As the frequency of an A.C. current rises, the apparent resistance of the wire increases. Thus if the wire is used in an inductor as part of a tuned circuit, the losses will increase with frequency, which is another way of saying that the 'Q' reduces. This reduction in 'Q' equates to a reduction in sensitivity and selectivity.
The reason that the apparent resistance of the wire increases is because the electrons tend to only move in the outer. As the frequency rises then the electrons are pushed further out towards the circumferance of the wire i.e. only utilise a fraction of the cross section of the wire.
So, what can we do about it ? Well, the most obvious solution may appear to be to use thicker wire. However this may result in an enormous winding which would in turn create other problems.
alternative is to use multiple strands of a thinner wire but with all the strands
connected in parallel. As the electrons tend to flow around the circumference
of the wire it follows that the resistance of the wire depends upon the circumpherance
of the wire (instead of its cross section area as would be the case at D.C.).
For the 3-strand example illustrated, it can be shown that the a 3 strand "Litz"
wire reduces the high frequency resistance by a factor of 1.6. However at D.C.
currents the resistance is dictated by the cross sectional area of the wire,
so at D.C. the Litz wire will have a higher resistance than the larger single
Although the example above shows three strands making up the Litz wire, having more strands reduces the losses still further. In the case of the Philips Super-Inductance range the wire consisted of an incredible 15 strands.
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