DESIGN OF INDUCTOR
FILTER DESIGN:
INVERTER LC FILTER
SWITCHING FRREQUENCY = 5KHz
OUTPUT CURRENT
= 10A RMS
LINE VOLTAGE
= 230V RMS
LINE FREQUENCY
= 50Hz
CAPACITANCE VALUE CALCUALATED = 10uF, 600V
INDUCTANCE VALUE CALCULATED
= 4.5mH
INDUCTANCE DESIGN PROCEDURE
Several factors need to be considered while designing an inductor, few of which are listed
below
1.
Frequency of Operation
2.
Core Material Selection
3.
Energy Handling Capability of the Inductor (determines the size of the core)
4.
Calculate Number of Turn
5.
Selection of Copper wire
6.
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 Capability
The 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 inductance
2. Applied voltage, (across inductor)
3. Frequency
4. Operating Flux density which will not saturate the core
5. 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, Pcu
2. Iron loss, Pfe
3. 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 STEPS
1 Design Spec
a
Inductance
b
Line Current
c
Line Frequency
d
Current Density
e
Efficiency goal
VL
L
IL
f
J
ef
Material
Magnetic
g
h
i
permiability
Flux Density
Window Utilisation
um
Bac
Ku
Temp Rise Goal
Tr
230
0.045
10
50
300
90
H
A
Hz
A/cm2
%
Iron Powder
Calculate Apparent power
2 Pt
Pt = VA = VL*IL
1200
1.4 Tesla
0.4
60 C
2300 A
3 Calculate Area Product
AP = VA*10^4/
AP
(4.44*Ku*f*Bac*J)
616.688116
7 cm4
4 Select Core
Iron Powder Core EI228
core Material
Magnetic Path Length
Mean Length Turn
Iron
Area
Window Area
Area product
coef
Surface Area
Material P
Winding Length
Lamination E
MPL
MLT
Ac
Wa
Ap
Kg
At
P
G
5 Calculate Number of Turns
6 Inductance Required
34.3 cm
2844 g
32.7 cm
31.028
24.496
760.064
288.936
1078
cm2
cm2
cm4
cm5
cm2
8.573
5.715
238.502559 turns
0.045 H
2.8KG + winding weight
7 Calculate required airgap
lg = (0.4piN2Ac10-4/L) (MPL/um)
8 Calculate Fringing flux F
Calculate New number of
9 turns
N1=sqrt(lg*L/0.4piACF108)
10 Calculate flux density
Bac = VL*10^4/
(4.44*N1*Ac*f
lg
lg
0.46404228
7 cm
1.30069975
1
N1
202.966702
7 turns
Bac
1.64511507
6 Tesla
11 Calculate Bare wire area
Awl
0.03333333
3 cm2
Awl=IL/J
12 Select wire from Wire table
AWG
14
Aw
uOhm/c
m
0.02 cm2
uOhm/c
82.8 m
Calculate Winding
13 Resistance
R
0.54954452
6 Ohms
PL
54.9544525
6 W
Calculate Watts per
15 kilogram
W/K =
0.000557*f^1.68*B^1.86
w/k
1.36544553
3 Ohm
16 Calculate Core Loss
Pfe =w/k *Wtfe
Pfe
0.92304118 W
R=MLT*N1*uOHm*10-6
14 Calculate Copper Loss
PL = IL2 * RL
17 Calculate Gap Loss
4.64042
3 mm
203
Pg = Ki*E*lg*f*B2
Pg
55.6247484
8 W
PL
111.502242
2 W
18 Calculate Total Loss
sum of losses
19 Calculate surface area watt density
psi = PL/At
psi
0.10343436
2 watts per cm2
Tr
69.0757599
5
Calculate the Temperature
20 rise
Tr = 450*psi^0.826
Calculate Window
21 utilisation
Ku = N1*Aw/Wa
0.16571416 watt
INDUCTOR WINDING DETAILS
210
200
3
2
I
I
0 1
Inductor
Termination
Winding Arrangement
WINDING DETAILS
N
o.
Winding
no.
Terminals
No of turns
1&2
200
Tapping
3
210
Core Details : EI 225
Wire
gauge
SWG
14
Insulation between
winding Layers
Nil
(Varnishing Reqd)
Remarks
CORE DIMENSIONAL DETAILS
WIRE TABLE
