Chapter 3

Inverters
Shahid Iqbal
Department of Electrical Engineering
COMSATS Institute of Information Technology
Email: shahidsidu@hotmail.com
 3.1 Introduction

• DC-to-AC converters are known as Inverters. The function of an inverter is to

change a DC input to a symmetric AC output voltage of desired magnitude
and frequency.
• The DC input could be a battery, fuel cell, solar cell, or other dc source.
• Inverter output voltage is normally controlled by Pulse-Width-Modulation
(PWM)
• Practical output voltage waveform of an inverter is non-sinusoidal and
contain certain harmonics.
• 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.
• Inverters are widely used in industrial applications, e.g. variable speed ac
motor drives, induction heating, standby power supplies, and
uninterruptible power supplies (UPS).
April 20, 2016 S. IQBAL 2
 3.2 Classification of Inverters
• Inverters can broadly be classified into Voltage Source (VSI) and Current
Source Inverters (CSI)
• Voltage source inverters (VSI) are supplied from ideal (or low impedance) DC
voltage sources. The DC source can be fixed or variable but very low source
impedance is assumed. With such a source, the output AC voltage is
determined completely by the DC source voltage and the states of the
switching devices in the inverter. Assuming that the load current does not
become discontinuous.
• Current source inverters (CSI) have a stiff dc current supply, usually through a
large inductor or a closed-loop controlled current source. The DC source
impedance of such inverters is very high. The output AC current is determined
completely from the magnitude of the source current and the states of the
switching devices used in the inverter. The output AC voltage is determined by
the load impedance and the inverter output current.
April 20, 2016 S. IQBAL 3
 3.2 Classification of Inverters

April 20, 2016 S. IQBAL 4

 3.3 Categories of Voltage Source Inverters

• Pulse-Width-Modulated inverters
– In these inverters, the input dc is constant. The magnitude and
frequency of the ac output voltage is controlled by PWM
• Square-wave inverters
– In these Inverters, the input dc is controlled in order to control the
magnitude of ac output voltage and the frequency is controlled by the
inverter
• Single-Phase inverter with voltage cancellation
– It combines the characteristics of the previous two inverters, in which
the input dc is constant and the output ac voltage is not controlled by
PWM.
– Voltage cancellation works only with single-phase and not with three-
phase.

April 20, 2016 S. IQBAL 5

 3.4 Basic Inverter Topologies

• Based on no. of phase, Inverter topologies can be

classified into
VP S1 D1
R L

– Single-Phase Inverter VP S2 D2
Half-bridge inverter
• Half-bridge
• Full-bridge P

• Push-pull
S1 D1 S3 D3
– Three-Phase Inverter VS R L

• Full-bridge +v_
S2 D2 S4 D4

Q
Full-bridge inverter
April 20, 2016 S. IQBAL 6
 3.4 Basic Inverter Topologies

1 3 5

S1 io
C

VS i
_ p + VS A B
+
vo
-
S2
4 6 2

Push-pull inverter
Three-phase full-bridge inverter

April 20, 2016 S. IQBAL 7

April 20, 2016 S. IQBAL 8
 3.5 Single-phase Half-wave Bridge Inverter with RL
Load

• A midpotential point is needed. (not practical)

• The midpotential point cab be replaced by connecting two identical
capacitors.
April 20, 2016 S. IQBAL 9
 3.5 Single-phase Half-wave Bridge Inverter with RL
Load

• If the capacitors are large, the voltage across them will be

constant.
• The output voltage is a square wave, and the output current
wave shape will depend on the type of load connected.
April 20, 2016 S. IQBAL 10
 3.5 Single-phase Half-wave Bridge Inverter with RL
Load
Gating Signals and Output Voltage waveforms

• For +ve voltage we turn On T1, and for –ve voltage we turn on
T2.
• The output voltage is a square wave with amplitude Vp=Vs/2
April 20, 2016 S. IQBAL 11
 3.5 Single-phase Half-wave Bridge Inverter with RL
Load
Load Current waveform

Io io
t
t t
-Io

io for pure resistive io for pure inductive load io for R L load

load

• Load current may not reverse at the same instant as the

load voltage. Therefore anti-parallel diodes are connected
to permit the load current to flow if necessary.
April 20, 2016 S. IQBAL 12
 3.5 Single-phase Half-wave Bridge Inverter with RL
Load
Voltage and Current Waveforms with R-L load
• T1 and T2 switched
in complementarily,
each at 50% duty
cycle.
• Turning off One
switch leads to turn
on the anti-parallel
diode of the second
switch. This is due to
the stored energy in
the inductance.
• The diode continues
to conduct until the
load current falls to
zero.
April 20, 2016 S. IQBAL 13
 3.5 Single-phase Half-wave Bridge Inverter with RL
Output Current at steady state Load
di VP
L  Ri  VP , i   I 0 at t  0 i I0 I0 I0
dt
The solution for this equation is
t t
VP   L T
i   I 0e  (1  e  ) ,  
 T
t0 2
R R -I0 -I0 -I0
At t = T/2, by symmetry -VP
consideration i = Io
T T
 VP  T

I 0   I 0e 2  (1  e 2  ) t
VP (1  e 2 )   V 
t
R i  e  P (1  e  ) for 0  t  12 T
R T R
T T /2 
  (1  e 2 )
VP (1  e 2 ) VP (1  e  )
 I0  T
 T /2 T
  
t t
R(1  e 2 ) R(1  e  ) VP (1  e 2 )   VP 
i e  (1  e  ) for 0  t  12 T
R T R

Thus, the expression for i is (1  e 2 )

April 20, 2016 S. IQBAL 14

 3.5 Single-phase Half-wave Bridge Inverter with RL
Output Voltage Equation Load
• The root mean square of the total output voltage can be found from
1/ 2
 2 T / 2  Vs  2  Vs
Vrms      dt    Vp
 T 0  2   2
• The instantaneous output voltage can be expressed in Fourier series as

a0
vo (t )    (an cos(nt )  bn sin( nt ))
2 n 1,3,5,.....
• Due to the quarter-wave symmetry along the x-axis, both ao and an are
zero. Therefore,
 2
1  Vs Vs  2V
bn  
  2
0

sin( nt )d (t )  

2 
sin( nt )d (t ) 


n
s


2Vs
 vo (t )  sin(nt )
n 1,3,5,....
n
• The fundamental r.m.s component 2Vs 2Vs
output voltage is V1rms    0.45Vs
2 
April 20, 2016 S. IQBAL 15
 3.6 Push Pull Inverter
• Topology and Operation

• Push-Pull inverter operates similar to the half-bridge with a square wave output
voltage.
April 20, 2016 S. IQBAL 16
 3.6 Push Pull Inverter

• Advantages and disadvantages

• Advantages
– presents only one switch voltage drop between the source
and the load, compared with the full bridge Inverter.
– The gate drive signals can be referenced to common ground,
which allow simple drive circuit.
• Disadvantages
– The transformer leakage inductance will have some stored
energy associated with it, which produces voltage overshoot
during a switch turn-off. Therefore a snubber circuit is
needed.
– the two halves of the transformer may not be perfectly
balanced. Therefore it needs very good coupling between the
two windings.
April 20, 2016 S. IQBAL 17
 3.7 Single-Phase Full-bridge inverter with RL load
• Topology

is

Vs

• The output voltage can be a square-wave or a quasi-square-wave.

• Switches T1and T2 are switched together while switches T3 and T4 are
switched together alternately to T1 and T2 in a complementary manner.
• The four feedback (freewheeling) diodes D1-D4 conduct when the
switches are turned off until load currents falls to zero.
April 20, 2016 S. IQBAL 18
 3.7.1 Single-Phase Full-bridge inverter with RL load
(Square-Wave Operation)
Square-wave Operation
Period I:
vg1=vg2>0 → T1, T2 on, current path:
VS+→T1→Load→T2→VS-
vo=VS
Period II:
vg3=vg4>0 → But io>0 → D3, D4 on,
energy stored in L is releasing to VS,
Vs current path:
Vs-→D4→Load→D3→Vs+
vo= -VS
Vs Period III:
vg3=vg4>0 → But io<0 → T3, T4 on,
current path:
Vs Vs+→T3→Load→T4→Vs-
Vo vo= -VS
Vs Period IV:
vg1=vg2>0 → But io<0 → D1, D2 on,
energy stored in L is releasing to Vs,
current path:
Vs-→D2→Load→D2→Vs+
vo=VS

April 20, 2016 S. IQBAL 19

 3.7.1 Single-Phase Full-bridge inverter with RL
load(Square-Wave Operation)
Output Current at steady state
• When T1 and T2 are on during 0< t < T/2, load current increases exponentially
according to
di
Vs  L  Ri
dt
• Solving this, R t
  t V
i Ae  L   s (1  e  )
R

• Assuming that at t = 0, the initial current is -Io, the solution for i is

t t
 Vs  L
i   I 0e 
 (1  e  ) ,... 
R R
• At t = T/2, by symmetry consideration i = Io T

T T Vs (1  e 2
)
I 0   I 0e

2

Vs 
(1  e 2 ) I0  T
R R 
(1  e 2 )
April 20, 2016 S. IQBAL 20
 3.7.1 Single-Phase Full-bridge inverter with RL
load(Square-Wave Operation)
• Thus for 0< t < T/2
Vs
vo
T
 io
Vs (1  e )  Vs 2 t t Io

i    t2
T
e (1 e )
R  R t1
(1  e 2 ) -Io T/2 T
t

-Vs
• Similarly when T3 and T4 are turned on
for T/2< t < T D1 D3 D1
D2 D4 D2
T

Vs (1  e )  Vs 2 t t
 Switch
i T
e  (1  e 
) ON T1 T2 T3 T4 T1 T2
R  R
(1  e )2
:
switches conduct current

April 20, 2016 S. IQBAL 21

 3.7.1 Single-Phase Full-bridge inverter with RL
load(Square-Wave Operation)
Output Voltage Analysis of Square wave
• The root mean square of the total output voltage can be found from
1/ 2
 2 T /2 2 
Vrms    Vs  dt   Vs
T 0 
• The instantaneous output voltage can be expressed in Fourier series as
a0 
vo (t )    (an cos(nt )  bn sin( nt ))
2 n1
• Due to the quarter-wave symmetry along the x-axis, both ao and an are zero.
Therefore,
 2
1  4V
bn  
 
0
 

Vs sin(nt )d (t )   Vs sin(nt )d (t ) 


n
s


4Vs
 vo (t )  sin(nt )
n 1,3,5,....
n

April 20, 2016 S. IQBAL 22

 3.7.1 Single-Phase Full-bridge inverter with RL load
(Square-Wave Operation)
The fundamental r.m.s component of output voltage is Vn, peak

4
VS n or plot
1 1.273 Vn, peak 1

1 4V 2 2VS 2
V1,rms   0 ( S sin t )2 d t    0.9VS
V1, peak n

    0.424
0.255

V 2
H , rms  V
n  2 , 3,...
2
n , rms V 2
rms V2
1, rms V 2
H , rms v(t)
1 3 5 7 9 11n  n
Fundamental 3rd harmonics
1

5th harmonics
VS

VH ,rms  Vrms
2
 V12,rms  Vs2  (0.9Vs ) 2
0  2 t
VH ,rms  VS 1  0.9  0.436VS 2
-VS

2
1  1  Vrms 
THD   Vn,rms 
2
Vrms  V1,rms  
2 2
 1
V1,rms n 1,3,... V1,rms  V1,rms 
 
April 20, 2016 S. IQBAL 23
 3.7.2 Single-Phase Full-bridge inverter with RL load
(Quasi-Square wave operation)
Quasi-Square wave Operation (PWM)
vo Vs

0
 2 3

-Vs
D  D  D  D

D is the pulse duty cycle of half-period and  is the waveform overlap angle

• Both output voltage and frequency can be controlled.

• Output voltage can be varied by changing the duty cycle D.
• The frequency can be varied by changing the switching frequency.
• The total rms value of the output voltage is given by
T /2
T /2
 
2 1
2
Vrms  (vo ) 2 dt  (Vs ) 2 dt  Vrms  DVs
T 0 
0

April 20, 2016 S. IQBAL 24

 3.7.2 Single-Phase Full-bridge inverter with RL load
(Quasi-Square wave operation)
• Switching Sequence of 1-ph full-bridge inverter with voltage
cancellation P
• The voltage cancellation (zero voltage) occurs when either
both top switches or both bottom switches are on.
S1 D1 S3 D3
• Two switching techniques can be used to achieve voltage
cancellation. VS R L
io
– fixed on-time (D) with variable overlap angle () (non-
+v_
PWM)
S2 D2 S4 D4
– variable on-time (D) (PWM).
• For periodic and symmetric output waveform, there are two
freewheeling intervals occur within one cycle, hence, there Q
are four switching patterns
(i) both freewheeling on the P side
(ii) both freewheeling on the Q side
t3 t4
(iii) t2-t3 freewheeling on the Q side during +ve half cycle
t1 t2 t5
and t4-t5 freewheeling on the P side during –ve half
D 
cycle
(iv) t2-t3 freewheeling on the P side during +ve half cycle and
t4-t5 freewheeling on the Q side during -ve-half cycle T/2

• (i) and (ii) can be obtained by PWM

• (iii) and (iv) can be obtained by non-PWM T

April 20, 2016 S. IQBAL 25

 3.7.2 Single-Phase Full-bridge inverter with RL load
(Quasi-Square wave operation)
Switching Pattern for PWM and Non PWM
vo vo
VS VS
0V 0 0V 0  2 3
 2 3
ON -VS
-VS
S1 D  D  D  D D  D  D  D
S2 (i) (iii)
S3
S4

D1
D2 (iv)
(ii)
D3
D4

April 20, 2016 S. IQBAL 26

April 20, 2016 S. IQBAL 27
April 20, 2016 S. IQBAL 28
 3.7.2 Single-Phase Full-bridge inverter with RL load
(Quasi-Square wave operation)
Output Current analysis at steady State • During t0 to t2 interval
di
L  Ri  VS
dt
VS • At t=0 , i= -I1 therefore
vo D T2
0
D T2 t t
 V 
t i   I1e 
 S (1  e  ) (1)
T/2 T/2 R
-VS
I2 • At t=t2=DT/2, i=I2 , therefore
I1 
DT

DT
io V
t0 t4 t6 I 2   I1e 2  S (1  e 2  ) (2)
0 R
t1 t2 t3 t5 t
-I1 • During t2-t3 , vo=0
di
-I2 L  Ri  0
dt
t

• If t2 =0 , i=I2 i  I 2e 

April 20, 2016 S. IQBAL 29

 3.7.2 Single-Phase Full-bridge inverter with RL load
(Quasi-Square wave operation)

• and substitute I2 by eq. (2)

DT DT t
 VS   (3)
 i  [ I1e 2
 (1  e 2
)]e 
R
• At t3, t = (1-D)T/2, i = I1, and substitutes into eq. (3)

DT DT t
 VS  
I1  [ I1e 2  (1  e 2  )]e 
R
(1 D )T T
 
VS e 2
e 2
 I1   T
(4)
R 
1 e 2

• Hence I2 can be found by substituting eq(4) into eq(2)

April 20, 2016 S. IQBAL 30

April 20, 2016 S. IQBAL 31
 3.7.2 Single-Phase Full-bridge inverter with RL load
Output Voltage Analysis (Quasi-Square wave operation)
Vs ....................... for0  t  D 
 
v(t )  0........................ forD  t    v(t) fundamental
 V .................... for  t    D 
 s 
VS
vn (t )  an cos(nt )  bn sin( nt )  +D
0 D 2
1 2 t
an   v(t ) cos nt dt
 0 -VS

  VS cos nt dt    VS cos nt dt 

1 D  D

 0  
 S sin nD  cos n sin nD
V
n
2VS
 an  (sin nD )........ for n  1,3,5...
n
Similarly
2VS
bn  (1  cos nD )............. for n  1,3,5.....
n
April 20, 2016 S. IQBAL 32
 3.7.2 Single-Phase Full-bridge inverter with RL load
Output Voltage Analysis (Quasi-Square wave operation)
The instantaneous output voltage v can be expressed using Fourier series is
  2VS
v   sin nD cos nt  (1  cos nD)sin nt 
n 1,3,5...  n 
As sin nD cos nt  cos nD sin nt  sin n( D  t )
and sin()   sin 
  2VS 
v    sin nt  sin n ( t  D ) 
n 1,3,5...  n 
A B A B
from sin A  sin B  2sin cos
2 2
 4VS nD D
v  sin cos n ( t  )
n 1,3,5... n 2 2
April 20, 2016 S. IQBAL 33
 3.7.2 Single-Phase Full-bridge inverter with RL load
Output Voltage Analysis (Quasi-Square wave operation)
• Thus the instantaneous output voltage

4VS nD D
vo  
n 1, 3, 5... n
sin
2
cos n(t 
2
)

• Expression for the fundamental component (n = 1),

4VS D D
v1  sin cos(t  )
 2 2
• The rms value of the fundamental component is
1

1  2  2 2 2VS D D
V1,rms    v1 dt   sin  0.9VS sin
 0   2 2
• The rms of the actual output waveform is
Vrms  DVS
• The rms value of the total harmonic components can be obtained from

2
V 2
V 2
V 2
 V V
2 2 8VS2 2 1
V rms 1, rms H , rms H , rms rms 1, rms VH ,rms  DV  2 sin 2 D
2


S

April 20, 2016 S. IQBAL 34

 3.8 Performance Parameters of Voltage Source
Inverters
• The output of practical inverters contains harmonics and the quality of an
inverter output is normally evaluated in terms of the following performance
parameters
– Harmonic factor of nth harmonic (HFn).
• It is a measure of individual harmonic contribution.
Vn,rms Vn, peak
HFn  
V1,rms V1, peak
– Total Harmonic distortion (THD)
• It is a measure of closeness in shape between a waveform and its
fundamental component (sinusoidal ).
2
VH ,rms V 
THD    rms   1
V1,rms  V1,rms 
– Lowest-order harmonic (LOH)
• It is that harmonic component whose frequency is closest to the
fundamental.
April 20, 2016 S. IQBAL 35
 3.8 Performance Parameters of Voltage Source
Inverters
• Voltage transfer ratio, TVV
– It is a measurement for voltage-mode (VM) control of VSI inverters.
V1,rms
TVV 
VS
• Voltage-to-current transfer ratio, TVI
– It is a measurement for current-mode (CM) control of VSI inverters.

I1,rms
TVI 
VS
• Output power (load power) P.
– In most applications the load is inductive and the output power due to the
fundamental current is generally useful power.
– The power due to harmonic currents is dissipated as heat and increases the load
temperature.
1 T 2
P  I rms R , I rms  
2
• Therefore the average power P i (t ) dt
T 0

April 20, 2016 S. IQBAL 36

 3.8 Performance Parameters of Voltage Source
Inverters
Vn
vn  Vn sin( n1t ), in  sin( n1t  n )
Zn
n1 L
n  tan 1 , Z n  R 2  (n1 L) 2
R
Vn  2 Vn ,rms

I n ,rms 
Vn ,rms
Zn
P  V I
n , rms n , rms
n 1, 2 , 3,...,
cosn

The VI ratings of the inverter is the apparent power (S) at the output
S = IrmsVrms
P
The power factor (PF) at the output is PF 
S
April 20, 2016 S. IQBAL 37
 3.9 Input Characteristics of Voltage Source Inverters
– Input current.
• iS(t) is a dc with ripple, IS,dc + iS,r(t). The average value IS,avg = IS,dc
t0 T
I S ,avg  T1  iS (t ) dt
t0
– Input current ripple
• It is defined as the difference of the instantaneous and the dc
(average) input current.
iS ,r (t )  iS (t )  I S ,avg

– Current reflection factor

• RII parameter (always <1) is load dependent and it provides the
measure of utilization of input source by the inverter. High
values indicate better utilization.
I S ,avg
RII 
I1,rms

April 20, 2016 S. IQBAL 38

 3.9 Input Characteristics of Voltage Source Inverters

– Harmonic of input current

• iS(t) can be represented by Fourier series.

iS   I Sn sin( n1t  n )
n 1, 2,3,...

• n is the phase angle of the n-th harmonic w.r.t. the

fundamental component of the output sinusoidal voltage.
• Harmonic content of the input current is the rms sum of all
harmonics.

I SH ,rms   Sn,rms
I 2

n 1, 2 , 3,...

 I S2,rms  I S21,rms

April 20, 2016 S. IQBAL 39

 3.9 Input Characteristics of Voltage Source Inverters
• Total harmonic distortion THDi
– THDi is the ratio of the harmonic content to the rms fundamental current.
I S2,rms  I S21,rms
THDi 
I S21,rms
• Peak-to-peak input current ripple IS
– the ripple current is responsible for resistive losses in the dc source and the
conductor wires.
I S  iS ,max  iS ,min
• Input power
– P = Vsavg Isavg
• Power efficiency 
– It is defined as the ratio of the output power in the fundamental harmonic to
the input power
V1,rms I1,rms cos 1 TVV
  
VS I S ,avg RII

April 20, 2016 S. IQBAL 40

 3.10 Harmonic Reduction
•Disadvantages of harmonic content:
•reduce the power factor
•interfere with the proper operation of the inverter and other equipment
by appearing as noise in control circuit.
•may excite mechanical resonance at detrimental frequencies or
generate acoustical noise in ac drive systems for electromechanical
loads, such as ac motors.
•Hence, harmonic content must be reduced.
•The Low-order harmonics such as the 3rd and the 5th are difficult to be
removed by a simple low-pass filter which is put at an inverter output. It is
because the fundamental component is also filtered partially. A complex
sharp-step low-pass filter can be used to avoid the problem but this
introduces additional cost and losses to the filter. The overall size of the ac-
dc system is also increased.
•Hence, an active harmonic reduction is more preferable.
April 20, 2016 S. IQBAL 41
 3.10 Harmonic Reduction
Harmonic elimination
•As discussed in the quasi-square wave generation
session, the 3rd harmonic is eliminated if D = 2/3. This
can be represented graphically as shown. v(t) 2
3 V3,peaksin(3
•The LOH is now the 5th harmonic, which has frequency VS t)
of 5 times of the fundamental’s. + +  2

•With fixed D = 2/3, the output amplitude is fixed which 0 6 - 5
6 t
is only suitable for fixed amplitude drive applications. -VS
•Note that all harmonics with multiples of 3 are also
eliminated.
•We should also check on the THD and TVV for this Vn, peak 4 nD
 sin
VS n 2
inverter performance at D = 2/3.
1.103 Harmonic profile
D
V1,rms  0.9VS sin  0.7794VS D = 2/3 (not in scale)
2
0.221
Vrms  DVS  0.8944VS 0.158
0 0 0.100

V 
2
V1,rms 1 3 5 7 9 11n
THD   rms   1  0.5629 TVV   0.7794
V  VS
 1,rms 

April 20, 2016 S. IQBAL 42

 3.10 Harmonic Reduction
Harmonic elimination by notching
•We can eliminate the 3rd harmonic yet can control the
amplitude by introducing a notch to the quasi-square D RW
wave.
• The Fourier expansion of TPW (thin pulse waveform center
at  ) is v p   4VS sin n sin n sin nt
n 2
 TPW
n 1,3,5...

• The Fourier expansion of RW is

 NW
v  4nV sin nD2  sin nt
S
+
n 1,3,5...
0   2
• The resulting NW (notched waveform) is v - vp. The amplitude of
the n-th harmonic of NW waveform is
v NW ,n 
4VS
n
sin nD2   sin n sin n2 
• The n-th harmonic can be eliminated by designing vNW,n = 0, resulting
sin nD2   sin n sin n
2
0
• The amplitude control parameter may be selected to be either the position angle  or the
notch-pulse width .

April 20, 2016 S. IQBAL 43

 3.10 Harmonic Reduction
Harmonic elimination by notching
nD n n
• Let fix  = /2, then sin 2  sin 2 sin 2  0
• Consider 0 < D < 1 and 0 <  < D, when n = 3, 7, 11, … , sin(n/2) = -1
sin nD2    sin n2  sin n2   v(t) Other set
V3,peaksin(3
nD n
2
 2
     ( D  n2 ) VS t)
 2
• when n = 5, 9, 13, … , sin(n/2) = 1 , hence
0 t
nD n n
sin 2
 sin 2
 sin 2
 2 -VS
nD
 n
 2    ( D  n4 ) 45º 75º 105º 135º
2 2
example
The elimination condition is
  ( D  23 ) for 2
 D 1 Vrms  D   VS  D  ( D  23 ) VS
3

All 3n-th harmonics are also eliminated. sin 2  sin 2 

VNW ,1,rms 
2 2VS

D 

This notching also reduces the fundamental 2

 Vrms 
amplitude: THD    1 TVV  2  2 sin D2  sin 2 
v NW ,1 
4VS D

sin

2

 sin 2    V1,rms 

April 20, 2016 S. IQBAL 44

 3.10 Harmonic Reduction
Harmonic elimination 3rd and 5th harmonic simultaneously
Method 2
Method 1
An alternative method uses two identical notch
pulses symmetrically placed around the center of the
main pulse, as shown graphically in figure.

sin 3 D2   sin 3 sin 32  0 v(t)

V3,peaksin(3t)

VS

sin 5 D2   sin 5 sin 52  0 

0
t
 
-VS
V5,peaksin(5t)
Sets of  and  must satisfy both
equations.

1 2
D

2 =  - 1

30º 54º 66º 114º 126º 150º

April 20, 2016 S. IQBAL 45

 3.10 Harmonic Reduction
Harmonic cancellation by stepping

•Adding of two or more inverter outputs io

+
in certain timing and switching sequence V
+ Full- + +
bridge vi2
S
leads to cancel certain harmonics. The 2 vo2
- - -
harmonics exist at individual inverter Vo1
output but disappear after output Full- + +
bridge vi1 vo1
adding (differ from the harmonic -
- -
elimination that the harmonics do not
exist at the inverter output due to
switching control). SW1
•A simple example is that by adding two SW2 t
equal- period square-wave outputs with 60º

phase angle difference of 60º produces a 120º 3rd HF

3rd harmonic free output, where the 3rd 0 150º
30º
harmonic exists at both individual
output.
April 20, 2016 S. IQBAL 46
 Three-Phase Inverters
is

Vs/2

Vs

Vs/2

• Three-phase inverters are normally used for high power applications.

• 3 single-phase half (or full) bridge inverters can be connected in parallel to form the
configuration of a three-phase inverter.
• The gating signals of single-phase inverters must be displaced by 1200 with respect to each
other to obtain three-phase balanced voltages.
• The load may be connected delta  or star Y.
• Two types of control signals can be applied to the transistors: 1800 conduction or 1200
conduction
April 20, 2016 S. IQBAL 47
 3-Phase Inverter with 1800 Conduction and -
Connected Load
• Each transistor conducts for 180. P

• Three transistors remain on at any instant

of time. +Vs/2 1 3 5
• There are six modes of operation in a
C
cycle and the duration of each mode is
600.
VS A B
• The switching sequence is T1T2T3 – T2T3T4–
T3T4T5 – T4T5T6 – T5T6T1 – T6T1T2 ....for the
positive A-B-C phase sequence and the -Vs/2
other way round for the negative (A-C-B) 4 6 2
phase sequence

April 20, 2016 S. IQBAL 48

 T1
 2
T2

/3
T3

2/3
T4

T5

T6

Vs

Vab
 2
Line to line
voltage -Vs

Vbc

Vca

April 20, 2016 S. IQBAL 49

 t1 t2 t3 t4 t5 t6 t7

1
+ 2
VS
0 vAN
1
- 2
1
VS
Phase
+ VS
2

0 vBN
voltages
1
- 2
VS
1
+ 2
VS
0 vCN
1
- 2
VS

+VS

0 vAB
Line to line
-VS voltage
+VS

0 vBC

The line-line output voltages

-VS are obtained by subtracting
+VS two square-wave phase
voltage waveforms which
0 vCA are 120 degree Displaced
from each other
-VS
April 20, 2016 S. IQBAL 50
 Sequential changes in power circuit configuration
S5 C

Vs S1 B
The voltage
A C S2
across each S6
phase is Vs/2 0°- 60°
S1
Vs A B
C S2 S6

Vs 60°- 120°
S1 A B S3

120°- 180° C
S2
Vs
S4 S3
C
S5 A B
180°- 240°
Vs S4 S3
S5
A B C
240°- 300°

A B
Vs
S4 S6

300°- 360°
April 20, 2016 S. IQBAL 51
 Waveform of Phase Current with -connected Load
v AB
+VS
iAB
I3
I4
I2

I1

-V S
+VS
v BC

iBC

-V S
+VS
v CA

iCA

-V S
April 20, 2016 S. IQBAL 52
t1 t2 t3 t4 t5 t6 t7
 Analysis of Output Voltage
Line to line voltage
Vs

-Vs

• Because of the 1200 phase shift between the waveforms, the triplen harmonics (of order which are
multiples of 3) of both will of the same phase and hence these cancel in the process of subtraction.
Consequently, the triplen order harmonic voltages are eliminated from the line – line voltage (this is true
for delta connection only)
• The line-line quasi-square output voltage waveform shown in the above figure has amplitude Vs and
duration  =1200. Fourier series representation of this waveform is given by


 n 

4Vs
vl  l  sin  cos nt
n 1,3,5,......
n  2 
2 3  1 1 1 
vl  l  Vs cos(t )  cos(5t )  cos(7t )  cos(11t )
  5 7 11 
April 20, 2016 S. IQBAL 53
 • The rms values of the fundamental and higher order output voltage are
 
Vrms ,l l  
4Vs
n
 n  1200 
sin 
 
4Vs
sin 
n  60 0

n 1,3,5,.....  2  n 1,3,5,..... n
 

 
2 3 6 3
 Vrms ,l l  Vs  Vs where sin(60) 
n 1,3,5,... n 2 n 1,3,5,...
n 2

6Vs
V1,l l  ;

6Vs
V5,l l  ;
5
6Vs
V7 ,l l  ;
7
6Vs
V 11,l l
11
.....and ....so...on
April 20, 2016 S. IQBAL 54
 3-Phase Inverter with 1800 Conduction and Y-Connected Load
• The switching patterns and timings of the P

inverter are the same, irrespective of

whether the load is  or Y connected. +Vs/2 1 3 5

• Therefore, with Y-connected load there C

C

are six different circuit configurations, N

VS
each one of 600 duration A
A B
-Vs/2
• The voltage across two phases is Vs/3 and B

the third phase is 2Vs/3 4 6 2

+ + + + + +

+ + +
+ 2
VS A + 2
VS B + 2
VS C
1 3 1 3 1 3
3
VS C A _ 3
VS A B _ 3
VS B C _
_ _ _

+ + +
+ 1
VS B C + 1
VS C A + 1
VS A B
2 3 2 3 2 3
3
VS_ B _ 3
VS_ C _ 3
VS_ A _

_ _ _ _ _ _
0-60 60-120 120-180 180-240 240-300 300-360
April 20, 2016 S. IQBAL 55
1 2 3 4 5 6
 Phase Voltages of 3-Phase Inverter with Y-Connected Load
t1 t2 t3 t4 t5 t6 t7

2
vAN
3
VS

1
3
VS

0° 60° 120° 180° 240° 300° 360°

1
3
VS

2
3
VS

2
vB N
3
VS

1
3
VS

1
3
VS

2
3
VS

2
vC N
3
VS

1
3
VS

1
3
VS

April 20, 2016 2

3
VS S. IQBAL 56
 Phase Current Waveforms

t1 t2 t3 t4 t5 t6 t7
vAN

I4 iAN
I3

I2

I1

April 20, 2016 S. IQBAL 57

 Output Voltage analysis
Line to neutral voltage
Vs
Vs

Vs
Vs

• The line-to-neutral voltage (phase voltage) is given by


4Vs  n 
vl  N   n sin  2  cos nt
n 1,3,5,......
  4 Vs n1800 4 Vs n600 
   n 3 sin 2  n 3 sin 2  cos nt
n 1,3,5,..... 
 
2  1 1 1 
vl  N  Vs  cos(t )  cos(5t )  cos(7t )  cos(11t ) 
  5 7 11 
April 20, 2016 S. IQBAL 58
 •The rms value of the fundamental and higher order terms of the phase
voltages are

2Vs
V1,l  N  ;

2Vs
V5,l  N  ;
5
2Vs
V7 ,l  N  ;
7
2Vs
V 11,l  N
11
.....and ....so...on

April 20, 2016 S. IQBAL 59

 3-Phase Inverter with 1200 Conduction with Y-Connected Load

• In this type of control,

each transistor conducts
for 1200.
• Only two transistors
remain on at any instant.
• The conduction
sequence of transistors is
12, 23, 34, 45, 56, 61.

April 20, 2016 S. IQBAL 60

 Vs/2
Vs/2
Vs
Vs

Vs/2
Vs/2

Vs/2

April 20, 2016 S. IQBAL 61

 Output Voltage analysis
Line to line voltage

Vs
Vs

Vs
Vs

• The line-to-line voltage is given by


4Vs  n 
vl l   n  2  cos nt
sin
n 1,3,5,......
 4 Vs n1800 4 Vs n600 
   n 2 sin 2  n 2 sin 2  cos nt
n 1,3,5,.....  
3  1 1 1 
vl l  Vs cos(t )  cos(5t )  cos(7t )  cos(11t ) 
  5 7 11 
April 20, 2016 S. IQBAL 62
 • The rms value of the fundamental and higher order terms

3Vs
V1,l l  ;
 2
3Vs
V5,l l  ;
5 2
3Vs
V7 ,l l  ;
7 2
3Vs
V 11,l l
11 2
.....and ....so...on
April 20, 2016 S. IQBAL 63
 Line-to-neutral voltage

Vs/2

Vs/2

The line-to-neutral voltage is given by


4Vs  n 
vl  N  
n 1, 3, 5,...... 2n
sin 
 2 
 cos nt

3  1 1 1 
vl  N  Vs cos(t )  cos(5t )  cos(7t )  cos(11t )
  5 7 11 

April 20, 2016 S. IQBAL 64