HIGH Q CAPS FOR VLF LOOP ANTENNA?
TRABEM wrote in message
...
I think you are heading down the wrong path with the series circuit
as your
fighting a loosing battle. Assuming a perfect coil and capacitor
you create
an infinite Q circuit. Now you hook it up in your circuit. First
there has
to be enough resistance to develop the voltage , and here is the
rub, as you
increase the resistance to develop a voltage you decrease the Q.
Yuk!
Go with a parallel circuit like the rest of the world uses and you
will be
going in the right direction.
I think I'm starting to get it. Am I cutting off my foot to spite my
face::
Comments made by you and a few others have nudged mein the right
direction.....
The higher I make the series resonant Q, the lower the impedance
goes,
hence it's almost impossible to get a lot of voltage out of it??
Not sure why it matters that much. But, I was under the impression
that a perfectly matched antenna and front end would only decrease
the
Q by a factor of 2.
Follow along with Richard's comments if you like and add your
comments
as I check here often and read everything, sometimes many mant y
times::
Regards,
T
PS:I had begun thinking that the higher imedance presented by a
parallel loop was harder to match with a balun, which is why I
started
thinking of a series loop. I'm gettin there, thansk again.
=======================================
Trabem,
This discussion is getting you nowhere very fast.
So let's summarise.
I don't have your exact dimensions but the following are good enough.
L = 27uH, Reactance = j10 ohms, Conductor loss = 0.05 ohms, ESR =
0.01 ohms, Radiation ohms = 0. Receiver input = 10 ohms, Ground
loss ohms = 0.01
The intrinsic Q of the loop is 10 / 0.05 = 200.
The working Q of the loop, when series connected, is Reactance divided
by the SUM of all resistances including the receiver.
Working Q = Reactance / ( 0.05 + 0.01 + 10 + .01 ) = 10 / 10.07 =
0.993
Take note of the ridiculous low value of working Q due to the loop
being in series with the receiver.
----
Reg.
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