skin effect

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I think it is useful to understand the cause of the skin effect.

Let me begin with debunking the often quoted, but incorrect, statement that there is no current flowing at the center of the wire carrying AC current. A slightly more correct statement would be that there is no useful current. Because the truth is, that there IS current flowing in the center, it's just going in the wrong direction.

When an AC current is flowing in a wire, it generates a magnetic field which rises and falls. If the wire is the primary winding of a transformer, this rising and falling magnetic field will induce a current in the secondary winding. But, even in the absence of a transformer core, or even a second wire, the magnetic field still rises and falls. And when moving magnetic field cuts through a conductor, a current will be generated, even if it's the same wire that is carrying the original current. These are called eddy currents. And it is the eddy current in a wire that causes the AC resistance to be larger than the DC resistance would be for the same wire.

For a given frequency, a larger diameter wire gives more area for the magnetic field to work in,making stronger eddy currents. And stronger the eddy currents give the wire a higher AC to DC resistance ratio.
 
Hi,

Skin effect occurs for all frequencies except DC. This includes the 60Hz line frequency, where a wire size greater than around 1/4 inch diameter starts to show signs of increased resistance from its DC value.

The basic idea is that the magnetic field pushes the major part of the current flow out near the surface of the wire. This makes the wire look more like a tube than a solid rod. Since a tube the same diameter as a rod has higher resistance, the skin effect causes a higher resistance for AC.

Since the skin depth is inversely related to frequency, the higher the frequency the more shallow the skin depth. This means wire diameters greater than a certain size will be less efficient, so smaller wire diameters can be used with almost the same success as the larger wire. For example, at 100kHz a wire size greater than AWG #26 doesnt reduce the AC resistance enough to make it practical unless the coil is very small.
 
So that is why skin effect takes place.
I didnt know that increasing the wire size worsens the effect.
 
dr pepper said:
So that is why skin effect takes place.
I didnt know that increasing the wire size worsens the effect.
Click to expand...

Hi,

I am not sure what you mean by "worsens the effect". Making the wire size larger by increasing the diameter lowers the AC resistance, but the efficiency goes down, and here efficiency has meaning more like "wire utilization" where we want to get the most from our wire size because we have to pay for the copper and also pay for the window area of a typical magnetic core. So when we talk about skin effect we have to be very explicit about everything we say because the comparison leads to different views. Let me try to elaborate a little.

For say 100kHz if we use wire size AWG 26 most of the wire diameter will be used for conduction. If we increase to AWG 23, we get a little lower AC resistance, but not by as much as we do when we use DC. So the extra copper that goes into making that 23 wire is partly wasted. If we use two strands of AWG 26 we'd be better off, and that would give us the same DC resistance as one strand of AWG 23 yet lower AC resistance than with AWG 23. So it's more about wasting copper than anything else. We can use AWG 10 gauge wire, but then it takes up a lot of room on the bobbin and the decrease in AC resistance wont be as good as if we make up that same wire area with insulated strands of AWG 26.

It's almost like buying a drum to do the rhythm in a musical number. If we hit a drum that is 18 inches in diameter with a certain force and we dont get a loud enough sound, we might try increasing the diameter of the drum, but two drums will probably make a louder sound so we use two instead of one bigger drum. The bigger drum does make a louder sound, but the two drums make an even louder sound yet (for this example).

Also, if the strands are not insulated then they tend to act like one solid wire with a little more surface area, and the wire strands inside the pack may end up not conducting much. Insulation on all wire strands works much better.
 
What we call 'skin effect' is caused by (usually a small fraction of) an electromagnetic wave propagating through a good conductor with exponential decay. If you calculate the speed of the EM wave in a good conductor you will see that it only travels at a snails pace irt its speed in air or vacuum (~300,000,000m/s) so the section of wire with current is only a few wavelengths into the surface from the EM fields that carry the energy surrounding the wire. The ohmic heating seen is a function of the conductor resistance as the energy directed into the conductor instead of outside the conductor (the Poynting vector) is dissipated.

The skin effect/depth is the typical distance a wave penetrates into a conductor.
https://pcwww.liv.ac.uk/~awolski/Teaching/Liverpool/PHYS370/AdvancedElectromagnetism-Part3.pdf
 
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I have been reading about skin effect. It is like having too many cars on the highway during rush hour traffic.

#10 copper wire measures .100" diameter. Distance around the wire is .100 x 3.14 = .314"

1/2" copper tubing measures .625" diameter. Distance around the pipe is .625 x 3.14 = 1.9625"

1.9625/.314=6.25 ratio

Surface area of the copper pipe is 6.25 times larger than the #10 wire.

Electrons travel on the surface so the discharge from a large capacitor bank will travel with less resistance along the 1/2" pipe than the #10 wire. It is like comparing a 1 lane highway to a 6 lane highway with the same amount of traffic.

Wire resistance of #10 copper wire is .998 ohms per 1000 ft. Resistance of .625" copper tubing should be .159 ohms per 1000 ft.
 
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Yes.
I meant its more economical copper wise to have more strands rather than a bigger strand.

I know rf travels in the average coax at 0.66 the speed of light, the reasons for that sound complicated.

My mrs is from Liverpool and knows the college well.
 
gary350 said:
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Wire resistance of #10 copper wire is .998 ohms per 1000 ft. Resistance of .625" copper tubing should be .159 ohms per 1000 ft.
Click to expand...

Resistance of #10 copper wire .998 ohms / 1000 ft. is the DC-resistance.
DC-resistance of .625" tubing depends on the thickness of the tube, for example with a standard type K the resistance is 0.105 ohms /1000 ft.

When the skin depth is much smaller than the diameter,
the resistance R ~ l*r/( pi*D*d ), where l is length, r is resistivity, D diameter, d skin depth
then the relation of resistances in your example are right, the tubing has 1/6.25 resistance of the #10 wire.
 
MrAl said:
Hi,

Since a tube the same diameter as a rod has higher resistance, the skin effect causes a higher resistance for AC.
Click to expand...

Why would a tube, and a rod, made of the same material, have different resistance unless saturated? Consider hollow wire. Why would someone go to that length only to end up with a higher resistance?
 
Hi there,

I think you may have taken that quote out of context. Let me quote the entire paragraph from which that came from here now:


The 'tube' here represents the outer shell of partial conduction that occurs with the skin effect, and that only happens with AC not DC, so we have to remember that we are not talking about DC resistance we are talking about AC resistance here. We are also not creating this 'tube' ourselves (yet) it is being created by the AC current.
For a solid copper rod of reasonable diameter relative to the frequency, if we could measure the current in each 'layer' of the copper rod starting from the surface and going deeper and deeper into the metal toward the center of the rod, we would see a decreasing level of current. If for example, we saw 1 amp flowing in the outer layer near the surface when we got to a depth equal to the skin depth we would see a current of only 1/e amps, which numerically is about equal to 1/2.7 which is close to 1/3 amps. So the current decreased by about 66 percent already, meaning the resistance is 3 times higher in that tubular layer.
In the next layer down, we'd see the current decrease to about 1/7 of it's surface value or about 0.14 amps.
One more layer down, a decrease of about 1/20 of it's surface value, or 0.05 amps.
Another layer down, a decrease of about 1/50 of it's surface value, or 0.02 amps.
One more layer down and we see a decrease of about 1/150 of it's surface value, or 0.0067 amps.

Looking at the above, we see most of the current was flowing in the first layer, then some in the second layer, then much less in the third layer, then less and less in the other layers. So we see the bulk of the total current flowing in the first few layers, where each layer is a distance equal to the skin depth for that frequency.

Now if we used a wire diameter that was only two skin depths in diameter, we'd see most of the copper in that wire being used for conduction, meaning the AC resistance is nearly equal to the DC resistance. But if we use a wire that is 20 skin depths in diameter, then we'll see a lot of the interior layers of the copper not carrying much current so in a way we are wasting that copper, and along with the increased cost comes the increased weight which means the shipping charge will be higher now too, as well as the weight for some applications like space flight.

We can calculate the AC resistance from this information, and come up with an equivalent "tube" or in real life it would be a pipe of copper that would have nearly the same resistance as a rod of copper at a certain frequency. We can make a chart then too. I might do this myself too just to refresh my own understanding a little.
 
Eddy currents do cause the skin effect and the direction of the eddy current is to oppose the current flow in the center of the conductor but I don't think it actually reverses the flow of current in the center, just impedes it.
 
Not in every case, but with certain combinations of frequency and conductor size, the current in the centre can indeed be in the opposite direction.

Edited to add diagram from N.W. McLachlan "Bessel Functions for Engineers" page 156. Note where it indicates the point where current reverses. As you can see, it's possible for it to change direction more than once.
 
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Your diagram demonstrates the decaying waveform traveling into the conductor and the resulting currents.
 

Electromagnetic waves do not penetrate a good conductor. The varying fields impinging on the surface of the conductor do penetrate, but the penetrating fields no longer obey the wave equation; rather, they obey a diffusion equation:

http://ilin.asee.org/Conference2007program/Papers/Conference Papers/Session 1A/Spreen.pdf

This gives an idea why the speed of the penetration is so slow. Diffusion generally tends to be a slow phenomenon. Think about the penetration of a temperature rise from a torch applied to the surface of a copper block. It's diffusion in action.
 
Hi,

If we look at the effect along the radius of a long wire the current density changes as an exponential. Integrating over the radius using this exponential as a weighing function provides us with the following formula:
r=2(g^2*(e^(-1/g)-1)+g)

where
g=1361/(6641*pi*sqrt(f)*R)

for copper wire, and
f is the frequency in Hertz and R is the radius of the conductor in meters,
and r is the ratio of DC resistance to AC resistance, so an r value of 0.5 for example means the AC resistance is 2 times higher than the DC resistance.

This formula is not like the formula where we assume that the penetration depth is only equal to the skin depth, and obtain the AC resistance from that. That formula is limited because the approximation falls short when the radius of the wire gets closer to the skin depth. For example, the approximation would give an estimate of close to 0.5555 for a wire with radius three times the skin depth, while the better approximation from the formula above yields an estimate closer to 0.4555.

The skin depth at 100kHz is 0.206mm, which is about equal to the radius of a #26 gauge wire, and using the older approximation this tells us that the AC resistance is about equal to the DC resistance. This is the largest wire size recommended by Magnetics Inc for use at 100kHz. They obviously used the single skin depth approximation to determine this, which isnt too bad really.

However, using the formula above, the DC to AC resistance ratio for an AWG #26 wire at 100kHz comes out close to 0.74, meaning the AC resistance is about 35 percent greater than the DC resistance. This tells us that #26 isnt that bad, but we can of course do better with a larger wire size.

(Checking out the Hayt solution to the 60 Hz wire size shows about the same results, Rac=1 with the single skin depth solution and Rac=1.36 with the solution above, for same length wires).

But how can we do better if a larger wire size results in more loss?

The key to understanding this is what i have been trying to get at. A larger wire size does not lead to more signal loss, it leads to more COPPER loss. Copper loss can be expressed as the equivalent copper used in conduction as compared to that not being used. Note that this statement is not strictly true because most of the copper is used, but some of the copper is not used as effectively so this view gives rise to the idea that there is an equivalent copper wire that is smaller in diameter that would do the same job as the larger diameter wire if the AC resistance was not higher than the DC resistance.

To understand this better we just have to look at some numbers, like the total AC resistance of a couple different size wires.
For AWG #26 wire we get a factor of 0.74 at 100kHz.
For AWG #22 wire we get a factor of 0.63 at 100kHz.
So initially the 22 wire looks worse than the 26 gauge wire, and it is, for the copper loss. But for the signal loss for a 1000 foot run, the AC resistances are:
#26: 55 ohms
#22: 26 ohms

Note that the 22 wire provides less resistance, and it is approximately half, and note that we went down in wire size by 4 wire sizes not 3.

Of course the diameter is larger for the #22 wire, about 60 percent higher, so it takes up more room on the bobbin. This may or may not be acceptable. For small air core coils, the larger diameter wins because of the higher Q.
 
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Ideally, students should be exposed to this range of development for skin effect – circuits through diffusion to wave propagation – to reinforce the sense of unity and relevance of lumped and distributed analysis.
Click to expand...

I'll just say there are more ways than one to skin a duck. The basis for all circuit theory is Electromagnetic field theory. The fact that students might be confused is a good reason for simplification but the fact that Magnetoquasistatic Approximation simplifies the calculations doesn't mean the underlying full electromagnetic field method is wrong in the general sense. I'm all for making it simple but don't make the mistake of thinking that's all there is to the effect.


http://www-num.math.uni-wuppertal.de/fileadmin/mathe/www-num/theses/ma_schoeps.pdf
 
I didn't refer to the Spreen paper for his discussion of approximations. I referred to it because he correctly points out that the diffusion equation controls the behavior of electromagnetic fields in good conductors like copper.

Immediately before the sentence you quote, we find: "...the diffusion equation, obtained directly from Maxwell’s Equations in MQS form, is necessary to obtain valid solutions for distributed problems."

In the paper you referenced, we find this (referring to the approximations discussed in the thesis): "It is obvious that such solutions are just approximations to the “exact” ones, which could be obtained from solving the full set of Maxwell’s equations, but there are error estimates and revisable conditions to validate the particular simplifications in each case."

Both authors are aware that for an exact solution, one must solve Maxwell's equations, which I assume is what you mean by "the underlying full electromagnetic field method."

What "...solving the full set of Maxwell's equations..." means in the case of skin effect in copper wires is to solve the diffusion equation for the fields inside the copper.

The case of a cylindrical copper wire admits of an exact solution, which can be found in most of the extant electromagnetics texts.

There is no propagation of electromagnetic waves through copper (at frequencies below X-rays); there is only diffusion of the fields.
 
The Electrician said:
There is no propagation of electromagnetic waves through copper (at frequencies below X-rays); there is only diffusion of the fields.
Click to expand...


As usually these arguments are not over facts but semantics and definitions. Electromagnetic wave (a transfer of energy from one state to another with a finite velocity) energy propagation methods includes the diffusion model that assumes infinite propagation speed for simplification. The simplification ( classical diffusion heat equation) is valid for some cases (where the wavelengths are much longer than the individual particle movements in response to the energy) but not for others. (like x-rays)
 
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Hi,

Interesting discussion here. I suppose we could solve that diffusion equation ourselves given the right boundary conditions, but do we really need that, and is that equation complete without some forcing function?
I am not sure if i want to do this myself, but i would like to point out one of the simpler concepts about the skin effect because i feel that i did not properly express the value of the simpler single skin depth view which i'll quickly run through right now. I think this gives a better general view for the practical issues such as AC resistance as compared to DC resistance, which is what we usually need to know in electrical equipment design.

The single skin depth approximation approximates the conduction part of the wire as an outer shell of the wire that resembles a tube. The tube has diameter equal to the wire, and wall thickness equal to the skin depth at the operating frequency. This gives a quick view to how the AC resistance gets higher than the DC resistance because the ratio of the area of the tube to the area of the wire tells us the ratio of how much copper cross section is being used for conduction, as an estimate.
The simple example where the radius of the wire is 3 times that of the skin depth of that frequency means we can only rely on the outer shell to conduct, and that shell is considered to conduct the same for AC is it does for DC (strange, but true). So for a very large scale example for simplicity, lets say the frequency is so low that the skin depth is actually 1 meter. This is never going to happen, but serves as a very simple numerical example which we can scale to any wire size.

First, since the skin depth is 1 meter, then the wire radius must be 3 meters. The total area of the wire cross section is pi*3^2=9*pi. The total area of the shell (1 meter deep from the surface) is pi*(3^2-2^2)=pi*5. The ratio of the shell area to total cross sectional area is then of course pi*5/pi*9 which is again 0.5555 approximately. The inverse of 0.5555 is about 1.8, so the AC resistance is considered to be 1.8 times higher than the DC resistance for this wire at that (low) frequency.

The copper loss is equal to 1 minus the ratio, which is 1-0.5555 which means we waste about 44 percent of the copper because of the skin effect.

This is the "single shell" approximation which gives us a quick idea what to expect. The other more complicated formula uses an infinite number of very very thin shells weighted by an exponential to come up with the approximation. The two solutions merge for conductors with large radius compared to the skin depth, but could be significantly different for small radius relative to the skin depth.

In the wave solutions there are other practical considerations too, such as how close the wire is to other wires, fields, etc.
 
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