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kl7aj
05-04-2007, 05:59 PM
Squint Flackwelder, NA9QXZ, has a problem. #(Besides his name, that is). #

Squint has a pair of really nice dipole traps....at least he THINKS they're nice. #They're hermetically sealed in opaque plastic. #Unfortunately, they're unlabeled. #He can only guess what band they're for.

However, he does have a nice grid dip oscillator. #He tries out several probe coils on his GDO, and finally finds a dip when he brings it near one of the traps....it's 14.2 MHz. #The other trap reads the same. #Cool, he says.#He can build a two band dipole, for #twenty meters and....and...

He realizes that he is still missing some information. #He can assume the trap was built for a 20/40 dipole, but it would be only a guess. #Even if he guessed right, he still doesn't know how long to make the ends on the outside of the traps. #He could take a stab at some "typical" trap dipole dimensions...but with the cost of copper, he really hates to do too much cutting and trying.

To really design this antenna, Squint knows he needs to know the L/C ratio of the trap just as much as he needs to know the resonant frequency!

Squint has a fairly well-stocked #supply of R.F. components, but not much in the way of test equipment.

With no more test gear than he has on hand, how can Squint determine the L/C ratio of the traps? #(No, he can't saw the thing apart!)

kr9d
05-04-2007, 07:26 PM
Quote[/b] (kl7aj @ May 04 2007,10:59)]With no more test gear than he has on hand, how can Squint determine the L/C ratio of the traps? (No, he can't saw the thing apart!)
I keep wanting to add a known capacitor in parallel to the trap, use the GDO to find the new resonant frequency, and relate the ratio of the two frequencies to the two total capacitances.

By my algebra, f2^2 / f1^2 = C1 / C2, where f1 is the frequency without the added cap, f2 is the frequency with the added cap, C1 is the cap in the trap, and C2 is the total capacitance with the added cap. I got here because the L in both cases is the same, and so I equated two conditions of the equation for a parallel LC circuit (f = 1 / (2*pi*sqrt(LC)). I solved for L in terms of both C1 and C2, and set those two equations equal to each other.

But then my algebra wants to cancel out C1 when I replace C2 with (C1 plus the added known capacitance), which doesn't help at all, heh, heh. I probably just got some term reversed and can't see it.

Rick "who frequently got tangled up in algebra even in engineering school" Denney

kl7aj
05-04-2007, 08:04 PM
Quote[/b] (kr9d @ May 04 2007,12:26)]Quote[/b] (kl7aj @ May 04 2007,10:59)]With no more test gear than he has on hand, how can Squint determine the L/C ratio of the traps? #(No, he can't saw the thing apart!)
I keep wanting to add a known capacitor in parallel to the trap, use the GDO to find the new resonant frequency, and relate the ratio of the two frequencies to the two total capacitances.

By my algebra, f2^2 / f1^2 = C1 / C2, where f1 is the frequency without the added cap, f2 is the frequency with the added cap, C1 is the cap in the trap, and C2 is the total capacitance with the added cap. I got here because the L in both cases is the same, and so I equated two conditions of the equation for a parallel LC circuit (f = 1 / (2*pi*sqrt(LC)). I solved for L in terms of both C1 and C2, and set those two equations equal to each other.

But then my algebra wants to cancel out C1 when I replace C2 with (C1 plus the added known capacitance), which doesn't help at all, heh, heh. I probably just got some term reversed and can't see it.

Rick "who frequently got tangled up in algebra even in engineering school" Denney
Actually, you have the right answer and didn't know it.

Without going into the actual numbers, the priciple's pretty easy to follow. If the coil has a very LARGE inductance, a small additional capacitance will have a large effect. If the inductance is small, you need a larger additional capacitor to have the same effect.

Very good.