On Wed, 29 Jan 2014 07:08:25 -0600, amdx wrote:

However, if I new what I was doing, I wouldn't be posting a question.

If I knew what you were trying to do, I wouldn't be asking you
questions.

Got an Al value for a single core?

I believe it is 9660 +/- 25%. This is for two core sandwiched. However,
does that have any use for calculations? I'm using the core in a manner
that is not normal. I don't remember how I placed the core halves, face
to face, back to back, mixed?

Argh. It's a pot core that's designed to have all its turns inside
the two halves of the pot core on a bobbin. What you've apparently
done is used it in a non-standard and unpredictable manner by shoving
a wired through the hole normally reserved for either a ferrite
adjustment slug, or a nylon mounting screw. The spec sheet Al value
is worthless.

However, all is not lost. You can take 1/2 of a core, shove a wire
through the hole, and measure the inductance. That's inductance as in
uH not the calculated reactance or guessed impedance. I presume that
there are 42 half-cores so just multiplying the measured value by 42
will give a tolerable approximation of the total inductance. You
could also measure the total inductance of all 42 cores to see if that
really works.

Note that I'm suggesting that you measure the inductance with a single
wire going through the cores, and not with a loop produced by shorting
one end of the coax cable. That takes some of the mystery out of the
measurements. You can see what shorting one end does later.

Ya, I found an exchange where I was asking for
that info back in back in 2007.
I got it privately from a friend with an old catalog.

I have several old Ferroxcube catalogs from the 1970's and 1980's.
You can't have them.

The vector sum of the reactances should give you the impedance.

Now, all you have to do is either supply a measured inductance or find
the Al of your cores.

I thought I had, but apparently it was in my secret code.
John S calculated "Therefore, R = 3072 ohms and X = 1336 ohms
As a sanity check, Z = sqrt(R^2 + X^2) = 3,350"

So if X = 1336 ohms at 3.85 MHz inductance is 55uH.

I'm a little surprised, I would have thought the reactance would have
been higher than the resistance.

The only thing I can be sure of here is that the resistive component
is *NOT* 3000 ohms because it can be directly measured with an
ohms-guesser. The number is (obviously) wrong because nowhere in your
circuit is anything resembling a resistor of that high a value. If
you work backwards and assume a DC resistance of zero, then Z = Xl.
I'm still not sure what's causing the 22 degree phase angle. My best
guess(tm) is that it's the capacitance of the coax cable, but that
should have disappeared when you shorted one end of the coax. Dunno.

Yes and more, just putting your hand by the coax causes the current
to change.

My hand is firmly attached to my arm and is nowhere near your setup.
Perhaps you meant your hand?

I had to stop eating cashews last night, I was moving in that green
feeling direction. Time to get ready for work.

I compounded my culinary error by eating about 1/3 a salted dark
chocolate bar before going to bed. It's now 7AM and I've had about 3
hours of erratic sleep due to the caffeine overdose. I've done this
before and should have known better, but it was sooooooo good. At
least I'm now caught up on paying my bills and reading various
reports. My next challenge will be to see if I can drive to the
office without falling asleep.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558