Piece of Wire Through a Toroid
D from BC
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?
D from BC
myrealaddress(at)comic(dot)com
British Columbia
Canada
Robert Baer
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?
Winfield Hill
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.
J.A. Legris
I think D from BC has a point, at least for the simple case of a
straight wire - the equation B = (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 = uo*ur*I*d*ln(ro/ri)/2*pi
That ri in the denominator foretells trouble in tiny toroids.
--
Joe
P.
<myreal...@comic.com> wrote:
>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.
This law does not apply here, see other postings.
And:
(4*pi*10e-7 * 4000 * 1) / (2 * pi * 1E-3) = 0,8 Gauss ?
P.
Bob Eld
"D from BC" <myreal...@comic.com> wrote in message
news:5ndam41rgdkqoogrm...@4ax.com...
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.
Bob Eld
"P." <dit3_werk...@hotmail.com> wrote in message
news:57acm45og4ibahtjm...@4ax.com...
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
D from BC
Like this?
* = wire
() = core
Side view:
I
^
* ri ro
* / /
* / /
( (*) ) ]<d
( (*) ) ]
*
*
*
(I think I have my text set right for ascii art.)
p = uo*ur*I*d*ln(ro/ri)/2*pi
Let's say ri = radius of #22AWG bare wire ~= 300 microns
Let's say ro = ri + 1mm. (Will this radius of ferrite saturate???)
say d = 1cm (use meters in equation)
say ur = 4000
I = 1Adc
total flux contained in core = 0.02 gauss
Huh...??
Is that equation complete???
Is the denominator supposed to be 2*pi*___ ??
J.A. Legris
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.
For your example:
p = uo*ur*I*d*ln(ro/ri)/2*pi
= 4pi*10^-7*4000*1 amp*.01m*ln(.0013/.0003)/2pi
= 2*10^-7*4000*1*.01*ln(.0013/.0003)
= 0.0000117 Wb
divide by cross-section: = .01m*.001m = 10^-5 m^2
= 1.17 T = 11,700 gauss
--
Joe
D from BC
<jale...@sympatico.ca> wrote:
<snip>
>
>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.
>
>For your example:
>
>p = uo*ur*I*d*ln(ro/ri)/2*pi
> = 4pi*10^-7*4000*1 amp*.01m*ln(.0013/.0003)/2pi
> = 4pi*10^-7*4000*1 amp*.01m*ln(.0013/.0003)/2pi
> = 0.0000117 Wb
>
>divide by cross-section: = .01m*.001m = 10^-5 m^2
>
>= 1.17 T = 11,700 gauss
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...
J.A. Legris
> On Thu, 8 Jan 2009 19:36:18 -0800 (PST), "J.A. Legris"
>
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 = d (ro-ri)
= d (1.65ri - ri)
= d*0.65ri
ri = A/0.65d
= 10^-5 m^2 / 0.65*10^-2m
= 1.54mm
ro= 1.65*1.54mm = 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
J.A. Legris
Correction: O.D. of 0.2"
So much for Karma.
--
Joe
D from BC
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 <<cancelled pi's
--------- = ------------------- = ---------------
ln(ro/ri) B 0.4
ro-ri
-------- = 0.002
ln(ro/ri)
ro/ri = e^(500*(ro-ri))
= (e^500)^(ro-ri)
e^500 ??? Stuck again :(
Anyways....
Here's what I'm attempting to do with all this theory.
Can a straight wire power inductor with stacked ferrite toroids have
less core loss than other inductor constructions.
Sideview:
============== <stacked toroids
--------------------- < wire
==============
f= 800khz
Ibias = 1A
Iac = 100mA
L ~390uH
Low core loss
According to online straight wire inductance calculator on:
http://www.consultrsr.com/resources/eis/induct5.htm
I'll need 80cm of 1mm wire in a material with u= 10000.
Thoughts....
mako...@yahoo.com
this may help you
http://www.coilws.com/index.php?main_page=page&id=104
and B = Ur*H
watch the units
note you need "effective path length"
Mark
Nobody
> p = uo*ur*I*d*ln(ro/ri)/2*pi
Is pi part of the denominator, i.e. should there be parentheses around 2*pi?
D from BC
>On Jan 9, 2:51 pm, D from BC <myrealaddr...@comic.com> wrote:
<snip>
>>
>> e^500 ??? Stuck again :(
>
>
>
>this may help you
>http://www.coilws.com/index.php?main_page=page&id=104
>
>and B = Ur*H
>
>watch the units
>note you need "effective path length"
>
>Mark
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.
Instead:
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.
D from BC
Here's a long link..
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.
Robert Baer
> 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.
Pardon the pun, but donut answer the question.
Robert Baer
Shhh...do not give away the secret!
Robert Baer
As Einstein said, it is all relative.
Increase I appropiately with larger Ro; should be able to keep P
constant..
D from BC
Correction..
(Huhhh.. )
H includes uo
when
0.4*pi*N*I
H= ----------
le
le in cm <<mean magnetic path length
So B = urH
B will be in gauss not tesla
I'm going to hammer some toroids now.. :P
Robert Baer
> On Fri, 09 Jan 2009 22:28:03 -0800, Robert Baer <rober...@localnet.com>
> When you went to the Christmas party and entered saying "I come baering
> gifts..." they knew it was you... The stripper almost said the same
> thing... "I come baring gifts..." The knew it was her too... and her
> little jugs too...
>
> You want turns count on a toroid, count the wires INSIDE the center.
> THAT is the right turns count.
YUP!
P.
Tim Williams
> 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.
Is that exactly true? I've actually cracked black (high mu) ferrite
toroids by saturating them. My crazy guess is, field is slightly
higher around the inner annulus, saturating it first and causing
magnetostriction that expands it, uniformly cracking it into about
fifteen pieces. If it saturated evenly, it would expand evenly.
Magnetic path length is always a factor in calculations, so I'm
guessing the ratio of B across the toroid goes something like ln(R/
r)? Logs always seem to show up in concentric things...
Tim
J.A. Legris
A straight-wire power inductor wastes space and core material compared
to a wound toroid. As you have seen, the field within a toroid on a
straight wire is not uniform so you'll have to choose your operating
point to avoid core saturation near the wire while wasting capacity
near the outside. This can be mitigated by using a bigger toroid, but
then you need even more space. A wound toroid has a nearly uniform
field inside, so its size can be optimized. Furthermore, inductance in
a wound coil is proportional to the SQUARE of the number of turns, not
just the length of the core. Core loss increases with maximum flux
density, so again, the wound toroid would be the way to go.
80 cm??? I rest my case.
--
Joe
D from BC
<jale...@sympatico.ca> wrote:
>On Jan 9, 2:51 pm, D from BC <myrealaddr...@comic.com> wrote:
>> On Fri, 9 Jan 2009 07:30:11 -0800 (PST), "J.A. Legris"
>>
>>
<snip>
Yup.. It's a SuperBead.. :)
A crazy 80cm long power inductor if the ur stays constant.
It's a chunky inductor but can still fit in a very long box.. :)
And still shorter than the power cord :)
I'll guess at some cored straight wire advantages:
-----------------------------
*) Lowest interwinding capacitance for high frequency.
*) 100% shielded for low EMI (No stray field)
*) High surface area for cooling.
*) Very easy construction.
*) Low profile
*) I can order factory direct cause so many cores are needed..
(Ok just joking with that last one.)
Consider:
B = (ur*uo*I)/(2*pi*r)
Say Bsat. ~= 0.4Telsa
Say I = 1A
Say wire diameter = 1mm
Since ferrite ur drops on creep to saturation.
Solving for ur at surface of wire.
ur = 0.4T*2*pi*0.5E-3/(uo*I )
ur = 1000
So..If I haven't messed up the physics and math, this means:
If the core ur drops to 1000, the surface of the core can handle the
4000 gauss flux density at the surface of the wire.
The core surface contributes to inductance and is not totally
saturated.
More core radius yeilds more ur yeilds more L.
That gives me a hint about deltaB and core loss.
So you say a straight wire in a core would have more core loss than a
coiled design..??
Is it worth trying an experiment??
J.A. Legris
It's definitely worth experimenting when someone tells you it cannot
work :-).
--
Joe
D from BC
<jale...@sympatico.ca> wrote:
>On Jan 11, 4:02 pm, D from BC <myrealaddr...@comic.com> wrote:
>> On Sun, 11 Jan 2009 08:24:13 -0800 (PST), "J.A. Legris"
,snip>
>>
>> Is it worth trying an experiment??
>>
>
>It's definitely worth experimenting when someone tells you it cannot
>work :-).
How about this test setup?
Objective: Just how lossy is a straight wire ferrite inductor using
the highest permeability material at 800Khz.
Say 50 toroids
=================
-------------------------- <1mm diameter wire
=================
ui = 10000 for Magnetic Inc material W
f = 800khz
Best predicted L ~63uH <<Will be less
Length of cores: 12.5 cm
Gen ------------+----->Scope x10 Ch1
|
|
L
|
Gnd ----Rsense- +----> Scope x1 Ch2
Display (Ch1 - Ch2) on scope is V
Display Ch1 on scope is I
V/I = Z /_ theta = (Rac + Rdc) + Xl
So if 90 degrees...no loss.
I^2Rac = core power loss
Rac copper losses will be lumped in and accidentally called core loss.
Then I have to account for scope loading...
See any bugs?
D from BC
Ooops..
I just noticed on a data sheet that ui of materials like W (ui =10000)
drop to ~2000 @800khz.
AFAIK there's no material with ui = 10000 at 800khz.
Looks like a straight wire inductor at 400uH will be much longer than
80cm.
More like 270cm with u=3000
Using calculator on:
http://www.consultrsr.com/resources/eis/induct5.htm
So no experiment.. Oh well.. Coils are good. :)
J.A. Legris
> On Sun, 11 Jan 2009 18:01:50 -0800, D from BC
>
>
>
Not so fast, D.
The inductance calculator you are using above is for a straight wire
made of a conductor with a particular permeability. That's not the
same as having a surrounding core.
It's easy to find the inductance of a toroid on a straight wire.
First, inductance is just the derivative of flux with respect to
current:
L=dp/di
We already know the flux in a toroid on a straight wire:
p = uo*ur*i*d*ln(ro/ri)/2*pi
Differentiating, we have:
L=dp/di = uo*ur*d*ln(ro/ri)/2*pi
For d=100cm, ri=1.5mm, ro=2.5mm, ur =3000
= 4pi*10^-7*3000*1m*ln(.0025/.0015)/2pi
= 2*10^-7*3000*1*ln(.0025/.0015)
= 306 uH
Getting closer, no?
--
Joe
legg
<myreal...@comic.com> wrote:
>On Sun, 11 Jan 2009 18:01:50 -0800, D from BC
><myreal...@comic.com> wrote:
>
>I just noticed on a data sheet that ui of materials like W (ui =10000)
>drop to ~2000 @800khz.
>
>AFAIK there's no material with ui = 10000 at 800khz.
>Looks like a straight wire inductor at 400uH will be much longer than
>80cm.
>More like 270cm with u=3000
I think you'll find the permeability drops considerably, for any DC
filtering application when you're using ungapped ferrite.
Your recent foray into marix (flat) transformers should have provided
some insight into single-turn multi-core structures.
RL
D from BC
<jale...@sympatico.ca> wrote:
<snip>
>
>Not so fast, D.
>
>The inductance calculator you are using above is for a straight wire
>made of a conductor with a particular permeability. That's not the
>same as having a surrounding core.
>
>It's easy to find the inductance of a toroid on a straight wire.
>First, inductance is just the derivative of flux with respect to
>current:
>
>L=dp/di
>
>We already know the flux in a toroid on a straight wire:
>p = uo*ur*i*d*ln(ro/ri)/2*pi
>
>Differentiating, we have:
>
>L=dp/di = uo*ur*d*ln(ro/ri)/2*pi
>
>For d=100cm, ri=1.5mm, ro=2.5mm, ur =3000
> = 4pi*10^-7*3000*1m*ln(.0025/.0015)/2pi
> = 2*10^-7*3000*1*ln(.0025/.0015)
> = 306 uH
>
>Getting closer, no?
Yup...
Geez...I suck at magnetics.. :P
Since it's just a ratio in ln and
if I use a stack of production toroids:
OD 9.53 mm
ln -------
ID 4.75 mm
L = 418uH <<near my L target
If per toroid height = 4.78mm
That's 209 stacked toroids.. Hilarious...:)
But that formula describes 1 loop of wire around a toroid.
Is that the same as 1 loop of wire shielded by ferrite?
Does this
| |
| |
(|)|
(|)| 100cm of cores
(|)|
(|)|
+-+
have the same inductance as this
| |
(|) (|) Total of
(|) (|) 100cm of cores
(|) (|)
(|) (|)
+----+
() = stacked toroids
About that wire inductance site...
Thanks for pointing that out that it's ur for the material not the
core on that site.
ur = 1 (basically) for air, copper, aluminum, gold, silver.
I could use that calculator if there's a benefit in choosing a wire
material.
Say 100% annealed Nickel urmax = 600 urmin= 100
(Ni Bsat ~ 6000 gauss)
L of 100cm of 1mm Ni wire = 31uH
It's bonus inductance if it doesn't cause other problems..
D from BC
I was just curious of the alternative if I didn't coil to get away
from interwinding capacitance, leakage flux and EMI.
Also, I thought it would be a nice start to learn some magnetics
theory...
I'll be getting back into researching matrix transformers after
research on planar mag and EFD cores.
I'm still a newbie at magnetics and I find it a pita..
flux
flux density
Al
current density
hysteresis loss
skin effect
core sizes
wire diameters
layering
resonance
heat sinking
heat rise
EMI
fringing
leakage flux
gaps
complex permeability
initial permeability
relative permeability
amplitude permaeability
+ more
...huhhh :P
I believe some understanding of all that is still needed to utilize
the magnetics development software.. :(
The program from EPCOS for example..
J.A. Legris
> On Mon, 12 Jan 2009 06:15:26 -0800 (PST), "J.A. Legris"
>
Yes, they are approximately the same because the most of the flux
follows paths entirely within the cores. Flux that passes partly
through air and partly through core has roughly the same relative
density as flux that passes entirely through air - i.e. a small
fraction, on the order of 1/ur.