DESIGN OF INDUCTOR

FILTER DESIGN: INVERTER LC FILTERSWITCHING FRREQUENCY = 5KHzOUTPUT CURRENT = 10A RMSLINE VOLTAGE = 230V RMSLINE FREQUENCY = 50Hz

CAPACITANCE VALUE CALCUALATED = 10uF, 600VINDUCTANCE VALUE CALCULATED = 4.5mH

INDUCTANCE DESIGN PROCEDURESeveral factors need to be considered while designing an inductor, few of which are listed below1. Frequency of Operation2. Core Material Selection3. Energy Handling Capability of the Inductor (determines the size of the core)4. Calculate Number of Turn5. Selection of Copper wire6. Estimation of Losses and Temperature Rise

In this application the Inductor has to handle large energy due to the RMS current is 10A maximum. At this current most of the ferrite core shapes does not support the design (computed from the Area Product). So we select Iron powder core for this design.

The design of the ac inductor requires the calculation of the volt-amp (VA) capability. In this applications the inductance value is specified.

Relationship of, Area Product Ap, to the Inductor Volt-Amp CapabilityThe volt-amp capability of a core is related to its area product, Ap, by the equation that may be stated as Follows.

From the above, it can be seen that factors such as flux density, Bac, the window utilization factor, Ku (which defines the maximum space occupied by the copper in the window), and the current density, J, all have an influence on the inductor area product, Ap.

Fundamental Considerations

The design of a linear ac inductor depends upon five related factors:1 . Desired inductance2. Applied voltage, (across inductor)3. Frequency4. Operating Flux density which will not saturate the core5. Temperature Rise

The inductance of an iron-core inductor, with an air gap, may be expressed as:

Final determination of the air gap requires consideration of the effect of fringing flux, which is a function of gap dimension, the shape of the pole faces, and the shape, si/e, and location of the winding

Fringing flux decreases the total reluctance of the magnetic path, and therefore increases the inductance by a factor, F, to a value greater than that calculated from Equation

Where G is winding length of the core

Now that the fringing flux, F, has been calculated, it is necessary to recalculate the number of turns using the fringing flux, Factor F

with the new turns, N(new), and solve for Bac

The losses in an ac inductor are made up of three components:1. Copper loss, Pcu2. Iron loss, Pfe3. Gap loss, Pg

The copper loss, Pcu, is I2R and is straightforward, if the skin effect is minimal. The iron loss, Pfe, is calculated from core manufacturers' data. Gap loss, Pg, is independent of core material strip thickness and permeability.

INDUCTOR DESIGN STEPS1Design SpecVL230

aInductanceL0.045H

bLine CurrentIL10A

cLine Frequencyf50Hz

dCurrent DensityJ300A/cm2

eEfficiency goalef90%

fMaterial Iron Powder

gMagnetic

permiabilityum1200

hFlux DensityBac1.4Tesla

iWindow UtilisationKu0.4

jTemp Rise GoalTr60C

2Calculate Apparent power Pt

Pt = VA = VL*IL2300A

3Calculate Area Product

APAP = VA*10^4/(4.44*Ku*f*Bac*J)616.6881167cm4

4Select Core

Iron Powder Core EI228

core Material

Magnetic Path LengthMPL34.3cm

2844g2.8KG + winding weight

Mean Length TurnMLT32.7cm

Iron AreaAc31.028cm2

Window AreaWa24.496cm2

Area productAp760.064cm4

coefKg288.936cm5

Surface AreaAt1078cm2

Material PP

Winding LengthG8.573

Lamination E5.715

5Calculate Number of TurnsN238.502559turns

6Inductance RequiredL0.045H

7Calculate required airgaplg

lg = (0.4piN2Ac10-4/L) - (MPL/um)lg0.464042287cm4.640423mm

8Calculate Fringing flux FF1.300699751

9Calculate New number of turnsN1

N1=sqrt(lg*L/0.4piACF10-8)202.9667027turns203

10Calculate flux density

Bac = VL*10^4/(4.44*N1*Ac*fBac1.645115076Tesla

11Calculate Bare wire area

AwlAwl=IL/J0.033333333cm2

12Select wire from Wire table

AWG 14Aw0.02cm2

uOhm/cm82.8uOhm/cm

13Calculate Winding Resistance

R=MLT*N1*uOHm*10-6R0.549544526Ohms

14Calculate Copper Loss

PL = IL2 * RLPL54.95445256W

15Calculate Watts per kilogram

W/K = 0.000557*f^1.68*B^1.86w/k1.365445533Ohm

16Calculate Core Loss

Pfe =w/k *WtfePfe0.92304118W

17Calculate Gap Loss

Pg = Ki*E*lg*f*B2Pg55.62474848W

18Calculate Total Loss

sum of lossesPL111.5022422W

19Calculate surface area watt density

psi = PL/Atpsi0.103434362watts per cm2

20Calculate the Temperature rise

Tr = 450*psi^0.826Tr69.07575995

21Calculate Window utilisation

Ku = N1*Aw/Wa0.16571416watt

INDUCTOR WINDING DETAILS

2210

13

200

II

0

Winding ArrangementInductor Termination

WINDING DETAILS

No.Winding no.TerminalsNo of turnsWire gauge SWGInsulation between winding LayersRemarks

1I1 & 220014Nil(Varnishing Reqd)

Tapping 3210

Core Details : EI 225

CORE DIMENSIONAL DETAILS

WIRE TABLE