[ARC5] 10 meter ARC-5s

[AKLDGUY .]

> What I don't understand is that if you are correct, why, then does
reducing
> the number of turns also result in the proper selectivity curve and "Q"?

Are we agreed that the coil can be regarded as containing all of
the resistive
losses and we can ignore those of all other components?

To get to 10m, you reduced the number of turns and also removed plates.

Xl = 2.pi.f.L.

That lowered the reactance of the coil and reduced the resistive losses in
it.
Since we have agreed that the coil is the only contributor to resistive
losses,
it follows that the coil is the only determinant of Q.

Let's assume you halved the number of turns. You immediately gained a
halving of the resistive losses, but the coil's reactance is a more complex
issue
and it will not be exactly halved.

As I understand it, the ARC-5 RF, mixer, and oscillator coils are single
layer,
air-cored coils. The formula for such coils is:

L (uH) = a^2.n^2 / 9a + 10b

where a=radius, n=number of turns, b=length of winding.

Here, a doesn't change but for our example of halving of turns, n and b are
both halved. Accordingly, the top line of that equation is quartered (by the
squaring of n). The "9a" does not change and the halving of "10b" means the
entire bottom line decreases by a factor that for small diameter coils such
as
found in the ARC-5, approaches half, ie. the bottom line will almost halve
for
small diameter coils.

The result is an L (uH) that, for our example of a halving of turns, is a
quartering
of the top line and an almost halving of the bottom line. The L (uH) will be
almost halved by halving the number of turns **for a small diameter coil**.
You can plug in values for n and b and check for yourself, as I did.

So what is the result? You halved the turns and thus the resistive loss in
the
circuit, and almost halved the inductance.

Q = Xl / r
Therefore, the Q rises slightly over what it was with no turns removed. The
one
constraint I place is that it applies only when coil diameter is small (9a
not grossly
greater than 10b). For large diameter coils, the Q will rise much more
sharply for
any reduction in turns.

In the method suggested by me, the coil is not modified, so its Q remains
the
same. I think you have achieved a small rise in Q, albeit with a lot of
trouble in
removing turns and plates.

73 de Neil ZL1ANM