# Power rating on Zero Ohm resistor

I was ordering some resistors online, and I saw that 0 Ω resistors have a power rating. Why is that? Power through a resistor is calculated with the equation $P = UI$ or $P = RI^2$. Since $R = 0\ Ω$, $P=0\ W$.

According to this post (How to calculate Power Rating for Zero Ohm Resistors?), a 0 Ω resistor has no power rating... But Farnell tells me the opposite:

• Don't trust Farnell , read the actual datasheet. May 27, 2017 at 20:31
• They are not perfectly 0 Ohms, so there will be some power loss across them. You might have to put a fairly high voltage on them, but at some point they will get hot and burn out. Imagine passing the starter current from your car's engine through that little jumper, and you can see what I mean.
– JRE
May 27, 2017 at 20:34
• 0R 1206 sometimes burn with a 8-10A current. Design fault of an old product of mine ;) May 27, 2017 at 20:58
• Note that even conductors like wire have maximum current capacities before they fail. May 28, 2017 at 0:48
• are there real 0 ohm resistors apart from SUPERCONDUCTOR? Don't trust the words 0 ohm May 28, 2017 at 6:40

While it may be true that distributors don't want to check every single part individually, in this case it is not down to laziness that the 0Ω resistor has a specified rated power of 125mW.

As pointed out by @BumsikKim's answer, the datasheet for the series does in fact specify this rating - the distributor product page is correctly representing the manufacturers specifications.

From Page 5, we have the following table entry:

Notice how for the entire RC0805 size series, there is a specified rating of 0.125W (1/8W). This includes the 0Ω resistors in that series.

There is also however crucially another specification - Jumper Criteria. This column specifies the rated current for an 0805 jumper (i.e. 0Ω resistor). We can see from the table your jumper is rated for 2A, with an absolute maximum of 5A (presumably short pulse).

So why might a "zero ohm" resistor have such ratings? Simple, it's not a 0Ω resistor. Unless the manufacturer of the resistor you are using have secretly made a room temperature superconductor, the jumper is actually still a resistor, just a very small one. According to the datasheet it is specified to be ~50mΩ or less.

Because the resistance is non-zero, some power will be dissipated. If we plug in the provided numbers, we actually find that the power rating is real and sensible:

$$P = I^2R = 2^2\times0.05=0.2W$$

So in the worst case resistance of 50mΩ, and at the rated current of 2A, it will be dissipating more than the 125mW rating.

Still think the rating is silly?

In a power supply design I had the pleasure of surge testing, the designer had added an 0805 0Ω resistor in series with a 24V DC input, just prior to a TVS diode. During the test, we charged a 10mF capacitor up to 200V and then connected the capacitor to the input of the power supply.

Naturally the TVS started conducting, and the 0Ω resistor turned quite literally into a firework...

• "the 0Ω resistor turned quite literally into a firework..." - I really wish you have a video of that :-D May 28, 2017 at 1:08
• The guys down in Non Destructive Testing may have cool toys, but they never get to have any fun.
– user39962
May 28, 2017 at 2:02
• A 10 millifarad capacitor at 200V? May 28, 2017 at 23:14
• shakes cane you kids with your metric system. In my day, we called that 10,000 microfarads! May 29, 2017 at 1:20
• @Harper: Microfarads are also metric. We do engineering here, and the convention is that you choose a power of 1000 multiplier to always have 1-3 digits left of the decimal point. Back in your day (I was doing electronics then too), they did other silly things too, like use micro-micro-Farads instead of pico-Farads. They also avoided nano-Farads. Good riddance to .001 and 20,000 microfarads! May 29, 2017 at 14:32

It is not really 0Ω. According to the datasheet, page 5, the resistance of the jumper (0Ω resistor) is less then 50mΩ, not the perfect 0Ω.

• Naturally - nothing is 0 ohms, is it May 27, 2017 at 21:52
• @TonyM right. So I am saying R=0Ω is wrong in the question. May 27, 2017 at 21:55
• The question text is correct, I think you're taking the words the wrong way. '0R resistor' is the term for that component, not a scientific statement. Same as '10K resistor' refers to a manufactured part, not a physical description of a theoretical entity with 10,000.0000000... ohms of resistance. They're nouns, not descriptors. May 27, 2017 at 22:03
• @TonyM Of course you are right but I am not sure why you argue that here. The OP thinks there will be no power rating since the resistance is zero so I pointed out there is resistance anyway and so is the power rating. May 27, 2017 at 22:10
• All your answer states is that the 0R (zero ohm) resistor actually has a resistance other than zero ohms. Which was pretty obvious - 'has a resistance of zero ohms' is not what '0R resistor' or 'zero ohm resistor' means. Sorry if that's taken as argumentative :-) May 27, 2017 at 22:15

The most likely explanation is that the resistor is part of a product series, and all product pages on Farnell have the same information for all values in the series.

I mean, if you're Farnell, you aint gonna pay someone to manually create each product entry for the E96 series into your database.

You would have a software tool which would create the product records according to a template. Like, enter the common data from the datasheet only once (brand, series, power, package, photo, etc), and then automatically create all values in the resistor series using these common datasheet values.

Since I once saw a mistake in a resistor manufacturer's part#, I guess the part# would be entered manually for each value too.

Now, 0R resistors aren't exactly 0 ohms, more like a couple tens milliohms, so yes, they do have a max current and max dissipation power.

• That makes sense. So my confusion come from Farnell's "laziness" May 27, 2017 at 20:43
• Also if you really want to nitpick, is the 0R really manufactured with "thick film" resistive compound just like the other values? I don't know. It could just as well be a metal plating on the ceramic resistor body. May 27, 2017 at 21:41
• I would hope they are using the manufacturer supplied values and not just using some template that may or may not be correct for a specific component. And indeed, as others have pointed out, the values do match what the manufacturer listed in the datasheet, and they are reasonable values. May 28, 2017 at 22:43
• Yeah, they would fill a template with the datasheet info, and use this to create the product records for the whole value range. I mean, who would want to manually copypaste the exact same info for each value in E96 series? May 29, 2017 at 11:29

It's really stating the power rating of the resistor family it belongs to.

Some 0R resistors are in place of a different value in future. If you place this 0R part on a board, that position will be able to accept any resistor in that family.

• ... if the chosen value results in a dissipation within the power ratings possible at that footprint, yes :) May 28, 2017 at 23:16
• @rackandboneman, it's the other way around, isn't it. You can place any value you like and dissipate up to 125 mW in it. The range of values are considered for when the circuit's designed. Series impedance resistor from clocks was 0R now 22R... Op-amp feedback circuit was unity gain (0R and N.F.) now x2 (10K and 10K)... Pretty common thing to do :-) May 28, 2017 at 23:26

0ohm resistor are not perfect. You can take 1mohm as value for you calculation. This will lead you to a very low power. You shouldn't bother much about it.

• I have actually measured a couple of 0402 0 ohm resistor. They were approximately 10 mOhms.
– Mike
May 27, 2017 at 21:00

As wikipedia says:

The resistance is only approximately zero; only a maximum (typically 10–50 mΩ) is specified.[*] A percentage tolerance would not make sense, as it would be specified as a percentage of the ideal value of zero ohms (which would always be zero), so it is not specified.

In the ideal world the 0ohm is the ideal wire. In this case the power is calculated as:

• $$P=RI^2$$ for current-driven applications and no power is consumed by the ideal wire.
• $$P=\frac{U^2}R$$ for voltage-driven applications and infinite power is consumed by ideal wire.

In the real world neither the ideal wire neither the actuall 0ohm resistor exists. That means some (little) power is consumed in current-driven applications.

That's why there are different 0ohm resistors with different power ratings; they do dissipate heat so they can be overloaded and burnt.

A physicist's perspective on a resistor $R = 0 \ \Omega$:

Applying a constant current source with finite current $I$, there is zero power dissipated.

In this case, $P = I^2 R = 0$, because $I$ is finite. Note that $V = I R$ is zero, and so $V^2 / R$ isn't infinite, even though $R$ is zero.

Applying a constant voltage source with finite voltage $V$, there is infinite power dissipated.

In this case, $P = V^2 / R = \infty$, because $V$ is finite. Note that $I = V/R$ is infinite, and so $I^2 R$ isn't zero, even though $R$ is zero.

More practically, if $R$ is small but nonzero, then by similar arguments:

• Applying a constant current source, the power dissipated is small
• Applying a constant voltage source, the power dissipated is large

The point isn't whether or not the resistance is exactly zero, but that applying a constant voltage source to a small [zero] resistance results in large [infinite] current in such a way that the final power dissipated is large [infinite and definitely nonzero].

Value of R should be rounded off and close to zero because that looks nothing like a superconductor. Safe to say that all electronical components have a non zero R value, even wires.