Sıfır Ohm rezistöründeki güç derecesi

30

Çevrimiçi olarak bazı dirençler sipariş ediyordum ve 0 dirençlerin güç derecesine sahip olduğunu gördüm. Neden? Bir rezıstoru aracılığıyla güç denklemi ile hesaplanır ya da . Yana , . $P = UI$ $P = RI^2$ $R = 0\ Ω$ $P=0\ W$

Bu yazıya göre ( Sıfır Ohm Dirençler için Güç Sınıfı Nasıl Hesaplanır? ), 0 Ω dirençli güç sınıflaması yoktur ... Ama Farnell bana bunun aksini söylüyor:

power resistors

— M.Ferru
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13

Farnell'e güvenme, gerçek veri sayfasını oku.

— Photon

11

Mükemmel 0 Ohm değiller, bu yüzden onların arasında bir miktar güç kaybı olacak. Üzerlerine oldukça yüksek bir voltaj koymak zorunda kalabilirsiniz, ancak bir noktada ısınırlar ve yanarlar. Marş akımını arabanızın motorundan bu küçük atlayıcıdan geçirdiğinizi hayal edin ve ne demek istediğimi anlayabilirsiniz.

— JRE

7

0R 1206 bazen 8-10A akımla yanar. Eski bir

— ürünümün

6

Kablo gibi iletkenlerin bile, arıza yapmadan önce maksimum akım kapasitesine sahip olduğuna dikkat edin.

— Todd Wilcox

4

SUPERCONDUCTOR dışında gerçek 0 ohm dirençler var mı? 0 ohm kelimelerine güvenmeyin

— JavaMan 28:17

45

Distribütörlerin her bir parçayı ayrı ayrı kontrol etmek istemedikleri doğru olsa da, bu durumda 0Ω direncinin belirtilen 125mW gücünde bir tembelliği yoktur.

@ BumsikKim'in cevabının işaret ettiği gibi , dizinin veri sayfası aslında bu derecelendirmeyi belirlemektedir - distribütör ürün sayfası üreticinin spesifikasyonlarını doğru şekilde temsil eder.

Sayfa 5'ten itibaren aşağıdaki tablo girişine sahibiz:

Tüm için nasıl dikkat edin RC0805 boyutu serisi, 0.125W (/ 8W 1) belirli bir derece vardır. Bu, o serideki 0Ω dirençlerini içerir.

Ancak çok önemli bir özellik daha var - Jumper Criteria . Bu sütun, 0805 jumper için anma akımını belirtir (örn. 0) direnci). Jumper'ınızın 2A olarak derecelendirildiği, mutlak maksimum 5A (muhtemelen kısa pulslu) tablosundan görebiliyoruz.

Öyleyse neden bir "sıfır ohm" rezistöründe böyle bir derecelendirme olabilir? Basit, 0Ω direnç değil. Kullandığınız rezistör üreticisi gizlice oda sıcaklığında bir süper iletken yapmış olmadıkça, atlatıcı aslında bir dirençtir, sadece çok küçük bir tane. Veri sayfasına göre ~ 50mΩ veya daha az olduğu belirtilmiştir.

Direnç sıfır olmadığından, bazı güçler tükenir. Sağlanan numaraları girersek, aslında güç değerinin gerçek ve mantıklı olduğunu görürüz:

P = I^{2} R = 2^{2} \times 0.05 = 0.2 W

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

Bu nedenle, 50mΩ'luk en kötü durumda direnç ve 2A'lık anma akımında, 125mW değerinden daha fazla dağılıyor olacak.

Yine de reyting aptal mı?

Güç kaynağı tasarımında dalgalanma testinden zevk aldım, tasarımcı, TVS diyotundan hemen önce 24V DC girişli bir seri 0805 0Ω direnç ekledi. Test sırasında, 200V'a kadar 10mF'lik bir kondansatör şarj ettik ve kondansatörü güç kaynağının girişine bağladık.

Doğal olarak, TV'ler yayılmaya başladı ve 0 quite direnci tam anlamıyla bir havai fişek gösterisine dönüştü ...

— Tom Carpenter
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7

"0Ω direnci tam anlamıyla bir havai fişek gösterisine dönüştü ..." - Gerçekten bunun bir videosunu diliyorum :-D

— John Dvorak

7

Tahribatsız Muayene Testinde çalışan çocuklar harika oyuncaklara sahip olabilirler, ancak hiç eğlenemezler .

— Sean Boddy

2

A 10 millifarad capacitor at 200V?

— rackandboneman

3

shakes cane you kids with your metric system. In my day, we called that 10,000 microfarads!

— Harper - Reinstate Monica

6

@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!

— Olin Lathrop

7

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.

— Bumsik Kim
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Naturally - nothing is 0 ohms, is it

— TonyM

@TonyM right. So I am saying R=0Ω is wrong in the question.

— Bumsik Kim

1

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.

— TonyM

1

@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.

— Bumsik Kim

1

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 :-)

— TonyM

4

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.

— peufeu
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That makes sense. So my confusion come from Farnell's "laziness"

— M.Ferru

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.

— peufeu

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.

— Johnny

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?

— peufeu

2

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.

— TonyM
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... if the chosen value results in a dissipation within the power ratings possible at that footprint, yes :)

— rackandboneman

@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 :-)

— TonyM

1

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.

— Mulet
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5

I have actually measured a couple of 0402 0 ohm resistor. They were approximately 10 mOhms.

— Mike

1

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:

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

— Crowley
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1

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].

— u54112
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0

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.

— See Jian Shin
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