Electronic – How to determine the current rating of Litz wire winding at high frequency

high frequencyinductormaximum-ratings

How can we determine the current rating of a litz wire winding at high frequency?

This is my try:
For example, if we use 660/38 litz at 85kHz, we can see the equivalent AWG is 10 (reference). Then, according to this table, the amperage for the equivalent AWG of 10 is 15A.

But is it same for high-frequency current?

Best Answer

This is not an easy question to answer, as there are many things going on in wire at high frequencies.

While skin effect causes the current to travel nearer to the surface of the strands, the size of the stranding will have been chosen to make that effect tolerable. There is also the proximity effect - the field from a conductor affects nearby conductors and increases their impedance. On a regular spiral lay stranded wire with insulated strands, this would have the effect of greatly reducing the current in the inner strands of the bundle, hence the need for Litz wire, which has the stranding arranged such that each strand spends a more or less equal proportion of its length in the interior and exterior of the bundle, thus evening out the currents carried by each strand. The specific lay pattern of the wire therefore has an effect on the current distribution. The thermal dissipation of the wire is also not easy to determine, since the strands are insulated, the dissipation from the inside of the bundle by radial conduction from contact with other strands is limited, and the heat has to mostly follow the length of the strand to its next turn at the outside layer where it can be cooled by convection/conduction.

There's good write-up here