On Thu, 27 Oct 2005 22:38:47 -0400, TRABEM wrote:

Hi Owen,

Hi Owen,

I am using large copper cable and relatively low loss capacitors. I
expect the 9 ohms of inductive reactance to cancel out the -9 ohms of
capacitive reactance leaving only the sum of the AC (or RF) resistance
of the copper and the ac resistance of caps to give me my net
impedance. Am I correct up to this point??

I'm trying to take this step by step so I can understand where I've
gone wrong.....clearly I must have made an error somewhere.

I don't know what the actual ac resistance of the caps is, but I do
know the ac resistance of 20 meters of #2/0 copper welding cable is
pretty damn low.

Yes, the loop is series tuned, so the output is taken on the
unconnected capacitor terminal and the unconnected wire end.

Rjeloop3 gives estimates my Q at 221 even though it thinks I'm
building a parallel tuned loop. But, I think Q is Q, and the Q of both
types of loops is the same provided the same materials have been used
in both loops.


Forget Rjeloop3 for the moment and think about what you have.

You focus on how low the resistance of the loop inductance is, and
whether or not the capacitor ESR is significant... neither is when you
jam a 2 ohms receiver in series with it all (ignoring the transmission
line).

You seem to be analysing your series circuit with part of it (the rx)
replaced with a s/c.


I am (for now) not considering the effects of hooking it to a receiver
and/or the transmission line. I confess I have not tried to quantify

Well, what good is it to know what the loop L and C do when not
connected to the receiver?

the actual values of the ac resistance of the copper and have only
rough estimates of what the esr of the caps is. I'd be pretty
surprised if the dc resistance of the cable is much more than .1 ohms
though, so the ac resistance should be a little higher at 60 KHz.

Can you estimate what the unloaded Q of the loop is (in round
numbers), and if so, can you agree that it might be around 221 (as
Reg's software predicts)? Can you estimate what the impedance of the
loop is (in round numbers)? Again, do not factor in the receiver input
impedance as we aren't sure whether I'll keep it as is or match it's
impedance with a preamp and/or toroidal transformer. For the moment,
assume the receiver is mounted at the loop (which is a very real
possibility since it's fairly small).

Read Richard's response, though it is more detailed and no doubt more
accuracy.

I think you will understand the problem when you analyse a three
component series circuit (your topology), the Loop L, the Loop C and
the Rx input z. (You can ignore radiation resistance, loop loss,
capacitor loss, they are all much less than rx input z so the loops
loss is dominated by the rx input z in your circuit.)

The place this will end up is that you will come to realise that
knowing how the L and C of the loop behave unloaded, and dwelling on
that behaviour ignoring the effect of loading is probably why you are
where you are (an assumption I know).

When you have worked that out, you may understand why others are
asking how you are going to couple to the loop. Your proposal to
insert the 2 ohms (or whatever) rx input in series with the loop
components isn't delivering what you wanted, and it won't matter how
thick the loop conductor is, or how low the ESR of the capacitor is.

Owen
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