from the .PDF page-5 and page-12switching frequency at about 165 kHz
massively over-expensive
far too expensive
such totally unsuitable
overly expensive
Obviously
To account for skin depth, we have to take into consideration the change in resistivity with frequency and depth, which means rho becomes a function of those two, so the formula becomes:
R=4*rho(f,r)*L/(D^2*pi)
where we can see that rho is now a function of frequency 'f' and depth along any radii 'r'. The function is a decreasing exponential so we see the change in resistivity decrease as we go deeper (r becomes larger).
The resistivity of copper in a wire does not change with depth.
FB:
Your also off by a factor of 10. 85 kHz is 0.00085 MHz; 0.00085 MHz * 1000 kHz/MHz = 85 kHz
https://chemandy.com/calculators/skin-effect-calculator.htm
thanks, in addition, I don't understand the concept of R(ac)/R(dc) = 1, because for any given wire, R(ac) is always greater than R(dc), surely?
What I'm uncomfortable with is that resistivity is a well defined property of matter: https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivityThere is in fact a change in conduction with depth (and frequency) and that can be viewed as a change in resistivity. If you are uncomfortable with this you can change that formula to contain a function of conductivity instead of resistivity but it will only come up with the same result, assuming the first result was correct to begin with of course.
More traditional methods use a decrease in area as the quantity that changes, which also results in less conduction. The actual copper wire area never changes, yet they use a decrease in area to account for the effect within the formula and might call it the 'effective' area. So if you like we can call the former the 'effective' resistivity.
What I'm uncomfortable with is that resistivity is a well defined property of matter: https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity
and you are misusing it.
Resistivity does not change with depth, and neither does conductivity, assuming conductivity is defined as the reciprocal of resistivity.
I found it more comfortable to deal with the change in conductance or resistance as the depth increases, and many other web sites quote the same.
To paraphrase a few:
"The skin effect causes the effective resistance of the conductor to increase at higher frequencies where the skin depth is smaller, thus reducing the effective cross-section of the conductor."
"Frequency dependence of resistance:
Another complication of AC circuits is that the resistance and conductance can be frequency-dependent. One reason, mentioned above is the skin effect."
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.
Well, the point i was making was, so is cross sectional area, The area is a well defined measurement yet traditionally no one has a problem with redefining the area to account for a change in properties even though it will not be the actual measured area anymore.
We can call it the AC resistivity then, if you feel more comfortable with that. Otherwise i'd have to change the formula to show the apparent (not real) change in area which would make it look like we were squeezing the wire from all radial directions
In any case though, the final result would come out the same.
A page I visited, called it the derating factor which is a typical term. The online calculators happen to use resistivity as an input.
I cant help if it you dont like the term "AC resistance", "AC conductance", "AC resistivity", or "AC conductivity", but that's how i choose to define it. If the numbers are skewed it is certainly not because of the fact that we 'call' the change in total resistance one thing or another.
There's no dispute over whether conductance (of the wire) changes with frequency; it does. But conductivity is a different thing and it doesn't change with depth or frequency.That one quote i posted previously actually states a change in conductance so that author agrees with my nomenclature convention.
The measurement was made with a Hioki IM3570. The instrument can resolve 1 milliohm and three feet is more than enough. The wire was attached to the standard fixture on the front of the instrument. The fixture terminals are close together, so the wire is in a loop with the largest possible diameter to keep the wire away from itself as much as possible.What is the make and model of the analyzer you are using to do the measurements?
I am not sure that 3 feet is enough length to test this with, but i will still look over the results and compare anyway. We probably also have to know what the shape of the wire was at the time the measurement was made, and also how it was attached to the measurement instrument.
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