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Query: design equations for tapped coil

I'm doing an analysis / simulation of a tube Hartley oscillator, shown below. (full circuit from Sept. 92 issue of QST, also in _QRP_Power_) My problem is that I don't know how to treat the coil, mathematically. I asked some of my professors, but they couldn't give a satisfactory answer. I was rereading some of my textbooks and found in the derivation of the usual formula for inductance that we need to assume the wire was uniformly wrapped about the core. Also, that only assumes a single uniform current flow in the core, whereas my circuit has taps that act as current sources... Finally, I am not even sure that the same value of L should be used for all three secions of the coil! Any help would be greatly appreciated.

Regards, Ross Tucker (NS7F) Arizona State Univ, Physics dept.

circuit schematic: (use mono font) +---|(----+ | | IN --)|--+------+--/\\/\\/--+-- -- -- -- | Ct Rg ________ +---|(--+ | | _|__ | || (_ / / / | Lt || (_ Cf | || (_/---|(--+ | || (_ | | || (_/-------)-----------+ || (_ | | _|__ | / / / | _|__ / / /

coil details: toroidal core, unknown type total 19 turns middle (Cf) tap at turns above ground bottom (cathode) tap at 1 turn above ground coils cover ~300deg of circle on core Wound on a

Reply to
rjtucke
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My guess is to treat it as 3 separate coils in series, and with inductive couplings between all of them. The Ls are not constant and definitely not the same in all three sections.

Reply to
linnix

Nope. Your garden-variety ferrite or powdered-iron toroid will give you coupling coefficients of at least 98% unless you saturate it. Refer Cf and the load resistance to the top of the tank by the inverse square of the turns ratio, and you'll be very close. Failing that, measure the coupling coefficient with a grid-dip meter--put a cap between the hot end and the tap, and measure the resonant frequency with the bottom open and then with the bottom shorted to the tap. You should see a pretty impressive difference if the coupling is high.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

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