Current Density meaning on magnetic’s design

Hi! My name is Raul, I am currently studyng masters degree in power electronics specialization. Personally I am very interested in learning more about magnetics for my projects related with power convertion.  

I am currently designing some trasnformers for it and personally I have some doubts regarding the real meaning of the current density. 

I have readen a lot of desinging guides were it specifies that the values should not surpass the 5 A/mm^2 but I am not really understanding it.  

I asume that the relation that the current density that is trying to remark is related with the power dissipation of the wire, capability of the wire to manage the winding losses. The current and the cross sectional area are two of the most relevant variable in determining the winding losses right? In one side current, in the other side the resistance.  

Therefore, we could asume that more at current density more losses, but it is not case. Cause the current density is not taking in consideration the real lenght of the wire which will determine the resistance.  

Is for the reason that I am assumming that it has to be related with an estimation for the power dissipation on the wire, but, and here is what I am having big times understanding it, the dissipation of the wire will depend on so many factors, heat transmision to the core, the type of convection, if the magnetic is potted or not, or with varnish, that it makes no sense to me to restrict the designs to such a low value as 5 A/mm^2. 

Please I would be very thankful of someone with more expertize than me to resolve this doubt.  

Edited by raul_stoica on 9 months before

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juangr says:

Hi, Raúl!

Of course, there are many things to take into account regarding copper losses. The limit of 5 A/mm^2 is generally valid for "handmade" designs, which use the methods in designing guides.

However, it's just used for comparison between different windings under the same conditions (same current, core, etc.), it can't be used to determine the final resistance, and many more things need to be considered as you commented. To keep it simple with just the DC resistance: P = R · I^2. But, as R = resistivity · length / area, then: P = resistivity · length / area · I^2 = resistivity · length · I · J, it's not just that, but if length is considered constant, then J gives a simple approach to comparing wires.

But it can be useful to see how well (approximately) the wire will dissipate heat: the larger the cross sectional area, the worse it will dissipate heat generated in the middle, even if its surface is also larger. So, again, it's just used as a rule of thumb. There are designs that will stand higher current densities and others that will necessarily have lower values.

Published: 9 months ago

Hi Juan,

thank you very much for answering me, I think I have understood what you mean and it has been very useful! Where have you learned about this?

Thank you very much!!

Published: 9 months ago

raul_stoica said:

Hi Juan, thank you very much for answering me, I think I have understood what you mean and it has been very useful! Where have you learned about this? Thank you very much!!
juangr says:

You're welcome! This was a simplified view of the problem. It's so much phenomena happening at the same time that a lot of testing and trial and error are needed. This is the way I learned it, but the knowledge can also be fed to algorithms to take everything into account automatically, as we have seen in the latest years in both academy and industry.

If you have any more questions, don't hesitate to ask!

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Published: 8 months ago

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