Hw2_solution - Problem 6 Simple UI Core Analyiss Dimensions etc 2 3 cm:= 1 10 mm:= 1.0 10 w:= 1 cm d s:= 2 cm g:= 1 mm ws:= 5 cm d:= 5 cm N:= 100

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Problem 6 - Simple UI Core Analyiss Dimensions, etc cm 110 2 := mm 1.0 10 3 := w1 c m := d s 2cm := g 1mm := w s 5cm := d5 c m := N 100 := B sat 1.3 := Point where saturation occurs u 0 4 π 10 7 := Now let's compute some reluctances. For flux densities that are not saturated u 1500 u 0 := R 12 w s w + wd u := R 56 R 12 := R 23 w 2w d u := R 81 R 23 := R 34 g u 0 := R 78 R 34 := R 45 d s w 2 + u := R 67 R 45 := R eff R 12 R 23 + R 34 + R 45 + R 56 + R 67 + R 78 + R 81 + := R eff 3.374 10 6 × =
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OK. Now, the cross sectional area we have is A wd := The flux where we saturate is Φ sat B sat A := The current to achieve this flux i sat Φ sat N R eff := i sat 21.932 10 0 × = L N 2 R eff := L 2.964 10 3 × = Inductance of UI Core with standard gap E 1 2 L i sat 2 := E 712.775 10 3 × = Energy stored in inductor Now lets change the gap to zero g 0 := R 34 g u 0 := R 78 R 34 := R eff R 12 R 23 + R 34 + R 45 + R 56 + R 67 + R 78 + R 81 + := R eff 190.986 10 3 × = i sat Φ sat N R eff := We know saturated with very little current i sat 1.241 10 0 × =
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L N 2 R eff := L 52.36 10 3 × =
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This note was uploaded on 09/22/2011 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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Hw2_solution - Problem 6 Simple UI Core Analyiss Dimensions etc 2 3 cm:= 1 10 mm:= 1.0 10 w:= 1 cm d s:= 2 cm g:= 1 mm ws:= 5 cm d:= 5 cm N:= 100

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