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I have a 0 Ohm resistor here.

Can someone tell me how the power dissipation will be calculated for this type of 0 ohm resistor.

The part says it has a power rating of 0.1W.

But, if we use either V squared over R formula or the I squared R formula, in both of the case, we would either get 0 or infinity for the power dissipation of the resistor.

So, how to calculate? If we cannot calculate, then why does the manufacturer provide a power rating value?

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    \$\begingroup\$ There is no such thing as a 0 Ohm resistor. Its like saying some object has a size of 0 meters. \$\endgroup\$ – kebs May 14 at 9:00
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Even nominally "0 ohm" resistors have some resistance.

The datasheet for the part you linked says that zero-ohm resistors have "< 20 mOhm" of resistance.

I don't know how you're using this part, but in most cases for a resistor that small, the exact value of the resistor won't affect the current flowing through it very much, so you can conservatively assume that the resistor value is actually equal to 20 mOhm and use P = I^2 R to find the maximum rated current.

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    \$\begingroup\$ @Newbie learn to read all the specs in a datasheet and don’t ignore the fine print. \$\endgroup\$ – Tony Stewart EE75 May 13 at 12:34
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This is certainly not really 0Ω, since I’m pretty sure it is not made out of superconductor material. Hence, although we don’t know its value, it will dissipate some power as soon as you give it some current. The datasheet should give you the maximum ohmic value of this resistor.

The power rating is determined by the package and how much power it can dissipate to the PCB and the air before it overheats and melts.

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The datasheet provides the real resistance (worst case) and current rating for the "jumper":

enter image description here

Edit: Moving my comment here (thanks to @ScottSeidman)

This component has a minimum value of 1 \$\Omega\$ (with 1% or 5% tolerance) but also offers a "jumper" ideally with 0 \$\Omega\$ (with only a worst case parasitic resistance of 20 m\$\Omega\$ specified).

If you perform the power calculation, you would be calculating the worst case power dissipation. You could just respect the current rating instead.

If you need smaller resistors with known values, there are specific products for that: current sensing

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  • \$\begingroup\$ Thank you for the answer. So, we can assume the resistor to be 20mohms and calculate the power dissipation? \$\endgroup\$ – Newbie May 13 at 12:07
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    \$\begingroup\$ @Newbie No need to calculate the power dissipation, you may simply consider the \$I_{\textrm{max}}\$ rating that is given in the table. \$\endgroup\$ – user2233709 May 13 at 12:15
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    \$\begingroup\$ You can see that this component has a minimum value of 1 Ohm and a "jumper", with worst case parasitic resistance of 20 mOhm. If you do that calculation, you would be calculating the worst case power dissipation. You could just respect the current rating. If you need to know the values, there are specific products for that: vishay.com/resistors-fixed/current-sensing \$\endgroup\$ – devnull May 13 at 12:19
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    \$\begingroup\$ @vangelo -- I read that table as showing a resistance under 20mOhm for the Jumper, and the 1Ohm number reflects the minimum of resistors in the series that aren't jumpers. \$\endgroup\$ – Scott Seidman May 13 at 14:28
  • \$\begingroup\$ @ScottSeidman This is exactly my understanding as well. I'm sorry my comment was unclear. \$\endgroup\$ – devnull May 13 at 14:44
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An ideal zero ohm resistor would have zero power dissipated in it for any finite current through it, on the other hand it would have infinite power dissipated for any finite voltage across it.

Ideal zero ohm resistors don't exist in the real world though. A real "zero-ohm" resistor will have some resistance.

Resistor datasheets usually cover a whole series of resistors. Distributors use these datasheets to fill in the parametrics on their websites. However this can lead to parametrics that are misleading and/or nonsensical, particularly for more extreme components in the range. To Digi-Keys credit in this case they have filled in the tolerance column with "jumper" rather than a meaningless percentage.

Generally a series of resistors will have a maximum power rating, but it may also have maximum voltage and/or current ratings. For medium resistances the power rating is normally the limiting factor, but for high-value resistors the voltage rating is often the limiting factor and for very low-value resistors the current rating may be the limiting factor.

The datasheet specifies a maximum resistance of 20mΩ for the jumper. If power rating was the only constraint one could use this to calculate a maximum safe current.

$$ P = I^2R$$

$$ 0.1 = I^2 0.02 $$

$$ I^2 = 5 $$

$$ I \approx 2.24 $$

However the row in the table also says "Jumper, Imax. = 2.0 A". To me that says you shouldn't go over 2A, even though the power rating would theoretically allow for slightly more.

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