On Mon, 15 Sep 2008 14:14:50 -0700 (PDT), w9cf wrote: >If the two are the same size, the inductance will be approximately the sum of >the inductances of the separate toroids wound with the same number of turns. >So for two identical cores the inductance will be about double for the same >number of turns. You can see this by calculating the H field from Ampere's >law making the usual assumption that the windings are equivalent to a >uniform current. The magnetic field is the same as in the separate toroids. >The magnetic energies then just add so the inductances add.
So does the loss component (resistance) inductively coupled from the core. There is also capacitance between turns, and from one end of the inductor to the other through the ferrite core (the ferrite core is a dielectric). It is a HUGE mistake to think of an inductor wound on a ferrite core as ONLY an inductor. Most cores contribute significant resistance and capacitance at HF. The approximate equivalent circuit of a ferrite inductor at HF and VHF is a parallel resonant circuit of fairly low Q. For #43 and #31 materials, a Q around 0.5 is common. For #61 material, a Q on the order of 10 is common below 10 MHz, but that Q drops with increasing frequency. For some ferrite mixes, the equivalent circuit is TWO parallel resonant circuits in series. See http://audiosystemsgroup.com/RFI-Ham.pdf Well below resonance, the equivalent circuit is a series R and L, and C drops out (because Xc is much much greater than XL). 73, Jim Brown K9YC _______________________________________________ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
