On 1/28/2014 10:29 PM, Jeff Liebermann wrote:
On Tue, 28 Jan 2014 21:05:07 -0600, amdx wrote:

Nope, almost zero ohms, shorting one end and measuring center to shield
on the other end at 3.58MHz.
To reiterate, measuring this 12ft of RG58/u with 42 half cores of 3B7
(material) (size 3019) slide over the coax, shows an impedance of about
3,350 ohms with an inductive phase angle of 22.5*.

You previously said "I have measured the R at 3.85MHz, It is 3,350
ohms." Now, you change it to impedance. Please make up your mind.

I did say that, I wrongly thought it was 3,350 ohms with 22.58 phase
shift, implying some as yet unknown amount of inductance. If you follow
the thread, I immediately followed my own thread with,
"The Total impedance is 3,350 ohms, this includes R and L."
I apologize for my screwups making it difficult for anyone reading to
follow along.
However, if I new what I was doing, I wouldn't be posting a question.


However, I screwed up in my previous message. I forgot that you're
working with coax cable, where the center wire is shielded from the
effects of the inductors by the outside shield. So, you'll see the
equivalent of what you would get with a single wire going through all
the cores.

Yes, a loss resistance and inductance of one turn.

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?


I can't find a data sheet on the
new

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.

and useless Ferroxcube web pile.

I thought the pile was ok, it just didn't have the 3B7,
must be old and obsolete.


Here's roughly how I would do it
if I had the Al of your cores.

L(mH) = Al * N^2 * n / 10^6

N = number of turns, which in this case is 1.
n = number of cores, which in this case is 42.

Xl = 2 * Pi * f * L
Xl = 6.28 * 3.85*10^6 * L(mH)/10^3


The DC resistance is so small that it can be neglected.


You obtained a 22.5 degree phase angle which might be the capacitance
of the coax cable.

The phase is inductive, E leads I.

I don't really know where it came from. That
angle would normally come from a resistance in the loop, but the coax
is nearly zero ohms. If there were any resistance, the phase angle
would be:
phase-angle = arctan(Xl/R)
If it is the cazapitance:

For RG-59/u 16.2 pf per foot for 12 ft would be 203 pf.
Xc = 1 / (2 * Pi * f * C)
Xc = 1 / (6.28 * 3.85*10^6 * 200*10-12)
Xc = 207 ohms

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.


Are you sure it's 3.85MHz and not the more common 3.58MHz? I smell a
transposition of numbers here.


I picked 3.85MHz to be in the 80 meter ham band.

The whole system is touchy, putting your hand on the coax changes the
current shown on the scope, also reorienting the coax will change the
current.

That's mostly because of the broken cores causing Al to change as they
move. More duct tape.

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


What are your thoughts, Jeff specifically and anyone else.

My thoughts are that I'm going to throw up. However, it's not your
questions or academic exercises. It's the junk food I excavated from
the back of the office fridge. Time to recycle everything.


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

Thanks, Mikek