LC FILTER FOR THREE PHASE INVERTER
A Project report submitted by:
MUTHURAJ P 13MQ37
ELDHO JACOB 13MQ81
Dissertation submitted in partial fulfillment of the requirements for the degree of
MASTER OF ENGINEERING
Branch: EEE
Specialization: POWER ELECTRONICS & DRIVES of PSG COLLEGE OF TECHNOLOGY
MARCH - 2014
ELECTRICAL & ELECTRONICS
PSG COLLEGE OF TECHNOLOGY (Autonomous Institution)
COIMBATORE 641 004
LC FILTER FOR THREE PHASE INVERTER
SPECIFICATION TO BE ACHIEVED:
INVERTER LC FILTER
SWITCHING FRREQUENCY = 5 KHz
OUTPUT CURRENT = 10A RMS
LINE VOLTAGE = 230V RMS
LINE FREQUENCY = 50 Hz
NECESSARY PROCEDURE IN FILTER DESIGN
The filter design is performed as follows:
1.Establishment of the necessary hypothesis of the filter specification and
progress.
2.Analysis of the filter and establishment of the design method.
3.Theoretical development according to the established method.
4.Presentation about the example of the filter design and comparison with the
simulation.
Filter specification and hypothesis
The filter design specification and hypothesis are performed as follows:
Specification:
1.Linear load is 1-0.9 lagging load.
2.Nonlinear load is a rectifier load with CF=3.
3.THD is below 5% in case of both linear and nonlinear
4.MI(Modulation Index) is 0.9 in the rated resistor load.
Assumption:
1.The plant with LC filter is linear.
2.The current is an ideal sinusoidal wave in case of the linear load.
3.The input voltage waveform of the filter is ideally same
4.The filter has no loss.
5.Each of the fundamental RMS value of the rated current as the output waveform
of the sinusoidal PWM-VSI and the rated voltage of the filter output is 1.0[p.u]
`
FLOW CHART TO DESIGN A PASSIVE(LC) FILTER
FORMULA USED:
(i)To find inductor,
= 1
8
1
iLmax
Where,
Vdc DC voltage of the inverter
iLmax Current ripple
(ripple current can be chosen as 10% - 15% of rated current)
Fs Switching frequency
(ii)To find capacitor,
= 15%
3 2 2
Where,
Reactive power rated
(reactive power is chosen as 15% of the rated power)
Vrated AC rated voltage
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, size, 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 VL 230
A Inductance L 0.045 H
B Line Current IL 10 A
C Line Frequency f 50 Hz
D Current Density J 300 A/cm2
E Efficiency goal ef 90 %
F Material Iron Powder
G
Magnetic
permiability um 1200
H Flux Density Bac 1.4 Tesla
I
Window
Utilisation Ku 0.4
J Temp Rise Goal Tr 60 C
2
Calculate Apparent power
Pt
Pt = VA = VL*IL 2300 A
3 Calculate Area Product
AP
AP =
VA*10^4/(4.44*Ku*f*Bac*J) 616.6881167 cm4
4 Select Core
Iron Powder Core EI228
core Material
Magnetic Path Length MPL 34.3 cm
2844 g 2.8KG + winding weight
Mean Length Turn MLT 32.7 cm
Iron
Area Ac 31.028 cm2
Window Area Wa 24.496 cm2
Area product Ap 760.064 cm4
Coef Kg 288.936 cm5
Surface Area At 1078 cm2
Material P P
Winding Length G 8.573
Lamination E 5.715
5 Calculate Number of Turns N 238.502559 turns
6 Inductance Required L 0.045 H
7 Calculate required airgap lg
lg = (0.4piN2Ac10-4/L) -
(MPL/um) lg 0.464042287 cm 4.640423 mm
8 Calculate Fringing flux F F 1.300699751
9
Calculate New number of
turns N1
N1=sqrt(lg*L/0.4piACF10-
8) 202.9667027 turns 203
10 Calculate flux density
Bac =
VL*10^4/(4.44*N1*Ac*f Bac 1.645115076 Tesla
11 Calculate Bare wire area
Awl Awl=IL/J 0.033333333 cm2
12
Select wire from Wire
table
AWG
14 Aw 0.02 cm2
uOhm/cm 82.8 uOhm/cm
13
Calculate Winding
Resistance
R=MLT*N1*uOHm*10-6 R 0.549544526 Ohms
14 Calculate Copper Loss
PL = IL2 * RL PL 54.95445256 W
15
Calculate Watts per
kilogram
W/K =
0.000557*f^1.68*B^1.86 w/k 1.365445533 Ohm
16 Calculate Core Loss
Pfe =w/k *Wtfe Pfe 0.92304118 W
17 Calculate Gap Loss
Pg = Ki*E*lg*f*B2 Pg 55.62474848 W
18 Calculate Total Loss
sum of losses PL 111.5022422 W
19 Calculate surface area watt density
psi = PL/At psi 0.103434362 watts per cm2
20
Calculate the Temperature
rise
Tr = 450*psi^0.826 Tr 69.07575995
21
Calculate Window
utilisation
Ku = N1*Aw/Wa 0.16571416 watt
INDUCTOR WINDING DETAILS
0
210
1
3
I
Inductor
Termination
Winding Arrangement
I
2
200
WINDING DETAILS
No
.
Winding
no.
Terminals No
of
turn
s
Wire
gauge
SWG
Insulation
between
winding Layers
Remarks
1 I 1 & 2 200 14 Nil
(Varnishing
Reqd)
2 I Tapping 3 210
Core Details : EI 225
CORE DIMENSIONAL DETAILS
WIRE TABLE
SIMULATION CIRCUIT:
SIMULATION RESULTS:
Without Filter:
With Filter:
