Based on my earlier question, since there supposed to be zero voltage (V=IR) drop across a 0 Ω resistor, how do we select the power rating of such a component?

For example, let's say I were to connect a zero-ohm resistor in between 5 V power source and a load (circuit variant) which take current ranges from 20-200 mA. What is the power rating of the 0 Ω resistor I should select?

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    \$\begingroup\$ That depends, is it 0 Ohms +-5% or +-1%? \$\endgroup\$ Dec 24, 2016 at 14:00

7 Answers 7


Yageo specifies both maximum current and maximum power, see page 5 of the datasheet:

\$P_{max}\$: 100 mW
\$I_{max}\$: 1 A

and you'll also see that for the jumper

\$R_{max}\$: 50 mΩ

That seems inconsistent: 1 A through 50 mΩ is only 50 mW, not 100 mW. In these cases you have to work with the lower value: 50 mW, since 100 mW would mean a 1.4 A current, which exceeds the 1 A limit.

EEs often scoff at the 5 % tolerance specification for the 0 Ω resistor. The engineers at Yageo know that that doesn't make sense, and if you take a good look at page 2 you'll see that they don't specify 5 % for the jumper at all:

F = ± 1 %
J = ± 5 % (for Jumper ordering, use code of J)

which should be read as "we use the same code for a jumper as the tolerance for other values". It does not implied that the 5 % tolerance would apply to the jumper.

Specifying maximum power isn't silly either: the part's weight and specific thermal capacity determine that, regardless of resistance value.


Zero-ohm resistors don't have power ratings, but they do have current ratings. You just need to select one that fits your needs.

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    \$\begingroup\$ actually they do have power rating, see here link \$\endgroup\$
    – Dennis
    Oct 11, 2012 at 4:30
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    \$\begingroup\$ Yes, it's pretty funny when a distributor makes a mistake like that (they also specify an error tolerance as a percentage). But if you drill down to the manufacturer's actual data sheet, you'll see the current rating for the zero-ohm resistor (jumper) that I'm talking about. \$\endgroup\$
    – Dave Tweed
    Oct 11, 2012 at 4:37
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    \$\begingroup\$ @Dave - it's not a mistake, and if you read the datasheet properly you'll see that they don't specify the 5 % at all; they just use the same code in the part's 12NC number. See also my answer. \$\endgroup\$
    – stevenvh
    Oct 11, 2012 at 7:14
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    \$\begingroup\$ Agreed. That 100mW rating is for 0603 resistors, not jumpers. Power rating does not apply for jumpers (or wire). Poor specmanship.. \$\endgroup\$ Oct 11, 2012 at 16:44

A zero ohm resistor (a.k.a. jumper) is a conductor. A piece of wire. A short piece of wire may have negligible resistance, but you can look at the resistivity: ohms per unit distance. If a wire is asked to carry too much current for its resistivity (and other attributes), then its temperature can rise, and that can happen to the point that it damages the circuit or even starts a fire. You wouldn't run small-signal hookup wire to a household socket, right? The conductor has to have the appropriate load carrying capacity for the current and for the application.

At only 200 mA, you do not have to worry about current, if you are using bare wire. According to the load capacity table in the Handbook of Electronic Tables and Formulas for American Wire Gauge, even 36 gauge wire can carry 200 mA when it's used for chassis wiring (not bundled into a cable for power transmission). This is only 5 mils thick. Some human hair is that thick, evidently.

Basically, you can use the clipped off terminal from just about any passive component as a jumper that will handle more than 200 mA.

22 gauge wire is about 25 mils thick and will take 7 amps. That's still thin enough to fit through 25 mil holes on a PCB, so why not use something close to that size. The less resistance, the better.

On the other hand, anything with significantly less resistivity than the PCB traces it is soldered to is overkill.

  • \$\begingroup\$ LOL 'some human hair...' +1 for content. \$\endgroup\$
    – kenny
    Oct 11, 2012 at 10:37

Some datasheets provide a power rating for the 0Ω resistors. From what I have seen, some companies use the maximum resistance value to calculate a power rating. Others use the same rating value as the low ohm resistors in their product range. Some will clarify that jumpers only have a current rating. Other datasheets might just be wrong.

This jumper datasheet from Vishay for example has current and power ratings for every component:

Excerpt of power and current ratings and resistance of some 0-Ohm resistors (from the Vishay datasheet)

This datasheet on the other hand, again from Vishay, only gives a current rating. The same is true for this from NIC Components.

If in doubt, it's probably best to contact the manufacturer and ask them to clarify.


Zero Ohm resistors are basically wires, packaged in a standard resistor package, and they mainly exist for easy handling in pick an place machines. Because the resistors come in a standard size package (foot print) the machine can grab and hold them properly, (bend the wires to the right pitch when through hole) and place them in or on the the PCB for soldering. My guess is that the power rating is more related to the package the resistor comes in, so the machines can be configured with standard component shapes.

Zero ohm resistors are commonly used for 'configuring' behaviour of a circuit in such a way that only a single PCB design is required to do two or more slightly differing tasks.

Zero ohm resistors can also be used when routing the PCB turned out to be impossible and an extra wire between two tracks is needed.


There is no Zero Ohm Resistor. That would contradict with the laws of Physics... Oooops, I mistakenly thought that even superconductors do have some resistance. Thanks @stevenvh for shedding light on this fact! (Though I still have a hard time accepting the fact that current can flow without inducing voltage, I need to catch up with the topic.)

But the rest still applies:

So your question is "how to calculate the power requirements of a very low resistance resistor". And mystery resolved.

If I had to do so, I'd assume the worst case, and assume that the 0Ohm resistor has the maximum actual resistance permitted by the datasheet, and calculate with that.

Also, why would anyone specify a 0Ohm resistor between a driver and a load? I would understand a small resistor (0.1-1Ohms) for overcurrent os short circuit protection, but for 0 Ohm resistors I can only think of some PCB layout, where two layers were not enough, and using some resistors, a "third layer" can be used for wires to jump over each other. Using that however isn't a clean design in my mind...

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    \$\begingroup\$ Superconductors are zero ohm resistors which, according to your answer, defy the laws of physics. To be clear, their resistance is not just very low, like 0.00000000001 ohm, it's absolutely zero. Also, a 0.1 ohm resistor in a 5 V line will limit the current, but then to 50 A. (FYI, the Yageo 0 ohm resistors I refer to in my answer are maximum 50 milli-ohm, that's nearly the 0.1 ohm of your current limiters. So you can use them as current limiters too, to 100 A?) \$\endgroup\$
    – stevenvh
    Oct 12, 2012 at 13:36
  • \$\begingroup\$ @stevenvh: Thanks, I corrected myself. For the second part: You are right, but I think that when a design is looked upon, it should reflect the intentions of the designer. If I want something to be as conductive as possible, I specify a 0 Ohm resistor. If I need a current limiting resistor, I specify its required value, with its tolerance... I'm a fan of clean code, and clean designs, and that would be misleading... (To be honest, in non-critical circuits, I did substitute specified parts for significantly different, but 'still within spec' ones myself, but I consider that hacking) \$\endgroup\$
    – ppeterka
    Oct 12, 2012 at 13:59
  • \$\begingroup\$ A Real Engineer has to be pragmatic. She paid good attention at school and knows the zero is not quite zero: "Yeah, OK, I know what you mean, I'll find the details in the datasheet." (Real Engineers know about datasheets.) Just like a 01005 size resistor. Nobody is going to complain that it isn't really 0.01 inch long, as the number suggests. (They're more like 0.016 inch.) A Real Engineer knows that, and she knows how to deal with it. -- HANWE! \$\endgroup\$
    – stevenvh
    Oct 12, 2012 at 14:09
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    \$\begingroup\$ A Real Engineer has to be pragmatic. That's what I say. A Real Engineer will never let a design leave his or her hands with component values not reflecting the intent of the design. (Mass producing is a whole different cookie though!) \$\endgroup\$
    – ppeterka
    Oct 12, 2012 at 14:33
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    \$\begingroup\$ "why would anyone specify a 0 Ohm resistor between a driver and a load" -- Testing. I'm building a prototype, I don't know if the power supply circuit will work, I don't know if the load circuit will work. So I'm going to put a 0 ohm resistor between them. I'm not going to fit it to start with, I'll temporarily solder wires onto the pads and check each part works individually. Then when I'm happy I will solder in the 0 ohm jumper. This can also be used to measure current flow, and to debug a broken board. \$\endgroup\$
    – user9876
    Aug 2, 2019 at 21:57

__ Also, why would anyone specify a 0Ohm resistor between a driver and a load? --

There are at least two reasons I use 0ohm resistors in my designs.

  1. First, I place them to allow power to be partitioned allowing for measurement/troubleshooting. In battery operated devices it is sometime difficult to understand where small currents are flowing. Adding a 0 ohm part to feed different paths allows measurement and troubleshooting. Since they cost almost nothing they can be useful in moderate volume designs (<10,000/year).

  2. Second, I use them to provide a well defined network tie point. The most common place would be a connection point between analog and digital ground. With each ground tied to one pin of a 0ohm resistor you can ensure that you get to completely control the tie point on a routed PCB. In a case like this it may also be useful to have the option of populating a ferrite bead in place of the 0ohm to inhibit noise.

  • \$\begingroup\$ I think you got the downvotes as this would be a separate question - and an answer too. You could post the question as a separate one (possibly with a reference to my answer, which originally had the question), and post your answer too - this way this content does not get lost, and you will have a chance to be rewarded... \$\endgroup\$
    – ppeterka
    Sep 11, 2019 at 1:48

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