A presentation on inverter by manoj

Submitted to :- submitted by :-
(Coordinator) manoj Kumar swami
Ujjwal kalla
A presentation on
INVERTER

Introduction
 Inverter is a device which convert a DC input supply
voltage into symmetric AC voltage of desired magnitude
and frequency at the output side. It is also know as DC-AC
converter.
 Ideal and practical inverter have sinusoidal and no-
sinusoidal waveforms at output respectively.
 If the input dc is a voltage source, the inverter is called a
Voltage Source Inverter (VSI). One can similarly think of a
Current Source Inverter (CSI), where the input to the circuit
is a current source. The VSI circuit has direct control over
‘output (ac) voltage’ whereas the CSI directly controls
‘output (ac) current.

Applications of inverter
 For low and medium power applications, square-
wave or quasi-square wave voltages may be
acceptable
 For high –power applications ,low distorted
sinusoidal waveforms are required.
 Using high speed power semiconductor devices
the harmonic contents at output can be reduced
by PWM techniques.
 Industrial applications:- variable ac motor,
induction heating , standby power supply ,UPS
(uninterrupted power supply)
 Inputs are (battery, fuel cell, solar cell, other dc

Classifications of inverter
Inverter can be mainly classified into two types-
1) Single-phase inverter
2) Three-phase inverter
 Turn-ON and turn-OFF controlling devices are-
a. Bipolar junction transistor[BJTs]
b. Metal oxide semiconductor field-effect
transistor[MOSFETs]
c. Insulated-gate bipolar transistor[IGBTs]
d. Gate turn-OFF thyristor[GTOs]

Classifications of inverter (Cant’d)
 Voltage fed inverter[for constant input voltage]
 Current fed inverter[for constant input current]
 Variable DC-link inverter[for controllable input
voltage]
 Resonant-pulse inverter :- If the output voltage or
current of inverter is forced to pass through zero
by creating an LC resonant circuit.

Single phase inverter:-
Principle of operation:-The inverter
circuit consists of two choppers. when
only transistor Q1 is turned ON for a time
T0/2 then instantaneous voltage across
the load is (V0=Vs/2). If the transistor Q2
only is turned ON for a time T0/2,-VS/2
appears across the load. This type of
inverter is called Half bridge inverter.
Root-mean square (rms) output voltage
24
2
2
1
2
0
2
0
0
0
S
T
S V
dt
V
T
V 










 

Single phase inverter:-
(Cant’d)

Single phase inverter (Cant’d)
Instantaneous output voltage in the form of Fourier
series is-
 Due to quarter wave symmetry along the X-axis, both
a0 and an are zero (or even Harmonic voltages are
absent ). then bn is
 Output voltage-
 v0 = 0 for n=2,4…..
    



1
0
0 sincos
2 n
nn tnbtna
a
V 
2
4
n
V
b S
n 
 





 

 ,...5,3,1
0 sin
n
S
tn
n
V
V 


 For an inductive load ,the load current can not change
immediately with the output voltage. diode D1 and D2 are
known as feedback diodes .Transistor can be replaced by
any switching device.
 There must be a minimum delay time between the
outgoing device and triggering of the next incoming device,
otherwise there would occur a short-circuit between
devices.
 Maximum conduction time of a device would be .
2
0






 don t
T
t

 For an RL load instantaneous load current i0 is
 Note-in most applications the output power due to the
fundamental current is generally the useful power ,and
the power due to harmonic current is dissipated as heat
and increase the load temperature.
   
 





....5,3,1
220 sin
2
n
n
S
tn
LnRn
V
i 


 EXAMPALE -1. A single phase half bridge inverter has a
resistive load of R=2.4 and the DC input voltage is
48V,Determine-
A. Rms value of output voltage at the fundamental frequency V01
B. output power
C. Average and peak current of each transisto
D. The peak reverse blocking voltage VBR.
E. The THD
F. The DF and
G. The HF and DF of the LOH.
Solution:- Vs=48V, R=2.4 .
A. V01 =
B.
V6.214845.0 
2
VS
0 V V24
2
48

SV45.0
2
2VS



The output power is
C. The peak transistor current is Ip=24/2.4=10A. because each
transistor conducts for a 50% duty cycle ,the average
current of each transistor is IQ=0.5*10=5A.
D. The peak reverse blocking voltage VBR=2*24=48V.
E. V01 =
And the RMS harmonic voltage
So THD =
F.
G. The LOH is the third ,V03 = V01/3
HF3=V03/V01 = 1/3=33.33% and
DF3=(V03/32)V01 =1/27=3.704%.
Because V03/V01 =33.33% which is greater than 3% ,LOH=V03.
R
V
S
2
0
V

WV 240
4.2
24 2
0 
SV45.0
  S
n
onh VvvVV 2176.001
2
0
2
2
1
...7,5,3
2






 


 
 
%34.48
45.0
2176.0
01

S
Sh
V
V
V
V
%382.5
45.0
024.0
.........
75345.0
11
2
2
07
2
2
05
2
2
03
2
1
...3,2
2
2
01









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




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




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



















 

 S
S
Sn
on
V
VVVV
Vn
V
v
DF

Single phase Bridge inverter
 It consists of four choppers. Full bridge converter is also
basic circuit to convert dc to ac. An ac output is synthesized
from a dc input by closing and opening switches in an
appropriate sequence. There are also four different states
depending on which switches are closed.
 When transistors Q1 and Q2 are turned on simultaneously,
the input voltage appears across the load. If transistor and
are turned on at the same time, the voltage across the is
reversed and is –Vs.
 Transistor Q1 and Q2 acts as switches S1 and S2. ,
respectively.

Single phase Bridge inverter(cont’d)
 The fundamental RMS output voltage obtained
from
 RMS output voltage is
 Fourier series of output voltage
= 0 for
n=2,4….
 For an RL load instantaneous load current i0 is   
 





....5,3,1
220 sin
4
n
n
S
tn
LnRn
V
i 

S
S
rms V
V
V 90.0
2
4
)(01 

SSrms VdVV  


 0
2
)(0
2
 





 

 ,...5,3,1
0 sin
4
n
S
tn
n
V
V 


 The output load voltage alternates between +Vs when Q1
and Q2 are on and -Vs when Q3 and Q4 are on, irrespective
of the direction of current flow. It is assumed that the load
current does not become discontinuous at any time. In the
following analysis we assume that the load current does not
become discontinuous at any time, same as for the half-
bridge circuit.
 Bridge inverters are preferred over other arrangements in
higher power ratings.
 With the same dc input voltage, output voltage is twice that
of the half-bridge inverter.

Three phase inverter :-
(VSI)
 Three-phase inverters are used for variable-frequency
drive applications and for high power applications
such as HVDC power transmission. A basic three-
phase inverter consists of three single-phase
inverter switches each connected to one of
the three load terminals. Gating signal of 1-phase
inverter should be delayed by 1200 with respect to each
other to obtain 3-phase balanced voltages.
 Two types of control signals can be applied to the
transistors:
A. 1800 conduction and
B. 1200 conduction.

 EXAMPLE -2. Q.2 A single phase full bridge inverter has a
resistive load of R=2.4 and the DC input voltage is
48V,Determine-
A. Rms value of output voltage at the fundamental frequency V01
B. output power
C. Average and peak current of each transistor
D. The peak reverse blocking voltage VBR.
E. The THD
F. The DF and
G. The HF and DF of the LOH.
Solution:- Vs=48V, R=2.4 .
A. V1 =
V1 =
SV90.0
2
4VS


V2.434890.0
2
4VS



B.
The output power is
C. The peak transistor current is Ip=48/2.4=20A. because each
transistor conducts for a 50% duty cycle ,the average
current of each transistor is IQ=0.5*20=10A.
D. The peak reverse blocking voltage VBR=48V.
E. V1 =
And the RMS harmonic voltage
So THD =
F.
G. The LOH is the third ,V3 = V1/3
HF3=V3/V1 = 1/3=33.33% and
DF3=(V3/32)V1 =1/27=3.704%..
,VS0 V VV 480 
,
V S
2
0
R
V  WV 960
4.2
482
0 
SV90.0
  S
n
onh VvvVV 4352.001
2
0
2
2
1
...7,5,3
2






 


%382.5
90.0
048.0
.........
75390.0
11
2
2
07
2
2
05
2
2
03
2
1
...3,2
2
2
01









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




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




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



















 

 S
S
Sn
on
V
VVVV
Vn
V
v
DF
 
 
%34..48
90.0
4352.0
01

S
Sh
V
V
V
V

1800 Conduction
 In this type of conduction each transistor conducts for
1800.3-transistor remain on at any instant of time.
 When transistor Q1 is switched ON ,terminal a is connected
to +ve terminal of supply and if Q4 is switched ON then
terminal a is connected to –ve terminal of supply.
 There are 6-mode of operation in a cycle and the duration
of each mode is 600.
 The load may be connected in star or delta form.
 Inverter switches of any leg can not be switched ON and
OFF simultaneously.

1800 Conduction(cont’d)
Circuit diagram
Figure: Three Phase Inverter circuit diagram

Circuit diagram and waveforms

 Transistor Q1 ,Q6 act as the switching
devices S1,S6 respectively.
 For states 1to 6 output voltage are non-zero
and states7 to 8 produce zero line
voltage.line currents freewheel through
freewheeling diodes.
 A modulating technique is used to select the
valid states only.

3-mode of operation of inverter 1800
Conduction(cont’d)
Mode-1.For transistors Q1,Q5, and Q6
conducts.
Mode-2- For transistors Q1,Q2, and Q6
conducts.
Mode-3- For transistors Q1,Q2, and Q3
conducts.
3
0   t
2
3
2
RR
RReq 
R
V
R
V
i S
eq
S
3
2
1 
32
1 S
cn
VRi
VVan 
3
2
1
SV
RiVbn


  t
3
2
2
3
2
RR
RReq 
R
V
R
V
i S
eq
S
3
2
3 
32
3 SVRi
VbnVan 
3
2
3
SV
RiVcn


  t
3
2
2
3
2
RR
RReq 
R
V
R
V
i S
eq
S
3
2
3 
32
3 SVRi
VbnVan 
3
2
3
SV
RiVcn



 In the form of Fourier series :-there is a0 and an are absent
so
 Instantaneous L-L voltages are given as:-
 Triple harmonics (n=1,3,5…) are absent in L-L voltage.
 RMS L-L voltage is













3
sin
2
sin
4 

nn
n
V
b S
n
















....5,3,1 6
sin
3
sin
4
n
S
ab t
n
n
V
v



 















....5,3,1 2
sin
3
sin
4
n
S
bc t
n
n
V
v




















....5,3,1 6
7
sin
3
sin
4
n
S
ca t
n
n
V
v




  SSSL VVtdVV 8165.0
3
2
2
2
2
1
3
2
0
2









  



 Fundamental line voltage is
 RMS L-N voltage is
 The diode across the transistors have no function with R-load .
 For an inductive load ,current of each arm would be delayed to
its voltage.
 The transistors must be continuously gated ,because the
conduction time of transistor and diode depends on load power
factor.
 
S
S
L V
V
V 779.0
2
60sin4 0
1 

S
L
P V
V
V 4714.0
3


1200 Conduction
 In this conduction
,each transistor
conducts for 1200.
 Two transistor remain
ON at any instant of
time.
 Conduction sequence
of transistors is as
61,12,23,34,45,56,61.

3-mode of operation of inverter 1200
Conduction(cont’d)
Mode-1.For transistors 1 and 6 conducts.
Mode-2- For transistors 1 and 2 conducts.
Mode-3- For transistors 2 and 3
conducts.
3
0   t
  t
3
2
  t
3
2
0,
2
,
2
 Vcn
V
Vbn
V
Van SS
0,
2
,
2
 Vcn
V
Vbn
V
Van SS
2
,
2
,0 SS V
Vcn
V
VbnVan 

 L-N ,phase voltages are as
 Instantaneous L-L voltages are given as:-
 There is delay of 300 between turning ON and OFF Q1
and turning of Q4.
 At any time ,two terminals are connected to DC supply
and third one remains open.
 Potential of open terminal depends on load
characteristics.
















....5,3,1 6
sin
3
sin
2
n
S
aN t
n
n
V
v




















....5,3,1 2
sin
3
sin
2
n
S
bN t
n
n
V
v



 















....5,3,1 6
7
sin
3
sin
2
n
S
cN t
n
n
V
v




VanVab 3

Why 1800 conduction is preferred
compared to 1200 ?
 In 1200 conduction ,one transistor conducts for 1200
,the transistors are less utilized as compared with that
of 1800 conduction for the same load condition.

 EXAMPLE-3. A three-phase bridge inverter delivers power to a
resistive load from a 450V dc source. For a star-connected load of
10 per phase,determine for both (a)1800 mode and (b) 1200
mode,
A. RMS value of load current
B. RMS value of thyristor current
C. Load power .
Solution :- For a resistive load ,the waveform of load current is same
as that of the applied voltage .
a) 1800 Mode :-
 Rms value of per-phase load current is
 Rms value of thyristor current is
 Power delivered to load is
2
1
222
0
3333
2
33
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 R
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I SSS
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AI r 213.21450
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AIT 15225
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2
0
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

b) 1200 Mode :-
 Rms value of per-phase load current is
 Rms value of thyristor current is
 Power delivered to load is
2
1
2
0
3
2
2
1
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
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



  KWRI r 125.10105.33733
2
0
2


Voltage control of Inverter
 VCI For 1-phase inverter
1. To control of output voltage of inverter is often
necessary .
2. To cope with the variation of DC input voltage.
3. To regulate voltage of inverter .
4. To satisfy constant voltage and frequency control
requirement.

Control techniques are
1. Single –pulse width modulation.
2. Multiple pulse width modulation.
3. Sinusoidal pulse width modulation.
4. Modified sinusoidal PWM.

Single –pulse width modulation
Single-PWM
 In single PWM control ,there is only one pulse per half
cycle and width of pulse is varied to control inverter
output voltage .
 The gating signals are generated by comparing
rectangular reference signal of amplitude Ar with a
triangular carrier wave of amplitude Ac.
 AM index :
 The ratio of Ar to Ac is the control variable ,called AM
index.
 M=Ar/Ac

Single –pulse width
modulation(cont’d)
 Fourier series output voltage is
 Time and angle of intersection is
 Pulse width d (or pulse angle )as
 Where TS = T/2
 






SS VtdVV 








 


2
1
2
2
2
0
2
2
   









....5,3,1
0 sin
2
sin
4
n
S
tn
n
n
V
tv 


 
2
11
1
ST
Mt 


 
2
12
2
ST
Mt 


SMTttd  12



Multiple pulse width modulation
(M-PWM)
 The harmonic contents can be reduced by using several
pules in each half cycle of output voltage .
 The frequency of reference signal sets the output frequency
f0, and carrier frequency determine the number of pulses
per half cycle p.
 The modulation index control the output voltage.
 Number of pulses per half-cycle is
 P=fc/2f0=mf/2

M-PWM(cont’d)
 Instantaneous output voltage is
V0= VS(g1-g4).
 The variation of M from 0 to 1 varies the pulse width d
from 0 to T/2p(0 to /p) and the rms output voltage v0
from 0 to vs.
 a0 , an and even harmonics are absent .
 so bn =
 






p
VtdV
p
V SS 








 


2
1
2
2
2
0
2
2
       


















 











m
m
m
m
tdtnintdtnin
2
2
2
ss

M-PWM(cont’d)
 The coefficient Bn is
for m=1,3,….2p
for m=2,4….2p
 Pulse width d (or pulse angle )as
























  4
sin
4
3
sin
4
sin
42
1






mm
p
m
s
n nn
n
n
V
B
 
2
Sm
m
T
Mmt 


 
2
11
Sm
m
T
Mmt 


Smm MTttd  1



Sinusoidal pulse width modulation
(Sinusoidal -PWM)
 In this type of modulation the width of each pulse is varied in
proportion to the amplitude of a sine wave evaluated at the
center of the same pulse.
 The DF and LOH are reduced significantly.
 Comparing the bidirectional carrier signal vcr with two
sinusoidal reference signals vr and -vr produces gating signals
g1 and g4 ,respectively .
 The output voltage is
V0=Vs(g1-g4)
 g1 and g4 can not be released at the same time.

Sinusoidal –PWM(cont’d)
 The number of pulses per half cycle depends on the
carrier frequency. Gating signals can also be
generated by using unidirectional triangular carrier
wave.
 The coefficient Bn is
for m=1,3,….2p
for m=2,4….2p
 Pulse width dm (or pulse angle )as
212
1
0 





 
p
m
m
SVV


























  4
sin
4
3
sin
4
sin
42
1
m
m
m
m
m
p
m
s
n nn
n
n
V
B






2
S
x
m
m
T
mtt 















2
sin
2
1 s
x
s
mT
tM
T
t














2
sin
2 s
x
s
mT
tM
T
t

Smm
m
m MTttd  1



Sinusoidal –PWM(cont’d)
 The output voltage of an inverter contain harmonics
.the PWM pushes the harmonics into a high –
frequency range around the switching frequency fc
and its multiples ,that is, around harmonics mf,2mf,3mf
and so on.
 The frequency at which voltage harmonics occur can
fn=
 peak fundamental output voltage for PWM and
SPWM control is
Vm1 = dvs for
 For d=1 ,Vm1(max) = Vs .
 The operation beyond d=1.0 is called over-
modulation.
 Over-modulation is normally avoided in UPSs
  cf fkjm  0.10  d

Modified
(Sinusoidal -PWM)
 In this PWM the widths of pulses nearer the peak of
the sine wave do not change significantly with
variation of modulation index.
 In this PWM the carrier wave is applied during ther
first and last 600 intervals per half cycle.
 The fundamental component is increased and its
harmonic characteristics are improved.reduces
switching losses .

Modified Sinusoidal –PWM(cont’d)
for m=1,2,3,….p
for m=1,3,….p
for m=2,4….p
 Intersection during the last 600 interval is
for m =p,p+1…..,2p-
1
 Pulse width dm (or pulse angle )as
 Instantaneous output voltage is
V0= VS(g1-g4).
2
S
x
m
m
T
mtt 















2
sin
2
1 s
x
s
mT
tM
T
t














2
sin
2 s
x
s
mT
tM
T
t

mp
m
m t
T
t 

  2
1
1
2

Smm
m
m MTttd  1



Modified Sinusoidal –PWM(cont’d)

Voltage control of 3-phase inverter
 The voltage control techniques are
1. Sinusoidal PWM
2. Third –harmonic PWM
3. 600 PWM
4. Space vector modulation.

Harmonic reduction
 The output voltage control of inverter requires varying
both the number of pulses per half –cycle and the pulse
widths that are generated by modulating techniques.the
output voltage contain even harmonics over a
frequency spectrum.
 Some harmonic reduction techniques are as:-
1. Phase displacement.
2. Bipolar output voltage notches.
3. Unipolar output voltage notches.
4. Transformer connections .

Phase displacement
 In this technique the nth harmonic can be
eliminated by a proper choice of displacement
angle  if
Cos(n) = 0
Or  = 900/n
 And third harmonic is eliminated if  =900/3=300.

Bipolar output voltage notches
 A pair of unwanted harmonics at the output of
single –phase inverter can be eliminated by
introducing a pair of symmetrically placed bipolar
voltage notches .
 the Fourier series output voltage is

 Bn =
 



....5,3,1
0 sin
n
n tnBV 
  
n
nn
n
Vs 21 cos2cos214 



Bipolar output voltage notches
wave-form

Unipolar output voltage notches
 Similarly to bipolar
notches symmetrical
unipolar notches can
also be introduced.
∞
vo (t ) = ∑ Bn sin(nωt)
n=1,3,5,.

Transformer connections
 The output of two or more inverters may be
connected in series through a transformer to reduce
certain unwanted harmonics.
 The second inverter is phase shifted by 600.
 So effective output of inverter is reduced by
13.4%.

Transformer connections
circuit diagram and wave-form

Transformer connections(cont’d)
vo1 (t ) = A1 sin ω t + A3 sin 3ω t + A5 sin 5ωt +
 Total output voltage is :-
....
3
sin
3
sin
3
sin 53102 























 tAtAtAV


















 ....
6
sin
6
sin3 5102010



 tAtAVVV

Current source inverter (CSI)
 In the CSI, the current is nearly constant. The
voltage changes here, as the load is changed. In
an Induction motor, the developed torque
changes with the change in the load torque, the
speed being constant, with no
acceleration/deceleration. The input current in the
motor also changes, with the input voltage being
constant. So, the CSI, where current, but not the
voltage, is the main point of interest, is used to
drive such motors, with the load torque changing

Single-phase Current Source Inverter
 The type of operation is termed as Auto-Sequential Commutated
Inverter (ASCI).
 A constant current source is assumed here, which may be
realized by using an inductance of suitable value, which must be
high, in series with the current limited dc voltage source.
 The thyristor pairs, Th1 & Th3, and Th2 & Th4, are alternatively
turned ON to obtain a nearly square wave current waveform.
 Two commutating capacitors − C1 in the upper half, and C2 in
the lower half, are used. Four diodes, D1–D4 are connected in
series with each thyristor to prevent the commutating capacitors
from discharging into the load.
 The output frequency of the inverter is controlled in the usual
way, i.e., by varying the half time period, (T/2).

Single-phase CSI
Circuit diagram

Operation of Single-phase CSI
 There are Two mode of operation:-
 Mode1:-The Starting from the instant, t = 0− , the thyristor pair,
Th2 & Th4, is conducting (ON), and the current (I) flows through
the path, Th2, D2, load (L), D4, Th4, and source, I. The
commutating capacitors are initially charged equally with the
polarity as given, i.e., vC1 = vC 2 = −VC 0 .
 At time, t = 0, thyristor pair, Th1 & Th3, is triggered by pulses at
the gates. The conducting thyristor pair, Th2 & Th4, is turned
OFF by application of reverse capacitor voltages.
 The voltage, vD1
 is obtained by going through the closed path,
 abcda
 as vD1 + Vco − (1 /(C / 2))⋅ ∫ I ⋅ dt = 0
 It may be noted the voltage across load inductance, L is zero.

Operation of Single-phase CSI(cont’d)
 The value of vC is
 vC1
= vC 2= vC= −Vco + (2 / C) ⋅ ∫ I ⋅ dt , which, if
computed at t = t1 ,
vC1
= vC2= vC (t1 ) = −Vco + (( 2 ⋅ I ⋅ t1 ) / C) = −Vco +
((2 ⋅ I ) / C)⋅ (C /(2 ⋅ I ))⋅VC 0= 0

 Mode2:- Diodes, D2 & D4, are already conducting, but
at t = t1 , diodes, D1 & D3, get forward biased, and start
conducting. Thus, at the end of time t1, all four diodes,
D1–D4 conduct. As a result, the commutating
capacitors now get connected in parallel with the load
(L).
 natural frequency:-
f0= 1/((2 ⋅π ) ⋅(L ⋅ C) ), ω0 = (2 ⋅π ) ⋅ f0= 1/(L ⋅ C)
 time period:-
T = 1/ f0 = (2 ⋅π ) /ω0 = (2 ⋅π ) ⋅ (L ⋅ C)

 The voltage across capacitor is
vC = vL = L ⋅ (di0 / dt) = (( 2 ⋅ I ) ⋅ (ω0 ⋅ L))⋅ sin (ω0 ⋅ t)
 The total commutation interval is,
tc = t1 + t2 = (1+ (π / 2))/ω0 = (1 + (π / 2))⋅ (L ⋅ C) .
 The procedure remains nearly same, if the load
consists of resistance, R only. The procedure in
mode I, is same, but in mode II, the load
resistance, R is connected in parallel with the two
commutating capacitors. The direction of the
current, I remains same, a part of which flows in
the two capacitors, charging them in the reverse
direction

Three-phase Current Source Inverter
 In this circuit, six thyristors, two in each of three
arms, are used, as in a three-phase VSI. Also, six
diodes, each one in series with the respective
thyristor, are needed here, as used for single-
phase CSI. Six capacitors, three each in two (top
and bottom) halves, are used for commutation. It
may be noted that six capacitors are equal, i.e. C1
= C2 = = C6 = C .
 The numbering scheme for the thyristors and
diodes are same, as used in a three-phase VSI.

circuit diagram

Wave-form

Three-phase CSI(cont’d)
 There are two mode of operation :-
 Mode:-The commutation process starts, when the thyristor, Th3 in the
top half, is triggered, i.e. pulse is fed at its gate. Immediately after this,
the conducting thyristor, Th 1 turns off by the application of reverse
voltage of the equivalent capacitor.
 It may be noted the equivalent capacitor is the parallel combination of
the capacitor, C1 and the other part, being the series combination of the
capacitors, C3 & C5 ( C′ = C / 2 ).
 value is Ceq = C / 3 ,
 When the voltage across the capacitor, C1 (and also the other two)
decreases to zero, the mode I ends.

Three-phase CSI(cont’d)
 After the end of mode I, the voltage across the diode, D3
goes positive, as the voltage across the equivalent capacitor
goes negative, assuming that initially (start of mode I) the
voltage was positive.
 when the mode II ends. The diode, D1 turns off, as the
current goes to zero. So, at the end of mode II, the thyristor,
Th3 & the diode, D3 conduct.
 This is needed to turn off the outgoing (conducting)
thyristor, Th3, when the incoming thyristor, Th5 is triggered.
The complete commutation process as described will be
repeated. The diodes in the circuit prevent the voltage
across the capacitors discharging through the load.

Variable DC Link Inverter
 Varying the modulation index (or pulse width) and
maintaining the dc input voltage constant has
shown that a range of harmonics would be
present on the output voltage.
 The pulse width can be fixed to eliminate or
reduce certain harmonics and the output voltage
can be controlled by varying the level of the dc
input voltage.

Variable DC Link Inverter (Cont’d)
Circuit diagram

Advantages of CSI
 The circuit for CSI, using only converter grade thyristor,
which should have reverse blocking capability, and also
able to withstand high voltage spikes during commutation,
is simple.
 An output short circuit or simultaneous conduction in an
inverter arm is controlled by the ‘controlled current source’
used here, i.e., a current limited voltage source in series
with a large inductance.
 The converter-inverter combined configuration has inherent
four-quadrant operation capability without any extra power
component.

disadvantages of CSI
 A minimum load at the output is required, and the
commutation capability is dependant upon load
current. This limits the operating frequency, and also
puts a limitation on its use for UPS systems.
 At light loads, and high frequency, these inverters have
sluggish performance and stability problems.

References
 M. Kojima, K. Hirabayashi, Y. Kawabata, E. C. Ejiogu and
T. Kawabata, “Novel Vector Control System Using
Deadbeat Controlled PWM Inverter With Output LC
Filter”, IEEE Transactions on Industrial Applications, Vol.
40, No. 1, January/February 2004, pp. 132-169.
 J. Nazarzadeh, M. Razzaghi, and K. Y. Nikravesh,
“Harmonic Elimination in Pulse-Width Modulated
Inverters using Piecewise Constant Orthogonal Functions”,
Electric Power Systems Research, Vol. 40,1997, pp. 45-49.
 Coker, J.O. (2004). Solar energy and its applications in
Nigeria. Global Journal of Pure and Applied Sciences 10:
223 – 225. Duncan, T. (1997). Electronics for today and
tomorrow 2nd Edition Hodder Education

A presentation on inverter by manoj

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