# Piece of Wire Through a Toroid

• posted

Here's a piece of straight wire through a ferrite toroid.

Toroid ------ | | wire======================== | | ------

I recall seeing an experiment where a wire is poked through a sheet of paper, some current applied and iron fillings revealing a concentric ring pattern. That's in air..

Replacing air with ferrite, the flux density at a radius from the wire is:

B = (uo*ur*I)/2*pi*r

say Idc = 1A say ferrite ur = 4000 B is in tesla r in meters uo= 4*pi*10E-7

Therefore the flux density at 1mm away from the wire is ...Yikes!...

8000 gauss. Well... that'll saturate a typical ferrite material.

At that radius, that part of the core is saturated and acts like air. The magnetic field continues to spread outward.

At r=2mm ..Still kinda saturated..Core material acts like air.

At r=3mm...Not saturated I supposed the field gets more contained here as opposed to fanning out beyond the core.

So..if I haven't botched the math and theory:

Can I conclude I need approximately >3mm toroid radius to get the most small signal inductance when there's a 1Adc bias point?

• posted

Why not be more complicated, and try calculating results when wire is against inner surface of a (say) 10mm toroid compared to same toroid, wire and current when wire is centerd?

• posted

Turns count on a toroid is the number of passes through the center.

If it is through the center, it represents one turn, and will behave as such.

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=3D=3D

No. When you place a ferrite or other flux concentrator in the system, all the in-air calculations and considerations are invalid. All the flux is captured, and distributed evenly throughout the ferrite's area. The flux density is the flux divided by ferrite cross-section area. Use the standard inductor and transformer equations to evaluate issues like core saturation.

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=3D=3D=3D

I think D from BC has a point, at least for the simple case of a straight wire - the equation B =3D (uo*ur*I)/2*pi*r applies. In a similar analysis of a current transformer (Chai, H. 1998. _Electromechanical_Motion_Devices_, p.119), the equivalent expression is integrated from the inner to the outer radii of the toroid (thickness d, inner and outer radii ri and ro) to obtain the total flux p:

p =3D uo*ur*I*d*ln(ro/ri)/2*pi

That ri in the denominator foretells trouble in tiny toroids.

-- Joe

• posted

This law does not apply here, see other postings.

And:

(4*pi*10e-7 * 4000 * 1) / (2 * pi * 1E-3) = 0,8 Gauss ?

P.

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Yes. In cgs units: H = .4(pi)NI/l = .4(pi)N I/ (2(pi)r). For N = 1, I =

1, r = 0.1, H = .4/(2(.1)) = 2 Oe

B = uH, So, B = 4000 X 2 = 8,000 Gauss!

Keep in mind that this flux density is at the radius of 1mm and no where else. As you progress outward through the thickness of the core, the flux density decreases so an accurate answer has to integrate through the total core thickness. Another poster posted an equation that takes this into account.

The relative permeability is a function of the flux density, B as illustrated by the B-H curve. Permeability is the slope of the B-H curve. Long before saturation occurs, the permeability drops off controlling the flux density for a given magnetization, H. This complicates the above analysis because "u" is changing and probably keeps the core from actually saturating

Also, when a magnetic material saturates, its relative permeability drops to unity. This has the effect of magnetically increasing the torroid's inner radius as though it were air. Since the dimensions in these examples are all very small, this phenomena is not particularly an issue in most cases.

• posted

cut....

Well, you're only off by 10,000, what the hell?? It's 0.8 Tesla not 0.8 Gauss. A Tesla is 10^4 Gauss. 0.8 Tesla = 8,000 Gauss

• posted

Like this?

• = wire () = core

Side view:

I ^ * ri ro * / / * / / ( (*) ) ]

• posted

=3D=3D=3D=3D

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The equation gives the total flux (Webers), not the flux density (Tesla or Webers/m^2). You are using SI units so if you want to talk about flux density in gauss instead of Tesla you must multiply by

10^4. To convert Webers to Tesla (averaged over the core), divide by the cross-sectional area.

p =3D uo*ur*I*d*ln(ro/ri)/2*pi =3D 4pi*10^-7*4000*1 amp*.01m*ln(.0013/.0003)/2pi =3D 2*10^-7*4000*1*.01*ln(.0013/.0003) =3D 0.0000117 Wb

divide by cross-section: =3D .01m*.001m =3D 10^-5 m^2

=3D 1.17 T =3D 11,700 gauss

-- Joe

• posted

Yikes.. 11.7kG I suppose all the ferrite material will saturate in the above example. The small signal inductance will be no different than an air core.

ok... How about I try to solve what toroid ID and OD is barely ok.

Say B = 4000 Gauss

0.4T = Webers/10^-5m^2

Webers = 4E-6

4E-6 = uo*ur*I*d*ln(ro/ri)/2*pi

ln(ro/ri) = 4E-6*pi*2/uo*ur*I*d

Say ur = 4000

ln(ro/ri) = 0.5

e^0.5 = 1.65

So if ri = 0.3mm then ro = 0.3mm*1.65=~ 0.5mm

The toroid got thinner??? I suppose the less core volume enveloping the wire.. the less flux density in the core.

Spray on ferrite...

• posted

No, you forgot that you set the core's cross-sectional area at 10^-5 m^2, which constrains ro and ri.

Solving for ri:

A =3D d (ro-ri) =3D d (1.65ri - ri) =3D d*0.65ri ri =3D A/0.65d =3D 10^-5 m^2 / 0.65*10^-2m =3D 1.54mm ro=3D 1.65*1.54mm =3D 2.54 mm

As you suspected, you need a bigger core. The fact that you managed to need one with an O.D. of exactly 0.1" must be some kind of cosmic proof that you're on the right track

-- Joe

• posted

Correction: O.D. of 0.2"

So much for Karma.

-- Joe

• posted

oops.. That's right. No wonder my electronics is so crappy, my math sucks.. Grrr... :(

ok...trying again...Perhaps I'll get your result...

(uo*ur*I*d*ln(ro/ri)*(1/(2*pi) B= _____________________________ Tesla d(ro-ri)

d cancels out. ur=4000, B=0.4T

ro-ri uo*ur*I*(1/(2*pi)) 4E-7*4000*1*0.5

• posted

• posted

Is pi part of the denominator, i.e. should there be parentheses around 2*pi?

• posted

Shouldn't that be:

B = ur*uo*H

H is in terms of Amps per length B is a density

ur is unitless..A factor above free space. (ferrite)

B =urH is missing a dimension.

B = uo*ur*H

amps B= ----- * uo * ur length

uo = free space 4*pi*E-7 amperes/meter^2

amps amps B= ----- * ------ * dimensionless length area

B = amps/length per area or amps/volume

It's B for flux density...I guess that's why it's called density. Like density of matter is weight/volume.

But B is in terms of webers/area=tesla

What I'm working on is: What are the benefits of a straight wire in ferrite. NO turns. (From a power conversion point of view.) Say 80cm of wire..

Note: No turns = the wire is not looped on the ferrite. However the wire does return to the current source and that's a loop so N=1.

• posted

's+law+H&source=web&ots=MG4FYgt_9D&sig=yL8X8-E4xa0zHrSMK3nvLbIZIvw&hl=en&sa=X&oi=book_result&resnum=4&ct=result#PPA119,M1

Handbook of Engineering Electronics By Rajeev Bansal Page 119 is viewable in Googele books.

Flux = (uNIh/2*pi) * ln(b/a)

So yes... parentheses around 2*pi.