- Open Access
- Total Downloads : 762
- Authors : A.Ganapathy , R.Rajesh, S.Sathishkumar
- Paper ID : IJERTV2IS111147
- Volume & Issue : Volume 02, Issue 11 (November 2013)
- Published (First Online): 27-11-2013
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License: This work is licensed under a Creative Commons Attribution 4.0 International License
Conversion of Single Phase to Three Phase AC Drive System Using Parallel Rectifiers
A.GANAPATHY, ME., R.RAJESH, ME., S.SATHISHKUMAR, ME.,
Assistant Professor, Assistant Professor, Assistant Professor, Department of Electrical and Electronics Engineering,
Chendhuran College of Engineering & Technology,
Lenavilakku, Pudukkottai-Tamil Nadu
ABSTRACT: This paper is a single phase to three phase AC drive system composed of two parallel single phase rectifiers, three phase inverter, and an induction motor. The main advantage of the paper is to reduce the rectifier switch currents, the harmonic distortion at the converter input side, improvements on the fault tolerance characteristics and an induction Motor run at any loaded conditions. Even with increase in the number of switches, the total energy loss of the system lower than the conventional system. The model of the system is derived and it shown that the reduction of circulating current is an improvement objective of the system design. It required output voltage for inverter using PWM technique. This paper is a single phase to three phase drive system composed of two parallel single phase rectifiers using MAT LAB Simulink model.
Keywords:- Power converter, Parallel Rectifier, Induction Motor, Boost Regulator.
-
INTRODUCTION
Several solution have been proposed when the objective is to supply a three phase motors from a single phase mains. It is quite common to have only a single phase power grid in residential, commercial, industrial, agriculture and mainly in rural areas, while the adjustable speed drives may request a three phase power grid. Single- phase to three-phase ac-dc-ac conversion usually employs a full bridge topology which implies in ten power switches as shown in fig .1. This converter is denoted here as conventional topology.
Parallel converters have been used to improve the power capability, reliability, efficiency, and redundancy. Parallel converter technique can be employed to improve the performance of active filters, uninterruptible power supplies, fault tolerance of doubly fed induction generators, and three phase drives. Usually the operation of converters in parallel requires a transformer for isolation. However, weight, size, and cost
associated with the transformer may make such a solution undesirable. When an isolation transformer is not used, the reduction of circulating currents among different converter stages is an improvement objective in the system. In this paper, a singlephase to threephase drive system composed of two parallel single phase rectifiers and a three- phase inverter is proposed as shown in fig .1
.The proposed system is conceived to operate where the single- phase utility grid is the unique option available. Compared to the conventional topology, the proposed system permits to reduce the rectifier switch currents; the total harmonic distortion (THD) of the grid current with same switching frequency (or) the switching frequency with same THD of the grid current; and to increase the fault tolerance characteristics. In addition, the losses of the proposed system may be lower than that of the conventional counterpart. Therefore mentioned benefits justify the initial investment of the proposed system, due to the increase of number of switches.
-
SINGLE PHASE PARALLEL RECTIFIER
Parallel rectifier means two single phase fully controlled rectifier are connected in parallel. In this rectifier circuit eight SCR are used and current divided into two rectifier circuit, so the
device switching loss is less. The main advantage of the project to reduce the rectifier switch currents, the harmonic distortion at the converter input side and improvements on the fault tolerance characteristics. Even with increase in the number of switches, the total energy loss of the system lower than the conventional system. The model of the system is derived and it shown that the reduction of circulating current is an improvement objective of the system design.
Fig.1 Proposed Single Phase to three Phase AC Motor Drive System
In this project mainly selected controlled rectifier using SCR .The specific reason for controlled rectifier circuit are not allowed the load current hence circulating current may be less but uncontrolled rectifier using diode are allowed the return current (or) circulating current ,so harmonics distortion to be created. Conclude that the PWM techniques are introduced in this project to reduce harmonics distortion and improve
the power factor. Uses of Parallel
a filter circuit q q
a2, a2
a2, a2
First arm of the
a1, a1
a1, a1
Converter: Power Capability, Reliability,
Efficiency, Redundancy.
rectifier A q q
Second arm of the
-
SYSTEM MODELING OF
rectifier A q
b1,
qb1
First arm of the
PARALLEL RECTIFIER
The system is composed of grid, input
rectifier B q q
b2, b2
b2, b2
rectifier B
Second arm of the
inductors (La, La, Lb and Lb.), and rectifiers
Inverter side using 6 switches
(A and B) Capacitor bank at the dc link,
q
q
s1,
qs 2
Inverter first arm q
s 2,
qs 2
Inverter
inverter, and induction machine. Rectifiers A and B are constituted of qa1, qa1, qa2,
second arm q
s3,
qs3
Inverter third arm
and qa2and qb1, qb1, qb2, and qa2 respectively. The inverter is constituted of switches qs1, qs1, qs2, qs2, qs3, and qs3, the conduction state of the switches is represented by variable Sqa1 to Sqs3, where Sq=1 indicates a closed switches while Sq=
0 indicates an open switch. The input inductor La, La (Rectifier A) and the
input inductor Lb, Lb Rectifier B). The
Switching states represented by ON and
OFF Sq=1 indicates a closed switching states Sq=0 indicates a open switching states Following equation can be derived for the front end of the rectifier
va10 va20 = Grid voltage voltage drop of the first arm Voltage drop of the second arm
va10 va20 =
eg ia ra Ldia / dt r ' i ' l ' di ' / dt
capacitor bank at the Dc link inverter &
induction motor. Here inductors connected
Where p= d\dt
a a a a
in series act as filter. qa1, qa1 are first
= eg (ra la p)ia (r ' l ' p)i'
(1)
a
a
a
a
a
a
arm of the rectifier A & qa2, qa2 are
vb10 vb20 = eg (rb lb p)ib (r ' l ' p)i'
(2)
second arm of the rectifier A. qb1, qb1
b b b
This equation from rectifier B
are first arm of the rectifier B & qb2,
qb2 are second arm of the rectifier B.
va10 vb10 = (rb
-
lb
p)ib
-
(ra
-
-
la
p)ia
(3)
va20 vb20 =
(r ' l ' p)i' (r ' l ' p)i' (4)
a a a b b b
System model
The input inductors =
L , L'
a
a
rectifier A
Here r, l represents resistance & inductance
a
a
inductors L L' rectifier B inductors. The
b, b
indicates a &b rectifier A & B.The capacitor bank at the dc link inverter & induction motor connected in series act as
(5)
Grid current equal to the sum of rectifier
current A & B Circulating current i0 can
The model equation (7) (9) can be
submitted to model given by,
be defined from ia and ia (or) ib and ib
V V0 e
r ' r ' (l '
-
l ' ) p i
(r ' l ' p)
a 2
V
V
(6)
0
g g g g g
' '
a g g
Va 2
eg 2rg 2lg p ia 0
Introducing circulating current i0 in
V V0 e
2r ' l ' p i
(13)
equation (1), (2), & (3)
a 2 g
g g a
v e (r
-
l p)i
(r ' l ' p)(i
-
i )
Similar expression for Vb
a= g
a a a
a a a 0
V V0 e
2r ' l ' p i
(14)
Va eg (ra la p)ia (r ' ia l ' pia r ' i0 l ' pi0 )
b 2 g
g g b
a a a a
a
a
a
a
a
a
a
a
Va eg (ra r ' la p l ' p)ia (r ' l ' p)i0
(7)
' '
' '
' '
'
V0
V0
rg
rg
rg
rg
(lg
(lg
lg ) p
lg ) p
i0
i0
rg
rg
rg
rg
(lg
(lg
lg ) p
lg ) p
0
0
Similar expression for vb
r ' r ' (l '
l ' ) p 0
g
g
g
g
g
g
g
g
b
b
b
b
b
b
b
b
V
V
2r ' 2l ' p
2r ' 2l ' p
i
i
Vb eg (rb r ' lb p l ' p)ia (r ' l ' p)i0 (8)
a
a
Where i' ia i0
0 g g 0
' '
Va10 Va20 Va , (9)
V0 2 rg lg p i0
(15)
Vb10 Vb20 Vb
(10)
Additionally, the equation for
Adding equation (3) & (4) find out values
i' ,i' andi' can be written
of circulating voltage v0
a b g
V V0 e 2r ' l ' p i'
(16)
V0 (ra r ' (la l ' ) p)ia (rb r '
a 2 g
g g a
a a b
(11)
(lb lb' ) p)ib (ra' rb' (la' l ') ) p)i0
V V0 e 2r ' l ' p i'
(17)
b
v0 va10 va20 vb10 vb2o
(12)
b 2 g g g b
' '
(18)
Therefore Va & Vb from rectifier A & and
Vab eg
rg lg p ig
rectifier B. V0 from rectifier A & B are to regulate currents ia ,ib and io respectively.
We are using four identical inductors; the circulating current can be reduced to zero.
a b g
a b g
Reference current i* & i* i* / 2 Reference
circulating current * In order to both
V0 Va10 Va20 Vb10 Vb20 0
When i0=0 i i' i i'
(19)
the system
i0 0
a a b b
facilitate the control and share equally current, voltage, and power between the
model (7) – (9) is reduced to
g
g
g
g
a
a
Va eg 2r ' l ' p i'
(20)
rectifiers, the four inductors should be
V e
2r ' l ' p i'
(21)
g
g
equal r '
l
l
,
g
ra
la
r '
a
a
a
a
l '
rb
lb
'
b
b
r '
b
b
l '
b g g g b
-
-
BOOST REGULATOR
Where I I 2 I1
is the peak to peak
Boost regulator means output voltage is greater than the input voltage. A boost
ripple current of inductor L from equation (22) & (23)
regulator using power MOSFET (or) IGBT shown in fig 4.1. The circuit operation can
I Vst1 / L (Va Vs )t2 / L
Sub suiting duty cycle k= t1/T
(24)
be divided into two modes.
Fig 4.1 Circuit diagram of Boost Converter Mode: 1
Mode 1 begins when transistor M1 is switched ON at t=0. The input current, which rises, flows through inductor L and MOSFET.
Mode: 2
Mode 2 begins when transistor M1 is switched OFF at t= t1 . The current that was flowing through the transistor (MOSFET) would now flow through L, C, diode (Dm), & load .The input current, which falls, until transistor. M1 is turn ON again in the next cycle. Assuming that the inductor current raises linearly from I1 to I2
in time t1
t1 =k T & t2 = (1-k) T yields the average output voltage
Va = Vs T/t2=Vs T/(1-k) T Va = Vs /(1-k)
(1-k) = Vs/ Va (25)
Sub suiting k=t1/T (duty cycle) into equation (4) yields
k1= t1 f where 1/T =f (1- t1f) Va =Vs
Va – Va t1 f = Vs Va t1 f= Va Vs
Gt1 = Va- Vs / Va f (26) Assuming lossless circuit VsIs = Va Ia
Vs = Va (1-k) VsIs= VsIa / (1-k)
The average input current is Is = VsIa/ Vs(1-k) = Ia / (1-k)
Is = Ia/ (1-k) (27)
The switching period T can be found from T = 1/f = t1 + t2 =
2 1 1 1
2 1 1 1
s
s
V LI I / t LI / t
IL / Vs IL /(Va Vs ) IL(Va Vs ) ILVs /(Vs (Va Vs )
t1 LI / Vs
(22)
T = ILVa /Vs (Vs Vs )
(28)
And the inductor current falls linearly from I2 to I1 in time t2
Vs Va LI / t2
And this gives the peak to peak ripple current
T = 1/f = ILVa /Vs (Vs Vs )
t IL /V V
f = Vs (Va Vs )/ ILVa
2 a s
(23)
I V (V V ) / fLV
(29)
s a s a
1-k = Vs/Va
k = 1- (Vs/Va) = (Va Vs)/ Va
I Vsk / fL
(30)
When the transistor is ON, the capacitor supplies the load current for t1= t2. The average capacitor current during time t1 is Ic =Ia and the peak to peak ripple voltage of the capacitor is
t1
t1
Vc Vc Vc (t 0)
t1
1 / C
0
t1
Ic dt 1 / C I a
0
1 / C[Ia
.t]0
Fig 5.1 Simulation of the proposed system and the PWM technique
Vc
Iat1 / C
(31)
Sub suiting equation (5)
t1 (Va Vs ) /Va f
from
Vc Ia (Va Vs )/Va fC
Vc Iak / fC
(32)
-
MATLAB/SIMULINK FOR THREE PHASE DRIVE SYSTEM
(WITH BOOST CONVERTER)
Fig 5.2 Input Voltage (Grid Voltage- Eg)
Fig 5.3 Output Voltage of the Controlled Rectifier
Fig 5.4 Output Voltage of the Boost Converter
Fig 5.5 Output Voltage of the VSI Inverter
Fig 5.6 Torque, Speed Characteristics
-
CONCLUSION
A single phase to three phase drive system composed of two parallel single phase rectifiers, a three phase inverter and an induction motor was included. This project combines two parallel rectifiers without use of transformer. The compared to the conventional topology, the proposed system permits to reduce the rectifier switch currents, the THD of the grid current with same switching frequency and increases fault tolerance characteristics. Thus the losses of the system may be lower than that of the conventional part.
-
REFERENCES
-
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Authors:
A.Ganapathy, ME., working as an Assistant professor in Chendhuran College of Engineering and Technology belongs to EEE Department.
R.Rajesh, ME., working as an Assistant professor in Chendhuran College of Engineering and Technology belongs to EEE Department.
S.Sathishkumar, ME., working as an Assistant professor in Chendhuran College of Engineering and Technology belongs to EEE Department.