How can I calculate Litz wire strands?

I am trying to calculate the Litz wire strands for a LLC ferrite transformer design.

  • IRMS = 2.4 A
  • J = 4 A/mm2

For these values:

  • A = 0.6 mm2
  • D = 0.9 mm
  • operating frequency 50 - 100 kHz

If I use 0.1 mm Litz wire, how many strands do I need to achieve the appropriate current density?


1 Answer 1


Simply, the total cross sectional area required for your wire is the number of strands times the cross sectional area of a single wire.

$$ N_{strands}=\frac{A_{tot}}{A_{wire}} $$

You have selected a strand diameter of 0.1 mm, so

$$ N_{strands}=\frac{0.6}{\pi \frac{0.1^2}{4}}=77 $$

So 80 x 0.1mm Litz wire should be enough.

I use this tool to have an idea about possible diameter of Litz wires, if it'll be a custom one. Using a diameter of 0.12mm (due to enamel coating) for inner circles, I got a total diameter of about 1.25mm for 80 strands.

If this wire will be used for primary winding and if the difference between input and output is high enough (or in other words, if the converter is an off-line low voltage converter e.g. 12V/10A) then, due to safety reasons and insulation requirements, you should use reinforced insulation such as triple insulation (TIW wires). And 0.1 x 80 is not a common TIW Litz wire. You can find 0.1 x 100 regular Litz wires and they have about 1.5mm total diameter.

NOTE: You can use a wire diameter of up to 0.5 mm. If the current density of 400A/cm² is not a strict requirement for you and if you can stretch it to 500A/cm² then you can use a 7x0.3mm TIW wire which is a very common TIW Litz.


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