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For example:A piece of steel wire of length of length 10um and of CSA 20E-10.

Can we calculate the resistance of such a tiny piece of wire (resistor) in the usual way?

thanks

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- Thread starter strokebow
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- #1

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For example:A piece of steel wire of length of length 10um and of CSA 20E-10.

Can we calculate the resistance of such a tiny piece of wire (resistor) in the usual way?

thanks

- #2

vk6kro

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However, the formula itself applies for all conductors provided they are uniform.

There will certainly be practical problems with making perfect contact with some very short or very wide conductors.

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There will certainly be practical problems with making perfect contact with some very short or very wide conductors.

Hi, thanks for your reply.

What sort of problems would your envisage for making contact with such small conductors? and how would you attempt to overcome them?

cheers

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I don't remember enough about inductance characteristics to be sure, but since the radius of your wire is greater than its length you might check what that means inductance wise as well.

Anyway, with a wire that short its going to be capacitance and inductance that cause you problems and not resistance.

- #5

vk6kro

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Hi, thanks for your reply.

What sort of problems would your envisage for making contact with such small conductors? and how would you attempt to overcome them?

cheers

It would vary with the situation, but one example would be a conductor 1 meter square and 1 mm thick. Just making contact with the entire surface area equally would be difficult, but just doing it would compress the conductor enough to reduce the thickness of the conductor.

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It would vary with the situation, but one example would be a conductor 1 meter square and 1 mm thick. Just making contact with the entire surface area equally would be difficult, but just doing it would compress the conductor enough to reduce the thickness of the conductor.

I take your point. So, ideally how would we want to take resistance measurements? and in practise, what would realistically be the best way to do this?

I don't remember enough about inductance characteristics to be sure, but since the radius of your wire is greater than its length you might check what that means inductance wise as well.

Anyway, with a wire that short its going to be capacitance and inductance that cause you problems and not resistance.

Are capacitances and inductances relevant? Since the supply will only ever be direct current??

cheers

- #7

sophiecentaur

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If you could, somehow, cool that small wire, then it would 'follow ohm's law' over a much wider range of currents than it would in air.

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vk6kro

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I take your point. So, ideally how would we want to take resistance measurements? and in practise, what would realistically be the best way to do this?

You just have to accept that some measurements will be difficult and live with it.

One approach is to not measure these extreme cases at all. The properties of metals are already well known, so you can extrapolate the resistance of common lengths of wire to oddly shaped ones and simply calculate the resistance.

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Ohm's law is limited in three ways:

1. The frequency of operation must be lower than wavelength of the signal - as you approach this limit so-called "distributed effects" start to dominate and Ohm's breaks down. Ohm's law is nothing but an approximation of Maxwell's equations and at this cusp, part of the wave natural of Maxwell's is taking over and you have to start deal with things as return loss and s-parameters instead of resistance, capacitance and inductance

2. The physical dimensions on the low end are dependent on the uniformity of statistical mechanics approximations. In condensed matter physics you are always using mixed classical-quantum approximations which as the "law of large numbers" breaks down results in quantum effects dominating.

3. As dimensions drop at the same time quantum effects like tunneling start to dominate as well separate from the statistical aspects. As these effects turn on they can appear as Ohm's Law (statistical aggregate) currents though the resistance cease to be linear (constant resistance vs. voltage). Like distributed, the wave nature of quantum mechanics starts to dominate.

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sophiecentaur

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Ohm's law is limited in three ways:

1. The frequency of operation must be lower than wavelength of the signal - as you approach this limit so-called "distributed effects" start to dominate and Ohm's breaks down.

I can't think what you mean here. How can a frequency be compared with a wavelength and what would the 'frequency of operation' be, if it was different from the 'signal'?

I stopped reading your post once I had read this. Can you clear it up?

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AlephZero

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I can't think what you mean here. How can a frequency be compared with a wavelength and what would the 'frequency of operation' be, if it was different from the 'signal'?

I guess he/she is referring to things like the http://en.wikipedia.org/wiki/Skin_effect , but soemthing got "lost in translation".

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sophiecentaur

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That makes sense. I must read the rest now . . . .

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I can't think what you mean here. How can a frequency be compared with a wavelength and what would the 'frequency of operation' be, if it was different from the 'signal'?

I stopped reading your post once I had read this. Can you clear it up?

Sorry. Brain going faster than typing. Causes a written traffic accident. :-)

The operating frequency's equivalent wavelength must be "sufficiently" larger than the physical dimensions. Once the wavelength is shorter than physical dimensions, the lumped model is no longer valid: the concepts of resistance, capacitance and inductance as independent, separable component qualities are no longer valid or useful.

The model that still works involves the concepts of power reflection and transmission of complex variables describing waves. This is what return loss and s-parameters are about. Strictly this "distributed model" is still only an approximation of Maxwells' equations. There are cases when it becomes Epic Fail also and you need to drop back and start from Maxwell's directly.

3 GHz is 1 cm, so any circuit element larger than 1 cm is 100% "distributed" and not "lumped" while at 300 MHz (1/10 3 GHz), 1 cm sized circuit elements are still "lumped" to a good approximation. In the in-between of 300 MHz-3Ghz, things get dicey because some aspects are lumpy enough while other aspects are distributed.

A case in point: all square wave or "pulse-y" waveforms have odd harmonics to make them "square". Basically all digital waveforms. So you need "sufficient" harmonics to get a square-ish edge on any digital waveform. The rule of thumb from this is digital waveform edges (rising or falling edge time) require bandwidths that 10x the clock rate.

So if your clock rate is 300 MHz, then you need 3 GHz for the edges. If you are at 3 GHz, you need 30 GHz for the edges. This is actually central to why microprocessor clocks hit a brick wall around 2000: how big is that average microprocessor die, diagonal corner to corner? ~1 cm. So you have edge issues already that are distributed rather than lumped.

And that's a problem: the

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sophiecentaur

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That is surely a transmission line issue, isn't it?

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That is surely a transmission line issue, isn't it?

Yes. When you have to worry about transmission lines, Ohm's law has broken down.

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sophiecentaur

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Ohm's Law only says what it says. It applies where it applies. Nothing more.

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vk6kro

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A case in point: all square wave or "pulse-y" waveforms have odd harmonics to make them "square". Basically all digital waveforms. So you need "sufficient" harmonics to get a square-ish edge on any digital waveform. The rule of thumb from this is digital waveform edges (rising or falling edge time) require bandwidths that 10x the clock rate.

So if your clock rate is 300 MHz, then you need 3 GHz for the edges. If you are at 3 GHz, you need 30 GHz for the edges. This is actually central to why microprocessor clocks hit a brick wall around 2000: how big is that average microprocessor die, diagonal corner to corner? ~1 cm. So you have edge issues already that are distributed rather than lumped.

And that's a problem: the entire concept of digital logic 0 or 1 is itself a lumped model on top of the analog lumped model that approximates Maxwell's equations. If the foundation turns to quick-sand, then the building itself will start having problems.

The wavelength of a 3 GHz radio wave in air is close to 10 cm or about 4 inches. (

Transmission line effects cause apparent Ohm's Law violations all the time.

If you connect a quarter wavelength of 50 ohm transmission line to a 100 ohm resistor, at the other end of the line it will appear to be 25 ohms.

Put 100 volts across it (at the right frequency for that wavelength) and it will draw 4 amps. That is 400 watts and the resistor will get 400 watts too, (ignoring losses), but at 200 volts and 2 amps.

Ohms Law still applies locally, but transmission line effects make it look as if it doesn't apply.

After all, you put 100 volts in and the resistor drew 2 amps, that is only 200 watts, so where did the other 200 watts go?

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sophiecentaur

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vk6kro

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That was the point I was making. You can get apparent anomalies, but Ohm's Law continues to apply.

If fact, I cringe when I hear talk of "non-ohmic" resistors. Just because the resistance of a resistor changes (due to temperature, for example) does not mean Ohm's Law no longer applies.

A small change in voltage still produces a small change in current and the ratio of these two still gives the resistance of the resistor.

- #21

sophiecentaur

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It would have been better if they had called it 'Ohm's Behaviour', perhaps.

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ohms law fails when the resistance of the conductor increases as it gets hot by electron friction

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ohms law fails when the resistance of the conductor increases as it gets hot by electron friction

That doesn't seem to make sense. All you are saying is that the resistance varies with temperature. Whatever the resistance is, Ohms Law should still hold.

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sophiecentaur

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ohms law fails when the resistance of the conductor increases as it gets hot by electron friction

You obviously haven't read what Ohm's Law says.

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