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Old 15th November 2010, 01:59 PM   #11
AndrewT is online now AndrewT  Scotland
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C=I
D=4I
E=2I
F=3I
B=F+2I=5I
A=D+2I=6I

This arrangement gives zero waste and more importantly gives equal flux per unit of width.

If you need more window area, then you cannot have zero waste. But bigger window area can still match all the flux flows per unit of width.
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Last edited by AndrewT; 15th November 2010 at 02:06 PM.
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Old 15th November 2010, 02:06 PM   #12
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Thank you so much for your clear answer.
I just wonder if those give the best solution why in many cases :
B#F+2I
A-D#B-F
thanks!
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Old 15th November 2010, 05:23 PM   #13
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Ferrite cores are not stamped. Ferrite cores can be made in lots of funny shapes without scrap.

But still, the middle leg of the E core is typically double the area of the outer legs, the I part and the back of the E to get the same flux density everywhere.

That gives:
E = 2I
B = F + 2I
A = D + 2I

EIR cores are not comparable though, having a round center leg. What about them?
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Old 15th November 2010, 06:35 PM   #14
AndrewT is online now AndrewT  Scotland
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oops, I did not read "ferrite".
Ignore all since it only applies to iron cored.
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Old 16th November 2010, 07:07 AM   #15
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yep. In EIR core: the cross-section area also =2xthat of leg.
I wonder if it is optimum solution for flux why in many cases
B-F-I# I
E=A-D#2I
(it is not difficult to shape the core in these condition)
and can you explain why ferrite core is different?
Thanks!
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Old 16th November 2010, 12:28 PM   #16
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Can anyone teach me how to calculate the effective magnetic path of a core? For PQ core, how can we determine it?
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