Richard Clark wrote:
On Sat, 15 Jan 2011 07:50:13 -0600, joe wrote:

joe wrote:

Saving a few cents on Lits wire will cost you almost half of the power
you could deliver to the tank.

This won't give you maximum power transfer.

Correction: You'll lose much more than half the power.

Hi Joe,

Care to go deeper? Would that "more than half the power" be 3.1dB?
6dB? 20dB?

Why "half the power" as a round number instead of a tenth?

73's
Richard Clark, KB7QHC

Why, because I chose not to go into detail.

If you look only at a resistive source impedance, the most power we can
get is

((Vsource/2)**2)/Rsource

Now the example starts with Rsource = 58 ohms and is trying to transform
that to match a 1.5 Meg load. The power available from a source with 1.5
M source impedance is much less than that from a 58 ohm source.

Hint : 58/1.5M

So, doing something to match the tank to the load could be much more
effective. A tapped coil or additional winding has been suggested.