Parallel LC Circuit

If this is for me I sent several days ago. In spam folder?

i have downloaded the FEMM software on my machine. My helmholtz coils has s= even turns. I gave their dimensions in my first original post in a drawing.= The pair is separated by 8.5 inches. The coil is generating a magnetic fie= ld in the area of 20 inches in length, 10 inches wide and 12 inches deep b= ox. I want to simulate the field. How can I do that in FEMM.

Thanks

jess

The Q of the coil is equal to (wL) / r

Where w = 2 * pi * F L = inductance of the coil

r = AC resistance

How can I find the AC resistance of my coil ?

jess

I thought a Helmholtz coil had the dimensions of separation =3D diameter/2

diameter of 18 inches means the two coils should be separated 9 inches, not 9.5 ?

I usually measure the ac R by resonating at the frequency of interest and find the Resr.

I calculate the expected Resr by using femm and the answer is part of the the table shown when you click on the little 'inductor' symbol while in answer

Hi,

Not in this case. The Helmholtz coils are rectangular. Plus I experiment with the placement and found the magnetic field strong at that distance.

Can I measure the AC resistance using resistance meter like measuring a resistance of a resistor? I guess, that I am little confused about measuring the AC resistance without using the software.

Thanks

jess

AKA "Yecch!"

By resonating it, and measuring its Q.

Unless you have an expensive HF LCR meter, like HP used to make for around $30k.

No.

What software?

FEMM won't *measure* it.

It might *estimate* it, if you get things dimensionally accurate.

AIUI, what was being discussed is the AC resistance of your secondary coil. Since nobody seems to know the characteristics of the ferrite, where it is all won and lost,(including the people who sell them), FEMM is not going to give meaningful results.

Resonate it, and measure its Q. Derive the AC resistance from that.

Hi,

One more thing that I do not understand. The Helmholtz coil is generating a uniform magnetic field at the frequency of 100KHz.

Why do we need the secondary coil to resonate at the frequency of 100KHz to receive that magnetic field? Because the magnetic field is everywhere and it will pass through the secondary coil and the voltage will be generated.

jess

You designed it. You should know ;-)

Bone up on coupled tuned circuits. There's lots of literature (and I don't mean on the Web).

Hi,

Yes, I am designing it by i really do not why the secondary has to be tuned in a hint would help?

I found a following ferrite core material

The Data sheet of the core 3C94 is as follows

I also found the following formula

Bop = ( E X 10^8 ) / ( (4.44 )x F x N x Ae )

Where

Bop = magnetic flux density in guass E = maximum rms voltage across inductor in volts f = frequency in hertz N = number of turns Ae = effective cross sectional area of the core in cm square

So, with 40 volts peak to peak at the across the inductor will give 28.2 rms voltage

F = 100KHz

N = 10

Ae = 0.6 cm

Bop = 1058.5 Guass which is equal to 105mT.

Now , data sheet is giving the value B = 470mT.

Am I right to conclude that this material will not heat up because 105mT is less than 470mT?

Thanks

jess

Datasheet is also giving you power density data at various B, F points. I think you'll find something like 100mT at 100kHz, or somewhere around that.

"Will not" is untrue because it has nonzero losses at any field or frequency.

In practical terms, you'll be lucky to even feel it warm up.

Tim

I'm sure you haven't measured the DC resistance of your inductor correctly.

Using the same ohmmeter, short the leads together while in ohms mode and see if you don't get about .2 ohms. That is the resistance of the ohmmeter leads. The actual resistance of your inductor should be much less, perhaps around .02 ohms.

I have an inductor with a similar core and similar dimensions. I don't know for sure that the ferrite material is the same, but it's probably similar. Looking at the photo of your core, I estimate it's wound with about 18 gauge wire, about 12 turns per layer.

I rewound my core with 18 gauge wire and about 2 1/2 layers to get about 20 uH inductance.

I then used an impedance analyzer to measure and plot the AC resistance of the inductor over the frequency range 100 Hz to 1 MHz.

This resistance is the combination of the increase in wire resistance due to skin and proximity effect and the ferrite losses. It would be nice to measure the wire loss alone to see how much of the resistance is due to wire loss and how much due to the ferrite. One way to do this if several inductors are available is to use a hammer to break up and remove the ferrite; then the resistance due to wire alone can be measured. Another way to do this with small inductors like this is to use a neodymium iron magnet to fully saturate the ferrite; then the effect is to have an "air core" inductor.

Here is an image of the analyzer scan of the AC resistance of the inductor.

The top curve is the AC resistance with the ferrite core in place. The bottom curve is the AC resistance with the ferrite "removed" by means of a magnet saturating the ferrite.

The resistance at low frequency is about .0137 ohms, and this is also the DC resistance.

At 100 kHz, the AC resistance with the ferrite in place is about .3 ohms. With the ferrite "removed", the AC resistance at 100 kHz is about .113 ohms; this is the wire resistance alone.

Here's another image showing the AC resistance versus frequency of a length of 18 gauge wire in 3 different configurations.

The lowest curve is the AC resistance of a straight piece of 18 gauge wire; its increasing AC resistance with frequency is due to skin effect only. The next curve up is the AC resistance with the wire wound into a single layer solenoid with a 1/4 inch ID; now we get some proximity effect. The top curve is the same length of wire wound into a 3 layer coil. Notice how the top curve exhibits a change of slope at about 50 kHz; this is due to the so-called layer effect, or "eddy current screening", as Snelling calls it.

Hi,

So, the following calculations that I did was right? I am heading in a right direction? The inductor will not heat up. I have no problem if it warms up a little and stay there. Please confirm !

jess

I also found the following formula

Bop = ( E X 10^8 ) / ( (4.44 )x F x N x Ae )

Where

Bop = magnetic flux density in guass E = maximum rms voltage across inductor in volts f = frequency in hertz N = number of turns Ae = effective cross sectional area of the core in cm square

So, with 40 volts peak to peak at the across the inductor will give 28.2 rms voltage

F = 100KHz

N = 10

Ae = 0.6 cm

Bop = 1058.5 Guass which is equal to 105mT.

Now , data sheet is giving the value B = 470mT.

ght direction? The inductor will not heat up. I have no problem if it warms= up a little and stay there. Please confirm !

rms voltage

You are confirming exactly why I never trust 'bottled' formulas.

Did you try modeling your structure using femm 4.2?

direction? The inductor will not heat up. I have no problem if it warms up a little and stay there. Please confirm !

voltage

Formulas are like Models/theories of how different aspects of science work that interact down to the atom level. It seems that for every field out there where atoms get into the mix, a formula has been created to work in that field and the explanation of how the ATOM interacts gives a totally different meaning of how the ATOM in general, works.

Since I have not seen a ATOM model, yet! that will fit every ones idea of how and why the atom interacts the way it does with the current subject at hand, One model that makes since for all aspects of science that is. I am convinced that we really don't know and it's been one big guessing game. As long as you can make it sound good and show some tricky effects that is hard to explain with mechanical theories, you have all kinds of funding coming your way.. Because some one always wants to be the first to say, "We got it and it's ours"

Of course we all know how first time discoveries has been handled over the years by big business, one theft after another.

Jamie

I thought I would reply to this old post because the thread contains = much=20 background information that helps explain some of the mystery of = Jessica's=20 circuit. But also, your information on skin effect and the AC resistance = of=20 wires at high frequency is something that I was not really very familiar =

with, and your explanation and images were very helpful. I usually deal = with=20 high currents (10s of thousands of amps) and line frequency, but even = there=20 skin effect has some influence, which is one reason for laminated bus = bars.

But I am starting to work on some projects that use higher frequency, = and I=20 was surprised to see how much effect there was even at audio frequencies = of=20 around 10 kHz, and of course if that's a square wave then harmonics will = be=20 affected even more. And at 50-100kHz or more, as I am considering, I = realize=20 that I will need to seriously take this into account.

Thanks!

Paul

much=20

Jessica's=20

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=20

with=20

there=20

bars.

and I=20

of=20

be=20

realize=20

Thank you Paul. I knew that i wasn't the only one who has had an interesting wake up call on this issue.

?-)