On 1/28/2014 8:34 PM, amdx wrote:
On 1/28/2014 8:11 PM, Jeff Liebermann wrote:
No need for that, This is just a curiosity.

Not sure where we have disagreement, do you disbelieve the 3,350ohms at
3.58MHz?

If I short one end shield to center pin, and measure the other end
would that double impedance?
I'm going to find out.
Mikek

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*.

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.

What are your thoughts, Jeff specifically and anyone else.

Thanks, Mikek

On Tue, 28 Jan 2014 20:34:21 -0600, amdx wrote:

Not sure where we have disagreement, do you disbelieve the 3,350ohms at
3.58MHz?

Yep. I am a heretic. You stated:
"I have measured the R at 3.85MHz, It is 3,350 ohms"
From where to where did you measure 3.35K, presumably with an ordinary
ohms-guesser?

If I short one end shield to center pin, and measure the other end
would that double impedance?
I'm going to find out.

It's like winding two turns on a toroid in opposite directions. The
inductances cancel and you get zero inductance. Replace the coax
cable with a single loop length of wire and see if it makes more
sense. Only turns that go AROUND the toroid in the same direction
provide useful inductances.

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

On Tuesday, January 28, 2014 8:11:33 PM UTC-6, Jeff Liebermann wrote:
The resistance from shield to shield should fairly close to zero.

Jeff, he is measuring the choking impedance of a w2du
choke-balun. It is like running one wire through a
number of ferrite beads/toroids.
--
73, Cecil, w5dxp.com

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.

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.

Got an Al value for a single core? I can't find a data sheet on the
new and useless Ferroxcube web pile. 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. 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.

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

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.

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.

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

On Tue, 28 Jan 2014 20:19:40 -0800 (PST), W5DXP
wrote:

On Tuesday, January 28, 2014 8:11:33 PM UTC-6, Jeff Liebermann wrote:
The resistance from shield to shield should fairly close to zero.

Jeff, he is measuring the choking impedance of a w2du
choke-balun. It is like running one wire through a
number of ferrite beads/toroids.

Thanks. Enlightenment arrived a few milliseconds after I clicked
send. I forgot that it was a coax cable and that the center wire was
shielded from the effects of the inductors by the coax shield. Sorry
for the muddle.


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

On 1/28/2014 9:51 PM, Jeff Liebermann wrote:
On Tue, 28 Jan 2014 20:34:21 -0600, amdx wrote:

Not sure where we have disagreement, do you disbelieve the 3,350ohms at
3.58MHz?

Yep. I am a heretic. You stated:
"I have measured the R at 3.85MHz, It is 3,350 ohms"
From where to where did you measure 3.35K, presumably with an ordinary
ohms-guesser?

The cable is 12 ft long with a PL259 on each end. I plugged each PL259
into an SO239 panel mount connectors. I then drove 2ma of current
through the shield from panel mount connector to panel mount connector
in series with a 100 ohm resistor. I measured the voltage across the
resistor and calculated the current. This was done at 3.58MHz.
The input voltage was divided by the current to get the impedance.

btw, at one point I took a measurement and got around 800 ohms, took my
a second to figure out I had lowered my frequency to 350 kHz.

Reread, (or read) my response (3:17pm) to Wimpie about my setup and
look at the pictures, I think I laid it out pretty clearly. I was
impressed when I first started using the "setup". Simple, obvious idea,
but the strays need to be worked to a minimum. The "setup" is cheap and
simple and I'd like to refine for those that have a scope and frequency
generator but no other way to measure complex impedances.


If I short one end shield to center pin, and measure the other end
would that double impedance?
I'm going to find out.

It's like winding two turns on a toroid in opposite directions. The
inductances cancel and you get zero inductance. Replace the coax
cable with a single loop length of wire and see if it makes more
sense. Only turns that go AROUND the toroid in the same direction
provide useful inductances.

You're correct, I measured very low ohms and no phase difference.

Curious to hear your input on the "setup" and what else I can do to
give you more confidence in my measurement, OR fix my "setup".

I'll measure some known parts tomorrow evening.

Thanks Jeff.

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

On Wednesday, January 29, 2014 7:08:25 AM UTC-6, amdx wrote:
I'm a little surprised, I would have thought the reactance would have
been higher than the resistance.

According to the graphs it should be. That's why I guessed it was #77 material.

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

On 1/29/2014 9:04 AM, Jeff Liebermann wrote:
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.

I have a feeling you didn't look at any of my pretty pictures, I think
at least one layer of fog should have dissipated if you did.
They're really pretty.

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,

I'm doing that now, only with 42 halves. I can try it with one, but I
question with my "setup" will resolve that inductance.

You can see what shorting one end does later.

I already did that, zero inductance or resistance.


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.

Do you need any potcores?

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.

Since you have used "ohms-guesser" I guess I need to ask, What is that?
and does it use ac or dc for the measurement?

The number is (obviously) wrong because nowhere in your
circuit is anything resembling a resistor of that high a value.

I'm sure you understand that ferrite beads on a transistor lead, show
up as a resistive and inductive. Why is this different?
Here's a pdf with a graph page 4 right side showing R, X, and Z.
Hand picked to show what I want it to! Although I should have secretly
altered the frequency range.
http://www.vishay.com/docs/ilb_ilbb_enote.pdf



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.

It's inductive.


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?

LOL, you're right, it was my 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.

Good luck.
I'm in Fl. where people are acting like the end is near.
We are slightly below freezing and businesses are closing, schools out.
The weather men are have fits of frenzy. If I was back in Michigan, we
would be happy it got warmer, because we were getting tired of starting
the car at 7* in 8 inches of snow.
Mikek