On Nov 9, 8:36*am, Grumpy The Mule wrote:
Howdy,

I'm guessing that it can be solved like this...
Consider the autotransformer action of the tapped inductor.
Then divide the tap capacitor (C2) value by the square of the
turns ratio (N) before adding it to the primary capacitance
of the parallel tuned circuit (C1.)

F=2Pi*sqrt(L(C1+C2/N^2))

I found a rigorous solution in chapter 8 of Alternating
Current Circuits by K.Y. Tang but it's too messy to type.

73,
Grumpy

exray wrote :

Hi,
Can somebody walk me thru the calculation of an LC circuit where the
capacitor is tapped down on the coil? *I see this often done for
bandspreading purposes.

Tnx,
Bill WX4A



One thing to be a bit careful about is including the coupling between
the coil sections. Since I don't know how those particular coils were
designed, I can't say for sure, but in general the coupling between
pieces of the coil isn't as high as you might think.

You can make a good estimate for typical HF single-layer air-core
solenoid coils just using your favorite coil calculation. For
example, consider a coil that's one inch diameter, two inches long, 20
turns per inch, and tapped at the 30th turn (1.5 inches) up from the
bottom. Then the whole coil is about 16.32uH, the 1.5" part is about
11.54uH, and the top 0.5" is about 2.63uH. If the coupling were
perfect between the sections, I believe the inductance of the whole
would be about 11.54uH + 2.63uH + 2*sqrt(11.54*2.63)uH = 25.19uH. At
16.32uH for the whole coil, the implied coupling coefficient between
those two sections is only about 0.20, and you need to be careful to
not think of the tapped coil as simply a transformer with a 3:1 turns
ratio, with implied close coupling between the sections. (This also
illustrates why you can short out turns of a tank coil without totally
killing the net inductance...)

I trust if I've hosed the calculation too badly, someone will point
out the error of my ways. ;-)

Cheers,
Tom