I have question about R L Mathematics

I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns.

I want to calculate the impedance of the reactance.

Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz.

My first step was to find the phase angle, 23.5*. Do we agree there?

Thanks, Mikek

  • it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it.
Reply to
amdx
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I had a thought, I measured the R by dividing Voltage by Current. So that means, my current was limited my the L also. The Total impedance is 3,350 ohms, this includes R and L.

Mikek

Reply to
amdx

The real resistance should not change with frequency so just measure it with an ohmmeter.

Total impedance is the square root of the sum of the squares of resistance and reactance.

The phase angle will tell you if the reactance is inductive or capacitive.

--
Jim Pennino
Reply to
jimp

I will assume that Z is 3350 ohms at 3.85MHz.

The impedance of the reactance (alone) IS the reactance (itself).

I will try.

We do (based on your numbers)...

Z = 3350 @ 23.5 degrees.

R = Z * COS(23.5) and X = Z * SIN(23.5)

Therefore, R = 3072 ohms and X = 1336 ohms

As a sanity check, Z = sqrt(R^2 + X^2) = 3350

Good!

HTH,

John S

Reply to
John S

Did you actually measured R (say Re(Z) ) , or |Z|?

Can you provide us some info on your setup?

--
Wim 
PA3DJS 
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Reply to
Wimpie

Measure it again. That's an awfully high resistance for a piece of coax cable of any length. Knowing the type of coax and the length would be handy. Hopefully, you're not measureing the resistance of teh broken pot cores. That won't work.

The inductance of gapped and non-gapped ferrites are quite different. Check to see if the inductance moves when you move the coax. Also, RG-59/u is not the best coax on the planet. Try to find some RG-6/u instead.

You might want to read through these papers on ferrites (especially the first):

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

I measured the voltage, the current and the phase relationship.

I can, it might turn into a word war, but have used it very successfully 100s of times at 600kHz, I'm not sure of the accuracy at 10 MHz. But with time I will refine it as needed.

Here's a diagram of the setup.

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Ignore the green lines and print for now.

Here's the board with scope probes attached.

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(Note the curly Qs holding the probes, very useful for measurements were the leads cause ringing.)

You connect a frequency generator and "Device to be Measured", adjust the frequency, and set the output level. I like to set the voltage at

1vpp, but it doesn't matter. On the scope you will see the voltage and the current as measured across the sense resistor. Say for your current you have 5 units pp on .02v scale, across a 100 ohm sense resistor. 5 x .02 / 100 = 0.001 amps pp. Then 1vpp / 0.001amps pp = 1000 ohms. Your 100 sense resistor is in series, so must be subtracted out, 1000 - 100 = 900 ohms. You can use the scope to see the phase. You can drop the pp, the numbers all come out the same.

Now the fun part.

I original used a device like this when I was measuring the R and C of bonded piezos in water. That brings me to the green lines in the first drawing, where it shows the capacitor, I had a variable inductor. I would adjust the inductor to tune out the capacitance of the piezo, (set the inductor so the scope shows zero phase difference.) I would then use the scope readings to calculate the R of the piezo. Then I would short the piezo connection and use the scope readings to calculate the impedance of the inductor, which is the same as the capacitance of the piezo with reverse sign. Then the higher functioning brains would calculate transformer and inductors for the amplifier. Any questions? Learned this from Henry. Mikek

Reply to
amdx

Thank you John. Mikek

Reply to
amdx

I'm not sure, it might change with frequency, this is a ferrite around a wire, so the ohm meter won't work. It's loss in the ferrites.

Yup, Eli the Ice man. Mikek

Reply to
amdx

It makes me feel good that I could assist. So, I thank you as well.

Cheers, John S

Reply to
John S

It is a lot. But when I put it together 15 years ago, I seem to remember about 4000 ohms. No, I'm measuring shield end to end. Here's a picture, maybe that will give you a different opinion.

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For sure, but I didn't care, I just wanted to see how it would act when I put RG-58u (said 59 before, wrong) thru a bunch of cores.

You will see in the picture I stabilized it and tapped the whole thing. The center hole is not big enough for RG-6/u.

Will do.

When I did this I was thinking about choke baluns hams use on coax driving antennas. The whole think is just a curiosity. Thanks, Mikek

Reply to
amdx

Correct, but if there is any significant loss in the ferrites at the frequency of interest you probably shouldn't be using that ferrite.

Something like an AIM 4170 is very handy for measuring this and a whole bunch of other things though a bit pricy.

--
Jim Pennino
Reply to
jimp

Ok.

Reply to
amdx

The resistance from shield to shield should fairly close to zero. Same with center pin to center pin. If you're not getting near zero, you have an open somewhere in the circuit. My guess(tm) would be the crappy old PL-259 connectors and UG-175 reducers. Check your continuity.

Perhaps you meant "taped" as in wrapped with duct tape?

Peel off the RG-6/u outer jacket and it will probably fit.

--
Jeff Liebermann     jeffl@cruzio.com 
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Reply to
Jeff Liebermann

My Kelvin resistance device says the the shield is 0.047 ohms end to end.

The center conductor is 0.105 ohms end to end.

Looks good here.

I just did.

That is what I meant.

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

Reply to
amdx

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

Reply to
amdx

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?

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     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

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.

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

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     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

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.

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.

Reply to
amdx

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.

Yes, a loss resistance and inductance of one turn.

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?

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

The phase is inductive, E leads I.

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.

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

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

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

Thanks, Mikek

Reply to
amdx

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