Source Impedance II

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For coefficient of coupling:

k =3D M/sqrt(L1*L2)

to determine M, use an inductance meter plus measurements for the series-aiding and series-opposing configurations.

series-aiding: Lta =3D L1 + L2 + 2M

series-opposing: Lto =3D L1 + L2 - 2M

therefore, M =3D (Lta - Lto)/4

Why do consider opposing? How to use inductance meter to measure M? I am using meter 875B.

jess

You use your inductance meter to measure the inductance of both coils connected in series, twice, once with the winding hooked up so that that the coils cooperate to create a magnetic flux (Lta), and once with leads to one coil swapped over so that the coils oppose one another (Lto).

As Simon Aysdie points out above, you can then calculate M from M =3D (Lta - Lto)/4

-- Bill Sloman, Nijmegen

1) "moisture sensitivity level" of three? Three _what_?? Furlongs per fortnight? Raccoons per garbage can? How about milligauss per %RH? 2) Treat the pair like a transformer and measure inductance of L1, L2, L1+L2, L1-L2, and since it looks symmetrical short L1 and measure inductance of L2. With that info, you can calculate the coupling coefficients M and K. 3) Then with a little math you can calculate the field inside center of them (series aiding is what i assume you want). A bit more and one can make a plot of field strength off-center. 4) That info can be used as calibration information for that fancy whatza doozie sensor.

Look at:

And: There are scads of other references; "jess" use the baby bird, Goo-Gull.

I used to have the "A" version of that, it was a nice tool, it worked and it was accurate enough for quick checks however, I lend it out to some one and you know the rest of the story.

The unit was replaced with the B version, in comparison, it is a big peace of junk! It is nothing like the original model. I am glade I don't rely on it, but, it works ok for repairing and doing basic stuff.

If you opening the limited manual that comes with it, there are some conversion formulas in there. But I am sure you have already looked there.

If two coils of the same value are tightly bound (Physically) together, phases the same, and connected in parallel. You'll get the same L as one coil when using that meter. Or any L meter for that matter.

That's because they are sharing the same B field and equalizing, to put it simply.

That meter will not tell you the (M12)/(M21)(M) directly, How ever, this all depends on the positioning of the coils from each other and how they merge the B fields. Try pulling them apart using the Radius as a measuring point and watch the L change on the meter.

With no change of L between 1/2 coils together normally depicts a M or 1 how ever, you also have to factor in the case where the 2 coils may not be exact to start with.

Jamie

l

-Gull.

I don't think much of that exposition - or most textbook expositions on transformers. I'd read everything that I could on transformers for years before I came across the transformer equation

V1 =3D L1. dI1/dt + M. dI2/dt

V2 =3D M. dI1/dt + L2. dI2/dt

Where V1 and V2 are the voltages developed across coil 1 and coil 2 by currents I1 and I2 respectively.

M =3D K. Square root of L1.L2 and is always less than 1 (though not much less than one for good transformers).

I found it to be a revelation.

-- Bill Sloman, Nijmegen

That's just the Fourier transform of the usual method.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510
845-480-2058

hobbs at electrooptical dot net
http://electrooptical.net

Whichever way I read that last bit, it always parses wrong. It's the dimensionless coupling constant K that's always less than 1, not the inductance M.

Jeroen Belleman

of

Can you give any advice on the math?

jess

You've been told all this before.

Connect the two coils so that the finish of the first goes to the start of the second. That is series-aiding, the field of one coil reinforces that of the other, which is normal Helmholz connection.

Measure the inductance of the pair between the free ends you have left. That will be L1+L2+2M, where M is the mutual inductance, one coil to the other.

Reverse the connections of one coil. The fields from each coil now oppose, hence M is negative for each coil.

Measure the inductance of the pair again, That will be L1+L2-2M

Subtract the two readings: (L1+L2+2m)-(L1+L2-2M)=4M

Simple (??) algebra.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Considerably less than 1 for coupled tuned circuits, such as IFTs.

K=1/sqrt(Q1.Q2) for critical coupling.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

So you can derive M.

You don't "measure M" via the method I and others explained.

You need to *derive* M via the two actual measurements of inductance, one in the series-aiding configuration, and one in the series-opposing configuration. That is, you obtain M indirectly.

r of

Are you an EE? No E&M course? Freshman Physics? Circular coils are easy, 'cause you can get an algebraic expression for the field along the center line. (Using Ampere=92s law) For a rectangular coil you=92ll have to do the integrals. (With messy limits)

George H.

I am trying to also understand that how can measure the flux inside the enclouser produced by these primary coils. The sensor HMC1043 is confusing. Is there another way?

The secondary coil has 30mm internal diameter. 15uH inductance, 408 turns, core : air. wire guage : 36. I am seeing 50 volts peak to peak across this secondary coil on the scope. But the voltage always changes if I slightly move any of the primary coil. I wish to make this helmholtz coil setup more stable.

I moved the secondary coil everywhere in the enclouser the field does look uniform because I am getting 50 volts peak to peak with the secondary coil. Is it possible to calculate the accurate position of the primary (helmholtz coil) based on the dimensions of the enclouser. Enclouser dimensions are : Length 18inches , width: 10inches and 8.5 inches height.

I am trying to calculate what should be the postion of the primary coils mounted on the enclosure based on the enclosuer 's dimensions and the amount of flux or magnetic field needed to get 50 volts or more peak to peak at the output of the secondary coil.

jess

html

Goo-Gull.

h

Absolutely. Sorry about that. Firefox crashed in mid-post, and I had to recreate what I'd written, and didn't spend enough time re-reading it.

-- Bill Sloman, Nijmegen

h

Coupled tuned circuits aren't good transformers, though they may be good demonstrations of transformer action.

-- Bill Sloman, Nijmegen

What length? That matters, too.

I am seeing 50 volts peak to peak across this

Of course it will.

I wish to make this helmholtz coil setup

Clamp the coils firmly to the enclosure.

What is the enclosure made of? How are the coils mounted?

Trial and error would probably be the easiest way to do it.

Post some accurately dimensioned drawings. DXFs are text, so you can post them here, just as you can post LTspice schematics. I can probably FEMM the setup.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Besides positioning the primary coils correctly. Sometimes it becomes impossible to achieve the resonance back If I untume the circuit for a second. Right now I am using a Function generator but later it will be problem with a microprocessor. I might have to build a feedback system to keep monitoring the primary coils. jess