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Magnetics, Inc. don't know how to calculate wire sizes

C

Chris Carlen

Jan 1, 1970
0
Hi:

Look at page 40 in this document on their powder cores:

http://www.mag-inc.com/pdf/2004_Magnetics_Powder_Core_Catalog.zip

The wire diameters and areas are way off from what you'd calculate using
the AWG formula:

d = 0.005*92^((36-AWG)/39) [in]

I thought maybe they were insulated diameters, from their use of the
term "heavy build".

But if that was the case, then the areas would still be right according
to the AWG formula. But they are too large as well.

For instance 22ga wire should be:

d = 0.02535 in = 0.6438 mm
A = 642.4 circ mils = 0.3255 mm^2

But Magnetics lists:

d = 0.701 mm
A = 810 circ mils = 0.411 mm^2


Notice, their area doesn't match the diameter, which would lead to 0.386
mm^2.

What are they doing?

Interestingly, their current capacities are close to, but still a bit in
error from the values one would calculate based on the correct areas
(the copper, not the insulated wire).


Good day!


--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
[email protected] -- NOTE: Remove "BOGUS" from email address to reply.
 
J

John Fields

Jan 1, 1970
0
Hi:

Look at page 40 in this document on their powder cores:

http://www.mag-inc.com/pdf/2004_Magnetics_Powder_Core_Catalog.zip

The wire diameters and areas are way off from what you'd calculate using
the AWG formula:

d = 0.005*92^((36-AWG)/39) [in]

I thought maybe they were insulated diameters, from their use of the
term "heavy build".

But if that was the case, then the areas would still be right according
to the AWG formula. But they are too large as well.

For instance 22ga wire should be:

d = 0.02535 in = 0.6438 mm
A = 642.4 circ mils = 0.3255 mm^2

But Magnetics lists:

d = 0.701 mm
A = 810 circ mils = 0.411 mm^2


Notice, their area doesn't match the diameter, which would lead to 0.386
mm^2.

What are they doing?

---
Dunno, but an inquiry to:

http://www.mag-inc.com/contacts/contact_form.asp

should get you an answer. :)
 
K

Ken Smith

Jan 1, 1970
0
Hi:

Look at page 40 in this document on their powder cores:

http://www.mag-inc.com/pdf/2004_Magnetics_Powder_Core_Catalog.zip

The wire diameters and areas are way off from what you'd calculate using
the AWG formula:

d = 0.005*92^((36-AWG)/39) [in]

I thought maybe they were insulated diameters, from their use of the
term "heavy build".

But if that was the case, then the areas would still be right according
to the AWG formula. But they are too large as well.

For instance 22ga wire should be:

d = 0.02535 in = 0.6438 mm
A = 642.4 circ mils = 0.3255 mm^2

But Magnetics lists:

d = 0.701 mm
A = 810 circ mils = 0.411 mm^2


Notice, their area doesn't match the diameter, which would lead to 0.386
mm^2.

What are they doing?

This is just a guess but I think they are including a packing factor in
the cross sectional area. The windings can never use up 100% of the
space.
 
J

John Smith

Jan 1, 1970
0
Chris Carlen said:
Hi:

Look at page 40 in this document on their powder cores:

http://www.mag-inc.com/pdf/2004_Magnetics_Powder_Core_Catalog.zip

The wire diameters and areas are way off from what you'd calculate using
the AWG formula:

d = 0.005*92^((36-AWG)/39) [in]

I thought maybe they were insulated diameters, from their use of the term
"heavy build".

But if that was the case, then the areas would still be right according to
the AWG formula. But they are too large as well.

For instance 22ga wire should be:

d = 0.02535 in = 0.6438 mm
A = 642.4 circ mils = 0.3255 mm^2

But Magnetics lists:

d = 0.701 mm
A = 810 circ mils = 0.411 mm^2


Notice, their area doesn't match the diameter, which would lead to 0.386
mm^2.

What are they doing?

Interestingly, their current capacities are close to, but still a bit in
error from the values one would calculate based on the correct areas (the
copper, not the insulated wire).


Good day!


--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
[email protected] -- NOTE: Remove "BOGUS" from email address to reply.


Hi, Chris -

According to my wire tables, your equation appears to give the bare diameter
of the wire. My wire table gives the following for #22:

Bare diameter .0253 nom
Insulated diameter .0277 nom
Insulated diameter .0284 max
Insulated design diameter .0291

My wire table does not indicate if the insulated diameter is single or heavy
film diameter, but I suspect the former. I can no longer find my complete
table which listed the diameter increases due to single and heavy film
coatings. Also, I seem to remember that the insulating film thickness varied
with the wire diameters.

I would urge you to use the wire data they provide. You will still need to
fudge numbers to get the wire to fit into a window.

Good luck.

John
 
C

Chris Carlen

Jan 1, 1970
0
John said:
According to my wire tables, your equation appears to give the bare diameter
of the wire. My wire table gives the following for #22:

Bare diameter .0253 nom
Insulated diameter .0277 nom
Insulated diameter .0284 max
Insulated design diameter .0291

My wire table does not indicate if the insulated diameter is single or heavy
film diameter, but I suspect the former. I can no longer find my complete
table which listed the diameter increases due to single and heavy film
coatings. Also, I seem to remember that the insulating film thickness varied
with the wire diameters.

I would urge you to use the wire data they provide. You will still need to
fudge numbers to get the wire to fit into a window.

Thanks for the input.

Actually, I am more interested in power loss calculations, so I will
rely on my AWG formula tables for that, since it is the real copper that
matters.

But then for winding arrangements, the larger diameters that account for
film thicknesses as well as some "packing factor" are more useful.

Too bad they don't specify what they are tabulating there.

Maybe I'll email them.


Good day!


--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
[email protected] -- NOTE: Remove "BOGUS" from email address to reply.
 
R

Rich Grise

Jan 1, 1970
0
For instance 22ga wire should be:

d = 0.02535 in = 0.6438 mm
A = 642.4 circ mils = 0.3255 mm^2

But Magnetics lists:

d = 0.701 mm
A = 810 circ mils = 0.411 mm^2


Notice, their area doesn't match the diameter, which would lead to 0.386
mm^2.

But, interestingly, 0.6438 ^2 = 0.41447844, which is pretty close. They're
just calculating to a square window.

Cheers!
Rich
 
K

Ken Smith

Jan 1, 1970
0
But, interestingly, 0.6438 ^2 = 0.41447844, which is pretty close. They're
just calculating to a square window.


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