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

Hi:

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

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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!

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_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov -- NOTE: Remove "BOGUS" from email address to reply.
Reply to
Chris Carlen
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Dunno, but an inquiry to: 

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

should get you an answer. :-)
Reply to
John Fields

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!

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_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov -- NOTE: Remove "BOGUS" from email address to reply.
Reply to
Chris Carlen

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.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

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

Reply to
John Smith

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