In message <00b001c75cc1$233a3810$154d4d0a@MmPc21>, dated Fri, 2 Mar 
2007, Piotr Galka <[email protected]> writes:

>My question is: What the voltage will be when the frequency is such 
>that ferrite is only resistive and 0 inductive ? Is it still trafo or 
>the wires are just two separate resistors ?

It's a very good question, but it's a bit wrongly-stated, I think. There 
is a long technical piece in the Fair-Rite catalogue 13th edition 
(1998), which is very helpful on the subject, but I don't know if it is 
in later editions.

What happens as you increase the frequency is that initially the 
inductance of the bead stays roughly constant, but the loss resistance 
increases roughly with the square of frequency. At higher frequencies, 
the loss resistance continues to rise, while the inductance decreases; 
there is a frequency range where it is inversely proportional to 
frequency and than at higher frequencies still it decreases very 
steeply.

Up to the frequency where the inductance begins to decrease, and a bit 
beyond, there is significant transformer action, but, depending on the 
circuit impedances, it becomes less effective as the frequency 
increases. However, the inductive effect doesn't disappear entirely 
until a much higher frequency.

What this means, as with so many things about EMC, is that what happens 
in your particular case can be calculated with difficulty but measured 
with relative ease.
-- 
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
There are benefits from being irrational - just ask the square root of 2.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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