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LC filter
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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
LC FILTER DESIGN:
A low pass LC filter is required at the output terminal of Full Bridge VSI to reduce harmonics generated by the pulsating modulation waveform. While designing LC filter, the cut-off frequency is chosen such that most of the low order harmonics is eliminated. To operate as an ideal voltage source, that means no additional voltage distortion even though under the load variation or a nonlinear load, the output impedance of the inverter must be kept zero. Therefore, the capacitancevalue should be maximized and the inductance value should be minimized at the selected cut-off frequency of the low-pass filter.
Each value of L and C component is determined to minimize the reactive power in these components because the reactive power of L and C will decide the cost of LC filter and it is selected to minimize the cost, then it is common that the filter components are determined at the set of a small capacitance and a large inductance and consequently the output impedance of the inverter is so high. With these design values, the voltage waveform of the inverter output can be sinusoidal under the linear load or steady state condition because the output impedance is zero. But in case of a step change of the load or a nonlinear load, the output voltage waveform will be distorted cause by the slow system response as the output response is non-zero. Figure 1 shows the power circuit of the single phase PWM-VSI with any linear or nonlinear load.
The load current flows differently depending on the kind of loads such as linear and nonlinear load. Therefore it is difficult to represent the transfer function of inverter output voltage to load current.The plant composed of L-C low-pass filter satisfies linear property, so it is possible to represent the system which has two inputs of inverter output voltage and load current.
R c
1k
T2
T5
n
T6
T1
R a
1k
T3
n
R b
1k
V 1100V dc
C 1
C 2T4
FIG 1
FIG 2
START
HARMONIC ANALYSIS OF PWM VOLTAGE AND NONLINEAR CURRENT
SELECTING CUT-OFF FREQUENCY
SELECTING MINIMUM CAPACITANCE BASED ON COST
SELECTING CONTROLLER RESPONSE
IS THE CONTROL RESPONSE REALIZABLE?
SELECTING CONTROLLER GAINS
SATISFING CONTROL RESPONSE
ANALYSING OUTPUT VOLTAGE HARMONICS UNDER
THE LINEAR AND NON-LINEAR LOAD
THD = 5%
SELECTING DC LINK VOLTAGE AN CALCULATE INDUCTANCE
STOP
SELECTING CUT-OFF FREQUENCY
FLOW CHART TO DESIGN A PASSIVE(LC) FILTER
FORMULA USED:
(i)To find inductor,
L=
18∗Vdc∗1
Δi Lmax∗Fs
Where,
Vdc – DC voltage of the inverter
Δi Lmax – Current ripple
(ripple current can be chosen as 10% - 15% of rated current)
Fs – Switching frequency
(ii)To find capacitor,
C=15%∗Prated
3∗2 πf ∗V 2 rated
Where,
Prated – 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 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 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
IWindow Utilisation Ku 0.4
JTemp Rise Goal Tr 60 C
2Calculate 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
5Calculate Number of Turns N
238.502559 turns
6 Inductance Required L 0.045 H
7Calculate required airgap lg
lg = (0.4piN2Ac10-4/L) - (MPL/um) lg
0.464042287 cm
4.640423 mm
8Calculate Fringing flux F F
1.300699751
9Calculate 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
11Calculate Bare wire area
Awl Awl=IL/J0.0333333
33 cm2
12Select wire from Wire table
AWG 14 Aw 0.02 cm2
uOhm/cm 82.8
uOhm/cm
13Calculate Winding Resistance
R=MLT*N1*uOHm*10-6 R
0.549544526 Ohms
14 Calculate Copper Loss
PL = IL2 * RL PL54.954452
56 W
15Calculate 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 Pfe0.9230411
8 W
17 Calculate Gap Loss
Pg = Ki*E*lg*f*B2 Pg55.624748
48 W
18 Calculate Total Loss
sum of losses PL111.50224
22 W
19Calculate surface area watt density
psi = PL/At psi0.1034343
62 watts per cm2
20Calculate the Temperature rise
Tr = 450*psi^0.826 Tr69.075759
95
21Calculate Window utilisation
Ku = N1*Aw/Wa0.1657141
6 watt
INDUCTOR WINDING DETAILS
210
1
3
II
2
200
WINDING DETAILS
No.
Winding no.
Terminals No of turns
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
0
Inductor Termination
Winding Arrangement
CORE DIMENSIONAL DETAILS
WIRE TABLE
SIMULATION CIRCUIT:
SIMULATION RESULTS:
Without Filter:
With Filter:
REFERENCES:
[1] Miss. Sangita R Nandurkar , Mrs. Mini Rajeev ,”Design and Simulation of three phase Inverter for grid connected Photovoltic systems,” Proceedings of Third Biennial National Conference, NCNTE- 2012, Feb 24-25
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