How to calculate rac for inductors when we know Rdc with

[f_t]

Friends,

Do you have any idea how to calculate rac of an inductor when you have measured RDC of it with considering skin effect? (The circuit is high current)

Thanks in advance

 

[schmitt trigger]

Rac is dependent on frequency, wire gage and type, winding methods, etc.

Rdc and Rac are unrelated.

 

[f_t]

Rac is dependent on frequency, wire gage and type, winding methods, etc.

Rdc and Rac are unrelated.

Hi
If you consider: R=pL/A which is valid for both ac and DC.
I think the only difference between rac and RDC is skin effect. In higher frequencies, area where current can pass through is decreased (because of the skin effect), so rac is increased. Now, I'm looking for the relation between them.

 

[Easy peasy]

skin effect & proximity effect both cause the current to flow in the outer surfaces of the wire at higher frequencies, skin depth (at which the current density, J, falls to 37%) is approx 0.066/sqrt(MHz) so at 60Hz it is about 8.5mm, 100kHz = 0.2mm, so round wires thicker than about 0.1mm at 100kHz (and not in a proper woven litz structure), will get hotter than their RDC suggests...

 

[albbg]

As you said Rdc=pL/A, then if the radius of the wire is "r":

Rdc=p*L/(pi*r^2)

while

Rac=2*p*L/(pi*r*delta)

from the first r*Rdc = p*L/(pi*r) substituting in the second:

Rac=2*r*Rdc/delta

where delta is the skin depth:

delta=sqrt[2*p/(2*pi*f*uo)]

uo=4*pi*1e-7

it should be correct but, please, check.

 

[f_t]

skin effect & proximity effect both cause the current to flow in the outer surfaces of the wire at higher frequencies, skin depth (at which the current density, J, falls to 37%) is approx 0.066/sqrt(MHz) so at 60Hz it is about 8.5mm, 100kHz = 0.2mm, so round wires thicker than about 0.1mm at 100kHz (and not in a proper woven litz structure), will get hotter than their RDC suggests...

I'm working at 6.78 MHz, thank you for the information though.

- - - Updated - - -

As you said Rdc=pL/A, then if the radius of the wire is "r":

Rdc=p*L/(pi*r^2)

while

Rac=2*p*L/(pi*r*delta)

from the first r*Rdc = p*L/(pi*r) substituting in the second:

Rac=2*r*Rdc/delta

where delta is the skin depth:

delta=sqrt[2*p/(2*pi*f*uo)]

uo=4*pi*1e-7

it should be correct but, please, check.

Thank you!

I found that equation in a book too.
Rac=Rdc*h/delta
Where h=diameter of the wire