AC CONTROLLERS

(EEL 744)
Prof. Bhim Singh

Department of Electrical Engineering,

Indian Institute of Technology, Delhi,
Hauz Khas, New Delhi-10016, India- 110016
email: bsingh@ee.iitd.ac.in, bhimsingh1956@gmail.com
Ph.:011-2659-1045
1

Lecture - 9

Single Phase Voltage Source

Inverters

Voltage Source inverters

The function of voltage source converter (VSI) is to
convert constant DC voltage (which can be realised by a
electrolytic capacitor or a battery) to AC voltage (which
are segments of input dc voltage).
An inverter is called a voltage source inverter (VSI) if the
input if the input voltage remains constant and current
source inverter (CSI) if the input current is maintained
constant.

Voltage Source inverters

Output of voltage waveforms of ideal inverters
should be sinusoidal.
However practical inverter waveforms are non
sinusoidal and contain certain harmonics.
For low and medium power applications square
wave or quasi square wave voltages are
acceptable.
5

Voltage Source inverters

A variable voltage can be obtained by varying the
input dc voltage and maintaining the gain of the
inverter constant.

On the other hand if the dc input voltage is fixed

then variable output voltage can be obtained by
varying the gain of the inverter. This can be
accomplished by PWM control within the inverter.
The inverter gain may be defined as ratio of the ac
output voltage to dc input voltage.
6

Applications of 1-Phase
Voltage Source Inverter

1-Phase Uninterrupted Power Supply

1-Phase Domestic Inverter

1-Phase VSI for driving 1-phase self excited

induction generators in wind power

1-Phase VSI for driving 1-phase self excited

induction generators in hydro power

Electronic Ballast

Welding

Power supply for small fans

Telecommunication Power Supply

Power Supply for Instruments

Power Supply for Medical

Equipment

Switch Mode Power Supply

Printer Power Supply

Laptop Power Supply

Desktop Computer Power Supply

Power Supply for Single Phase

Motors

1-phase 115V-Emergency Power

Supply for Aircraft Cockpit

Photo Voltaic Array

Single-phase Shunt Active- Filter

Z sa
isa
v sa

1- phase
Non -linear
Load

Single-phase Active-Series Filter

Zs
Transformers
is

AC
Mains

vs

Nonlinear
Loads

Performance parameters
Harmonic factor of nth harmonic, HFn : The harmonic factor,
which is a measure of individual harmonic contribution, is
defined as
HFn =Vn/V1
where V1 is the rms value of the fundamental component and
Vn is the rms value of the nth harmonic component.

The total harmonic distortion THD is a measure of closeness in

shape between a waveform and its fundamental component,
is defined as
THD = 1/V1(Vn2) for n=2,3.
Distortion factor DF indicates the amount of harmonic
distortion that remains in particular waveform after the
harmonics of that waveform have been subjected to second
order attenuation(i:e divided by n2)
27

Performance parameters

contd

Distortion factor of an individual (or nth) harmonic

component is defined as
DFn =Vn/(n2V1)
Lower order harmonic LOH The lowest order harmonic
is that harmonic component whose frequency is closet
to the fundamental one, and its amplitude is greater
than or equal to 3% of the fundamental component.
THD = (Vh2)1/2)/V1
DF = V1/Vrms
THD = (Vrms2-V12)/V1 = ((Vrms/V1)2 1)1/2
((1/DF)2 1)1/2

28

Classification
Voltage Source Inverters are classified as:
Single

Phase Inverter

Three

Phase Inverter

29

Single Phase Inverter Classification

Single Phase

Half Bridge Inverter

Single Phase

Full Bridge Inverter

Push

Pull Inverter

30

Single phase half bridge inverter with

different loads
Vs /2

Vs /2

S1

D1

S1

D1

S2

D2

S2

D2

Vs /2

Vs /2

Vs /2
S1

D1

Vs /2
S1

D1
IM

S2
Vs /2

D2

S2

D2

Vs /2
31

Single phase half bridge inverter

Q1

Vs
2

Vs
2

D1

i1

vao = vo

+
i2

Q2

D2

32

Single phase half bridge inverter

Load current with highly
inductive load

Vo

2
To

V1

2Vs
2

Waveforms with resistive load

To / 2
0

2
s

V
dt
4

1/ 2

Vs
2

0.45Vs
33

Example: The single phase half bridge has a resistive load of

R = 2.4 and the dc input voltage is Vs = 48 V. Determine (a)
the rms output voltage at the fundamental frequency V1, (b)
the average output power Po, (c) the average and peak
currents of each transistor, (d) the peak reverse blocking
voltage VBR of each transistor (e) the THD (f) DF, and (g) the
harmonic factor and distortion factor of lowest order
harmonic.
Solution: Vs = 48 V and R = 2.4
(a) V1 = 0.45 * 48 = 21.6 V
(b) Vo = Vs /2 =48/2 = 24V ; Po = 242 / 2.4 = 240 W
(c) Ip = 24/2.4 = 10 A, since each transistor conducts for a
50% duty cycle, the average current of each transistor is
ID = 0.5 * 10 = 5A
34

Solution: contd.
(d) The peak reverse blocking voltage VBR = 2*24 =48V

(e) Rms harmonic voltage Vh

Vh = (Vn2)1/2 for n=2,3. = (Vo2-V12)1/2 = 0.2176 Vs
THD = 0.2176 Vs /0.45 Vs = 48.34%
(f) (Vn2/n2)1/2 for n=2,3. = 0.01712 Vs
DF = 0.01712 Vs / 0.45 Vs = 3.804%

(g) The lowest order harmonic is the third, V3= V1/3 and HF3=
V3 / V1 = 33.33% and DF3 = (V3/32) / V1 =1/27 =3.704%
since V3 is 33.33% which is greater than 3%
LOH is V3

35

Example: A single-phase, half bridge IGBT based voltage source

inverter (VSI) has source voltage +120V and 120V. The inverter
operates with the square voltage waveform at a frequency of
50Hz. The load resistance is 10 ohms. Calculate (a) rms output
voltage, (b) rms of fundamental output voltage, (c) output power,
(d) average, rms and peak current of IGBT used in VSI, (e) THD of
output voltage, (f) distortion factor, (g) harmonic factor.

Solution: Vs = 120-(-120) V=240V and R = 10

(a) Vrms = 0.5* 240 = 120V;
(b) V1 = 0.45 * 240 = 108 V
(c) Po = 1202 / 10 = 1440 W
(d) Ip = 120/10 = 12 A, since each transistor conducts for a
50% duty cycle, the average current of each transistor is
ID = 0.5 * 12 = 6A ; rms current = 12/2 = 8.84A
36

Solution: contd.
(e) Rms harmonic voltage Vh
Vh = (Vn2)1/2 for n=2,3. = (Vo2-V12)1/2 = 0.2176 Vs
THD = 0.2176 Vs /0.45 Vs = 48.34%
(f) (Vn2/n2)1/2 for n=2,3. = 0.01712 Vs
DF = 0.01712 Vs / 0.45 Vs = 3.804%
(g) The lowest order harmonic is the third, V3= V1/3 and HF3=
V3 / V1 = 33.33% and DF3 = (V3/32) / V1 =1/27 =3.704%
since V3 is 33.33% which is greater than 3%
LOH is V3

37

Single phase full bridge inverter with

different loads

38

Single phase full bridge inverter with

different loads
Vdc/2

Q1

Q3

Vdc/2

Q1

Q4

Load

Vdc/2

Q2

Q4

Load

Vdc/2

Q2

Q4

39

Single phase full bridge inverter

Q1

Q3

Vdc/2
Load

Vdc/2

Q2

Q4

40

Single phase full bridge inverter

Vo

2
To

V1

4Vs
2

1/ 2
To / 2
0

Vs2 dt

Vs

0.9Vs

Waveforms
Load current with highly
inductive load
41

Analysis of Single phase full bridge inverter

If g 1 and g 4 are gating signals for switches Q1 and Q 4 ,
respectively, the switching function is
S( ) = g 1 - g 4
= 1 for for 0
= -1 for for
2
If f o is the fundamental
frequency of inverter
t 2 f ot
42

Analysis of Single phase full bridge inverter

S( ) can be expressed in a Fourier series as
Ao
S( ) =
( An cos n Bn sin n )
n 1,2...
2
2
4
Bn
S( ) sin n d
for n 1,3.....
n
0
Dueto half

wave symmetry , Ao

An 0

Substituting Ao , An and Bn inthe above eqn. yields

S( ) =

4
n

sin n
1,3,5...
n
43

Analysis of Single phase full bridge inverter

If the input voltage, which is dc is Vi ( ) = Vs
gives the output voltage as
4Vs
Vo ( ) =S( ) Vi ( )=
n

sin n
1,3,5...
n

44

Example: The single phase full bridge has a resistive load of R =

2.4 and the dc input voltage is Vs = 48 V. Determine (a) the rms
output voltage at the fundamental frequency V1, (b) the average
output power Po, (c) the average and peak currents of each
transistor, (d) the peak reverse blocking voltage VBR of each
transistor (e) the THD (f) DF, and (g) the harmonic factor and
distortion factor of lowest order harmonic.

Solution: Vs = 48 V and R = 2.4

(a) V1 = 0.9 * 48 = 43.2 V
(b) Vo = Vs =48 = 48V ; Po = 482 / 2.4 = 960 W
(c) Ip = 48/2.4 = 20 A, since each transistor conducts for a 50%
duty cycle, the average current of each transistor is ID = 0.5 *
20 = 10A
45

Solution: contd.
(d) The peak reverse blocking voltage VBR = 48V
(e) Rms harmonic voltage Vh
Vh = (Vn2)1/2 for n=2,3. = (Vo2-V12)1/2 = 0.4352 Vs
THD = 0.4352 Vs /0.9 Vs = 48.34%

(f) (Vn2/n2)1/2 for n=2,3. = 0.03424 Vs

DF = 0.03424 Vs / 0.9 Vs = 3.804%
(g) The lowest order harmonic is the third, V3= V1/3 and HF3=
V3 / V1 = 33.33% and DF3 = (V3/32) / V1 =1/27 =3.704%
since V3 is 33.33% which is greater than 3%
LO H is V3
The peak blocking voltage of each transistor and quality of
output voltage for half bridge and full bridge are same.
However for full bridge inverters. The output power is four
times higher and the fundamental component is twice that
of half bridge inverters.
46

Push Pull Inverters

It requires a centered tapped primary.
Push-pull inverter can operate in PWM or a square
wave mode and the wave forms are identical to
those for half bridge and full bridge inverters.
Theoutput voltageequals
^

V o1

Vd
ma
(m a `1.0)
n
^

Vd
and
n

V
o1

4 Vd
(m a `1.0)
n
47

Push Pull Inverters

id

n:1
io

vo

vd
T1

D1

T2

D2

48

Push Pull Inverters

Advantages
The main advantage of the push-pull circuit is that
no more than one switch in series conducts at any
instant of time.
Efficiency is increased significantly.
Gate drive circuit is simplified

Disadvantage
It is difficult to avoid dc saturation of the
transformer

49

Ripple in the inverter output voltage

square-wave switching

50

Ripple in the inverter output voltage

PWM bipolar voltage switching

51

Voltage Control of Single-Phase Inverters

Commonly-used Technique

Single-Pulse-Width-Modulation
Multiple-Pulse-Width-Modulation
Sinusoidal-Pulse-Width-Modulation
Modified-Sinusoidal-Pulse-Width-Modulation
Transformer connections
52

Single-Pulse-Width-Modulation
Carrier signal
Reference signal

/2

g1

Gate signal for S1

/2-/2 /2 /2+/2

g4

Gate signal for S4

vo
/2-/2 /2 /2+/2

3/2

Single pulse width modulation

53

rms value of the output voltage

vo
/2-/2 /2 /2+/2

3/2

1
2

2
2

Vo

Vs2 d ( t )
2

Vo

Vs

0
0 Vo

180
Vs
54

Single-Pulse-Width-Modulation
Compare the Reference Signal with the Carrier
Frequency of the Reference Signal determines the
frequency of the Output Voltage
Modulation Index = M = Ar/Ac
vo (t )
n

4Vs
n
n
sin
sin
sin n t
2
2
1,3,5,... n

55

Times and angles of the intersections

vo
/2-/2 /2 /2+/2

t1

t2

3/2

TS
(1 M )
2
TS
(1 M )
2

Pulse width d (or pulse angle )

d

t2 t1

TS = T/2

MTS
56

Harmonic Profile

57

Single Pulse-Width-Modulation
There is only one pulse per half cycle and width of the
pulse is varied to control the inverter output voltage.

The gating signals are generated by comprising a

rectangular reference signal of amplitude, Ar with a
triangular carrier wave of amplitude Ac.
The ratio of Ar to Ac is the control variable and defined as
amplitude modulation index.

The dominant harmonic is the third, and the distortion

factor increases significantly at a low output voltage.

58

Example: A single-phase, full bridge IGBT based voltage source

inverter (VSI) is operated single-pulse width modulated. The
inverter operates at a frequency of 50Hz. The dc source potential
is 400V, and each pulse has duration of 6.0 ms. Calculate rms load
voltage, rms value of fundamental component of output voltage
and THD of the output voltage waveform.

Solution:
f = 50 kHz, t = 20msec, pulse duration d = 6 ms ; Vs = 400V

or = 1.884

rms load voltage

= 314; d=

Vo =Vs

= 400 * 0.7745 = 309.83 V

59

Solution:

4Vs
n
v o (t)=
sin
sinnt
2
n=1,3,5,... n

Vo1 =
sin = 291 V
2
2
THD of the output voltage waveform
4Vs

THD =

2
o

V
2
o1

2
o1

*100

3092 2912
*100
2
291

35.71%

60

Multiple-Pulse-Width-Modulation
Carrier signal
Reference signal
0

+ m

61

Multiple Pulses per Half-Cycle of Output Voltage

62

Gate Signal Generation

Compare the Reference Signal with the Carrier

Frequency of the Reference Signal determines the Output Voltage Frequency

Frequency of the Carrier determines the number of pulses per half-cycle

Modulation Index controls the Output Voltage

63

rms value of the output voltage

(

2p
2

Vo

Vo
0

Vs
M

0
0
0

Vo

1
2

)/2

Vs2 d ( t )
(

)/2

p
1
T
2p
p
Vs
64

Fourier Series of the Output Voltage

vo (t )

Bn sin n t
n 1,3,5,...
2p

Bn
m

4Vs
n
sin
sin n(
4
1 n

3
) sin n(
4

4
65

The times and angles of the intersections

tm

tm

(m M )

Ts
2

TS
(m 1 M )
2

The pulse width d (or pulse angle )

tm

tm

TS = T/2p

MTS

66

Harmonic Profile of Multiple Pulse-Width-Modulation

67

Multiple Pulse-Width-Modulation
The harmonic content can be reduced by using several
pulses in each half cycle of output voltage.
This type of modulation is also known as uniform pulse
width modulation(UPWM)
p=fc (carrier frequency) /(2fo (frequency of reference
signal) = mf /2, where mf is defined as frequency
modulation index ratio.
Larger the value of p, the lower order harmonics would
be lower, but the amplitudes of some higher order
harmonics will increase.
However, such harmonics produce negligible ripple or
can be easily filtered out

68

Sinusoidal Pulse-Width-Modulation
Instead of maintaining the width of all pulses the same
as in case of multiple pulse modulation, the width is
varied in proportion to the amplitude of a sine wave
evaluated at the centre of same pulse.
The distortion factor and lower order harmonics are
significantly reduced
The number of pulses per half cycle depends on the
carrier frequency.
This type of modulation eliminates all harmonics less
than or equal to 2p-1 for p=5, the lowest order harmonic
is 9th.

69

Single Phase PWM Voltage Source Inverter

TA+

TB+
DA+

Id
Vd
TA-

DB+

Ls

Load

B
DA-

TB-

Ls

DB-

70

VSC PWM Control

-

Sine-triangle PWM can be used but it may

produce high losses;

During transients better to use sine-triangle method

A small percentage
of 3rd harmonic can
be added to the
reference voltage to
increase
fundamental
component and use
better the converter.
-

71

Sinusoidal-Pulse-Width-Modulation with low

frequency carrier wave

72

Sinusoidal-Pulse-Width-Modulation with very

high frequency carrier wave

73

PWM with bipolar voltage switching

74

PWM with bipolar switching

Diagonally opposite switches (TA+, TB-) and (TB+, TA-) from two
legs are switched as switched as switch pairs 1 and 2
respectively.
With this type of PWM switching, the output voltage
waveform of leg A is identical to the output of the basic one
leg inverter.
The output of inverter leg B is negative of leg A (shown in next
slide).

The output voltage vo switches between Vd and +Vd voltage

levels. That is the reason this type of switching is called a
PWM with bipolar switching.
75

PWM with unipolar voltage switching

76

PWM with unipolar voltage switching

77

Output waveform and Fourier analysis

78

PWM with unipolar switching

In PWM with unipolar voltage switching, the switches in the

two legs of the full bridge is not switched simultaneously.

Here legs A and B of the full bridge inverter are controlled

separately by comparing vtri with vcontrol and
vcontrol
respectively.

The output voltage changes between zero and +Vd or between

zero and Vd voltage levels. This is the reason this type of
PWM scheme is called as unipolar switching scheme.

This scheme has a advantage of effectively doubling the

switching as far as the output harmonics are concerned, as
compared to the bipolar switching scheme.
79

PWM with unipolar switching

The voltage jumps in the output voltage at each
switching are reduced to Vd, as compared to 2Vd in the

previous scheme.
The advantage of effectively doubling the switching
frequency appears in the harmonic spectrum of the
voltage waveform.
Where the lowest harmonics (in idealized circuit) appear
as sidebands of twice the switching frequency.
80

rms value of the output voltage

The rms output voltage
2p

Vo

Vs

1
2

m 1
th
is
the
width
of
m
pulse
m

Fourier coefficient of output voltage

2p

Bn
m

4Vs
n
sin
4
1 n

sin n(

sin n(

3
)
4
m

where n 1,3,5.....
81

The mth time tm and angle m intersection

Ts
tm
tx M
2
where t x can be solved from
m

2t
1
M sin
Ts

tx

Ts
M
2

for m = 1,3, .....2p

2t
M sin
Ts

tx

Ts
M
2

for m = 2,4, .....2p; where Ts =T/2(p+1)

The m th pulse width d m (or pulse angle
dm

tm

tm
82

Harmonic profile of sinusoidal pulse width modulation

83

Analysis of Single Phase Full-Bridge Inverters With

Sinusoidal Pulse-Width-Modulation

84

Analysis of Single Phase Full-Bridge Inverters With

Sinusoidal Pulse-Width-Modulation

The switching function of single-phase full bridge

inverter with sinusoidal pulse width modulation.
If g 1 and g 4 are gating signals for switches Q1 and Q 4 ,
respectively, the switching function is
S( ) = g 1 - g 4
S( ) can be expressed in a Fourier series as
Ao
S( ) =
2

n 1,2...

( An cos n

Bn sin n )
85

Analysis of Single Phase Full-Bridge Inverters With

Sinusoidal Pulse-Width-Modulation
An

S( ) cos n d

1 sin n

m 1,2,3...

Dueto half

wave symmetry , Ao

Bn 0

Substituting Ao , An and Bn inthe above eqn. yields

p
4
m
S( ) =
1 sin n m cos n
n 1,3,5...
m 1,2,3...
n
If theinput voltageisVi ( ) Vs the above equations

givethe output voltage as

4Vs
Vo ( ) =
n

p
n 1,3,5...

m 1,2,3...

1 sin n

cos n
86

Modified Sinusoidal Pulse-Width-Modulation

87

Modified Sinusoidal Pulse-Width-Modulation

The carrier wave is applied during the first and last 60
intervals per half cycle (e.g., 0 to 60 and 120 to 180 )

This type of modulation is called modified sinusoidal

pulse width modulation.
The fundamental component is increased and its
harmonic characteristics are improved.
It reduces the number of switching of power devices also
reduces switching losses.

88

Harmonic profile of modulation Sinusoidal Pulse-WidthModulation

89

Advanced modulation Techniques

The sinusoidal pulse width modulation,
which is most commonly used, suffers from
drawbacks (e.g. low fundamental output
voltage). The other techniques that offer
improved performances are:
Trapezoidal modulation
Staircase modulation
Stepped modulation
Harmonic injection modulation
Delta modulation.
90

Trapezoidal modulation
The gating signals are generated by comparing a
triangular carrier with a modulating trapezoidal wave.
The trapezoidal wave can be obtained from a
triangular wave by limiting its magnitude to Ar, which
is related to the peak value Ar(max) by Ar = Ar(max) ,
where is called the triangular factor, because the
waveform becomes a triangular wave when = 1. The
modulation index M is

Ar
M
Ac

A r(max )
Ac

for 0 M 1
91

Trapezoidal modulation

92

Staircase modulation
The modulating signal is a staircase wave
The levels of the stairs are calculated to eliminate
specific harmonics.
The modulating frequency ratio mf and the number of
steps are so chosen to obtain the desired quality of
output voltage.
This is an optimised PWM and is not recommended for
fewer than 15 pulses in one cycle.
The optimum number of pulses in one cycle is 15 for
two levels, 21 for three levels and 27 for four levels.
This type of control provides a high quality output
voltage with a fundamental value of up to 0.94Vs.
93

Staircase modulation

94

Stepped modulation
The modulating signal is a stepped wave.
The stepped wave is not a sampled approximation of
sine wave.
It is divided into specified intervals, say 20, with each
interval controlled individually to control the
magnitude of the fundamental component and to
eliminate specific harmonics.
This type of control gives low distortion, but higher
fundamental amplitude compared with that of normal
PWM control.

95

Stepped modulation

96

Selective injected modulation

The modulating signal is generated by injecting
selected harmonics to the sine wave.
This results in flat-topped waveform and reduces the
amount of over modulation.
It provides a higher fundamental amplitude and low
distortion of output voltage .
vr = 1.15 sin t + 0.27 sin 3t 0.029 sin 9t.
The modulating signal with third and ninth harmonic
injections is shown in figure in next slide.
The injection of 3nth harmonics does not affect the
quality of the output voltage, because the output of 3phase inverter does not contain triplen harmonics.
97

Selective injected modulation

98

Delta modulation
In delta modulation a triangular wave is allowed to
oscillate within a defined window V above and below
the reference sine wave vr . The inverter switching
function , which is identical to the output voltage vo is
generated from the vertices of the triangular wave vc.
It is known as hysteresis modulation.
If the frequency of the modulating wave is changed
keeping the slope of the triangular wave constant, the
number of pulses and pulses widths of the modulated
wave would change.
The delta modulation can control the ratio of voltage
to frequency, which is a desirable feature, especially in
ac motor control.
99

Delta modulation

100

Transformer Connections

Circuit

Waveforms

Transformer Connections

The output voltages of two or more inverters may be

connected in series through a transformer to reduce
or eliminate certain unwanted harmonics

The output of first inverter can be expressed as

vo1 = A1sint + A3sin3t + A5sin5t +

The output of second inverter is vo2 delayed by /3

vo2 = A1sin(t- /3) + A3sin3(t- /3) + A5sin5(t/3) +

Transformer Connections

The resultant voltage vo is obtained by vector

addition.
vo = vo1 + vo2 = 3[A1sin(t- /6) + A3sin3(t- /6)
+ A5sin5(t- /6) + ]

The effective output has been reduced by (1-0.866)

= 13.4%.

Third harmonic is been eliminated by addition

Comparison of Single phase inverters

To compare the utilisation of switches in various single-phase
inverters, we will initially compare them for a square -wave
mode of opeartion at maximum rated output
Push-Pull Inverter
VT = 2Vd.max ;
Vo1.max

Io.max
IT = 2
n

4 Vd.max
whereq
2 n

2(noof switches in inverter)

1
n turns ratio;Maximumswitch utilisation ratio
0.l6
2
104

Comparison of Single phase inverters

Half Bridge Inverter
VT = Vd.max ;

IT = 2Io.max

4 Vd.max
Vo1.max
where q 2(noof switchesin inverter)
2 2
1
n turns ratio;Maximumswitch utilisation ratio
0.l6
2
105

Comparison of Single phase inverters

Full Bridge Inverter
VT = Vd.max ;
Vo1.max

4
2

IT = 2I o.max

Vd.max where q

4(noof switchesin inverter)

1
n turns ratio;Maximumswitch utilisation ratio
0.l6
2
Thisshows that in each inverter, the switch utilisation is the
1
same with maximumswitch utilisation ratio
0.l6
2
106

Comparison of Single phase

PWM SWITCHING
Usin g the PWMswitching with m a 1.0
This ratio would be smaller by a factor of

ma ascompared

tosquare waveswitching
1
same with maximumswitch utilisation ratio
2
(PWM,m a 1.0)

ma

1
ma
8

107

Example: In a single-phase, full bridge based PWM voltage source

inverter (VSI), input dc voltage varies in a range of 295-325V. The
output voltage is required to be constant at 200V (rms) and the
maximum load current (assumed to be sinusoidal) is 10A (rms).
Calculate the combined switch utilization ratio.
Solution:
In this inverter
VT = Vdmax = 325V
IT =

2I o

2*10 =14.14A

q = no of switches = 4
Maximum Volt Amperes = 2000VA
V I
2000
SUR = o1 omax
0.11
qVT IT
4 *325*14.14
108

Inverter Circuit Design

The voltage and current rating of power devices in
inverter circuit depends on the type of inverter, load
and method of voltage and current control.
The design requires
Deriving the expressions for the instantaneous load
current.
Plotting the current waveforms for each device and
component.
Voltage ratings requires establishing the reverse voltages
of each device.
109

Inverter Circuit Design

To reduce the output harmonics, output filters are
necessary.
C filter is very simple , but it draws more reactive power.

An LC-tuned filter can eliminate only one frequency.

A properly designed CLC filter is more effective in reducing
harmonics of wide bandwidth and draws less reactive power.

110

Output Filters
Ce

Le

Le
Load

Ce
Load

Load
Ce
LC- Filter

C- Filter

Le
C1

Ce
Load

LC- Filter

CLC- Filter

111

Example: A single-phase, PWM voltage source converter feeds 5

kW to 230 V rms, 50Hz, single-phase ac mains from a constant
voltage DC bus. The switching frequency is 20kHz and ac inductor is
5.0 mH. The power-factor is corrected close to unity and PWM
modulation index is 0.9. Determine (a) an input dc voltage, (b) rms
ac current and phase shift in fundamental component of PWM
voltage and supply voltage.
Solution: P = 5kW, 230 V, 50 Hz, fs = 20kHz, L-s = 5 mH, ma = 0.9
Is1 = 5000/(230)*1 = 21.74 A
XL = 314*5*10-3 = 1.57
Vcon1 = V2s +(Is*Xc)2 = 2302+(21.74*1.57)2 = 232.5 V
Vdc = 2* Vcon1 /m
(a) Vdc = ( 2*232.5)/0.9 = 365.4V
(b) Is = 21.74A
= sin-1(P*Xc/Vs*Vc) = sin-1(5k*1.57/3*232.5*230) = 8.44
112

References
N. Mohan, T. M. Undeland and W. P. Robbins, Power
Electronics, Converter, Application and Design, Second
Edition, John Willey & Sons, 1995, New York.

M. H. Rashid, Power Electronics, circuits, Devices and

Applications, Second Edition, Prentice-Hall, 1995, India.
J. Vithayathil, Power Electronics: Principles and Applications,
McGraw-Hill,1995, New York.
P. C. Sen, Modern Power Electronics, Wheeler Publishing,
1998, New Delhi .
113

References
B. K. Bose, Power Electronics and Variable Frequency Drive,
Standard Publishers Distributors, 2000.
V. Subrahmanyam, Power Electronics, New Age International
Publishers, India, 1996.
B. D. Bedford and R. G. Hoft, Principles of Inverter Circuits,
John Wiley & Sons, 1964, USA.

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