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# How to calculate ac resistance of the coil?

Apr 30 at 13:24

I have a 324uH coil with 80 turns of thickness(2r1) is 0.44mm. I am running the resonant circuit at 500kHz and 1MHz. I want to calculate the ac resistance(due to skin effect) of the coil. The formula I derived is:

pL/A = p(2πr N)/π(r1^2 - (r1- δ)^2) = r N p/( r1 δ) where r , r1 , p , δ = radius of the coil, cross section radius of the wire, resistivity and skin depth.

Assuumption taken is 2r1 >> δ hence 2r1 - δ = 2r1.

However with this value, the resistance comes out much lesser than 1 ohms at 500k/1000k Hz, though I am expecting much more than that. Whats wrong with the formula? Any other formula online/in books anyone aware of?

I've never paid much attention to these formulae.

Practical experience says that it is difficult to make coils, even for limited frequency ranges, with Q = Xl / R of more than 100 or 200. Hence at 1 MHz your 324 uH coil has Xl of 2036 ohms, so a total AC resistance of 10 to 20 ohms would be expected, for a well designed coil.

In part this can be due to Leakage Inductance (poor magnetic circuit) and the fact that Not All of the flux links all of the turns, and other factors.

0.44mm diameter wire is extremely thick for use at 1 MHz unless carrying a Colossal current. 78 pF does not seem enough, suggesting the inductance value is too large.

Normal practise (small signal) at these Medium wave frequencies might be multiple, fine stranded "Litz wire". In 1 layer your 80 turns are some 35 mm long.

It may well be that although skin effect is Minute in your coil, other physical factors will conspire to dominate the unloaded Q available.

Apr 30 at 9:38

I wouldn't think skin effect would matter even at 1 MHZ.

The XL is ~ 2K ohms.

Apr 30 at 13:24
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