On 24 Oct 2005 16:57:07 -0700, "K7ITM" wrote:

I'm puzzled. My copy of rjeloop3 suggests the Q will be about 200 at
60kHz with a 9mm wire diameter, and you'll see about 2kohms when it's
resonated. Are you not taking the output across the ends of the loop
(across the capacitor)?


No, you are describing a parallel tuned loop, aren't you Tom,

That is NOT what I'm trying to build.

I am planning a series tuned loop, which is C in series with L and the
output is taken across the unused loop terminal and the unused cap
terminal.

I think it should be called a series tuned loop, shouldn't it?

I know I suggested a whole bunch of times that your 2K loop impedance
sounded like a parallel tuned loop value and you keep insisting that
my series tuned loop will have an impedance of 2K ohms.

You also told me that "Tom has already carried the water describing
what your antenna Z looks like". I suggested that perhaps Tom and you
thought I was referring to a parallel tuned loop and said several
times that it was a series tuned loop.

Then you ranted on and on or maybe I ranted........

Did Tom and you not hear me when I said it was a series tuned loop or
did I not make it plain enough.

Isn't the impedance of a series tuned circuit LOW at resonance??? It
was when I went to school.

If I've err'd, please let me know how.

Thank you.

T




PS:

And, yes......I expect the Q to be cut in half if I attach a receiver
and a loop with identical impedances to each other. I call it loaded Q
and it's a necessary evil if one doesn't want to resort to electronic
(active component) impedance matching.

Put another way, if my receiver had a 50 ohm input impedance and my
loop had a 50 ohm output impedance (with a Q of 100, unloaded), I'd
expect to have a (net) Q of 50 after the receiver was connected to the
antenna.

Reg's software tells me I have a Q of around 221. I assume that's net
Q for the loop itself (unloaded). If my receiver is made to have the
same Q as the loop, then I expect the loaded Q to be around 110 after
they are connected together.

I know you mentioned an active buffer amp to transform impedances. No
doubt this would help to keep the loaded Q up, but I'd like to avoid
any active antenna preamp/rf stage if possible......as previously
discussed.

And with a skin depth of about 0.01" at 60kHz
in copper, certainly 3" diameter soft copper pipe would have the lower
resistance. You might have some trouble finding soft copper pipe,
though. But even hard copper pipe should have a low RF resistance.
"Reference Data for Radio Engineers" (or "Reference Data for Engineers"
in newer incarnations) has lots of good info for figuring out things
like RF resistance of copper wire. I assume your welding cable doesn't
have strands that are insulated from each other like Litz wire.


I thought about litz, and it probably would have been cheaper than the
copper welding cable I bought. But it's fragile in the outdoors and
breaks easy when the wind blows it especially in long spans like I am
going to have. Rather than encase it in some sort of protected sheath,
I decided to use the welding cable.


Consider that Q is energy stored divided by energy dissipated per
radian (1/2pi of a cycle). Then the net Q will be 1/(1/Q(inductor) +
1/Q(capacitor)). So if the cap and inductor have the same Q, the net Q
will be half that. And if you put a resistive load across the
coil+cap, that will dissipate power and lower the Q further.


I think I understand that now and understood it before you explained
it. But, thank you.