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A variable output voltage may be
obtained in two ways :
1. By varying the input dc voltage
2. By adjusting the gain of the inverter,
usually done by PWM control.
dc to ac converter (Inverter)
It is a variable voltage ac
source obtained from a
fixed voltage dc source.
Theoretically the output is a
SINE wave.
In real life the output is non-
sinusoidal and contains
harmonics.

Single phase half bridge inverter
Vi
t
VS/2
VO
t
VS/2
T/2
T
R
Chopper
VO
+
_
Chopper
2
V
S
2
V
S
T/2
VS/2
























 
 dt
dt
T
2
T
2
s
2
T
0
2
s
s
m
r
2
V
2
V
T
1
V
The output rms
voltage is given by
2
V
V s
o 

VO
t
VS/2
T/2
T
T/2
VS/2
Here we see that the output
voltage wave is not a pure Sine
wave.
It can be represented as sum of n numbers of
sine waves by Fourier’s series.
t
n
n
n


sin
V
2
v
5
,
3
,
1
s
o 


 ,..
4
,
2
for
,
0
vo 
 n

Vo
t
t
n
n
n


sin
V
2
v
5
,
3
,
1
s
o 




S
2V

3
2VS
Fundamental
component
Third
harmonics
component

VO
t
VS/2
T/2
T
T/2
VS/2
t
n
n
n


sin
V
2
v
5
,
3
,
1
s
o 



For n=1, the
fundamental
component is
s
s
1 V
45
.
0
2
V
2
V 



t


sin
V
2
v s
1 
the rms fundamental component is

VO
t
VS/2
T/2
T
T/2
VS/2
In the output the power due
to fundamental component is
the useful power.
s
s
1 V
45
.
0
2
V
2
V 



The power due to harmonic components
is dissipated as heat and increases the
load temperature.

The quality of the inverter output is evaluated
by some parameters.
Harmonic Factor of
nth harmonic, HFn
1
n
n
V
V
HF 
Total Harmonic
distortion, THD




2,3
n
2
n
1
V
V
1
THD
Distortion Factor, DF 



2,3
n
2
2
n
1 n
V
V
1
DF

Numerical Example :
The dc input voltage of a single phase half bridge
rectifier is 48 volt. It is supplying power to a 2.4
resistor. Find (i) output voltage (ii) Fundamental
component of output voltage (iii) output power (iv)
peak and average current in each transistor (iv) peak
reverse blocking voltage of each transistor (v) THD
(vi) DF
Given VS =48V and RL =2.4 
Output voltage Vo=Vs/2=24V
Output current Io=24/2.4=10A
Output Power VoX Io=24X10=240W
Each transistor conducts for 50% of time
Transistor current = 10X0.5=5A

The peak reverse blocking
voltage for the transistor
= 24 + 24 = 48V
R
Chopper
VO
+
_
Chopper
2
V
S
2
V
S
Fundamental component
V1 = 0.45 X 48=21.6V
rms harmonic voltage
10.46V
21.6
24
V
V 2
2
2
1
2
o 




R
VS
VO
t
V
Q1
Q2
Q3
Q4
VB
t
Q3 and Q4
Single phase bridge inverter
VB
t
Q1 and Q2
Vi
t
VS
H Bridge

Vi
t
VS
VO
t
VS
T/2
T
T/2
VS
   












 
 dt
dt
T
2
T
2
s
2
T
0
2
s
s
m
r V
V
T
1
V
The output rms
voltage is given by
s
o V
V 
Students must do the
integration in detail
and find the result

VS
VO
t
V
Q1
Q2
Q3
Q4
VB
t
Q3 and Q4
VB
t
Q1 and Q2
Vi
t
VS
Inductive Load
RL
IL
t
IL

Vi
t
VS
VO
t
VS
T/2
T
T/2
VS
The frequency of
the output voltage
is given by
T
1
f 
Frequency can be chosen
from the time period of
the base signal

Vi
t
VS
VO
t
VS
T/2
T
T/2
VS
s
o V
V 
Control of output voltage
Single pulse width modulation
Multiple pulse width modulation
Sinusoidal pulse width modulation
Modified Sinusoidal pulse width modulation
Phase displacement control

Vi
t
VS
Control of output
voltage
Single pulse width modulation
VO
t
V
VB
t
Q3 and Q4
VB
t
Q1 and Q2
VO
t
Vs
VB
t
VB
t
Q1 and Q2
Q3 and Q4
(-)/2
0  2
The rms output voltage
   












 





dt
dt
2
3
2
3
2
s
2
2
2
s
s
m
r V
V
2
1
V













S
s
m
r V
V 
The output voltage can
be controlled by
controlling the value of 

Vi
t
VS
Control of output
voltage
Multiple pulse width modulation
VO
t
Vs
VB
t
VB
t
0  2


p
V
V S
s
m
r 

 Is the width
of single pulse
p Is the number of
pulses per half
cycle
 and p both can be
adjusted to control
the output voltage

Vi
t
VS
VO
t
VB
t
VB
t
0  2
Q1 and Q2
Q3 and Q4



3
,
2
,
1
S
s
m
r V
V
m
m


The duration of individual
pulse can be adjusted to
different desired values, so
that the lower order
harmonics are eliminated
and a pure/near sine wave
could be obtained

Modified PWM inververter
VO
t
VB
t
10ms
10ms
50Hz
f 
Suppose
Carrier
frequency
=20kHz
No of pulses per
10ms = 200
VB
t
VB
t
All the pulses
are present
11111111
Alternate pulses
are present
1010101010
pulses at
desired pattern
1011011101101
Keep on trying with
different patterns of pulses
until a (i)sine wave of
desired (ii)frequency and
(iii)voltage magnitude is
obtained.

Push-Pull inverter
Q1
Q2
Q1
t
Q2
t
VB
Output
Easy design
Near sine wave
t
VO
Low TUF
Transformer
saturation
Low
frequency
operation

Three phase inverter
Inverter
1
Inverter
2
Inverter
3
A
B
C

Three phase inverter
Q1
Q2
Q3
Q4
Primary A
Primary B
Primary C
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12

Three phase inverter firing sequence
0 1800 3600
T
t
t
t
t
t
t
VB
t
Q3 and Q4
1200
1200
Q5 and Q6
Q9 and Q10
Q7 and Q8
Q11 and Q12
Q11 and Q12
600
Q1 and Q2
Sequence :
Q1 and Q2
Q11 and Q12
Q5 and Q6
Q3 and Q4
Q9 and Q10
Q7 and Q8

Three phase inverter output voltage
Primary side
0 1800 3600
T
t
t
t
VP
t
Phase A
Phase B
Phase C
1200
1200

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