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