE
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schnedan
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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

marsheng
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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.

Dutchman Electronics
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Lets think of fuse as small value resistor that burns with too much power.

If we move 1 Ω resistor from 12 V, 1 A circuit to be part of 230V, 1 A circuit the heating power of resistor will not change much. Low resistance of resistor and much higher resistance of high voltage circuit keeps the voltage and power of resistor close to the same.

In practice fuse is only affected by current.

Example of circuit

Cmazay
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