2. CONTENTS
۩ Introduction
۩ Basic Principles of Inverter
۩ Single-phase Half-Bridge Square-Wave Inverter
۩ Single-phase Full-Bridge Square-Wave Inverter
۩ Quasi Inverter
۩ Three-phase inverter
۩ Fourier Series and Harmonics Analysis
۩ Pulse-Width Modulation (PWM)
3. Definition:
Converts DC to AC power by switching the DC input
voltage in a pre-determined sequence so as to
generate AC voltage.
INTRODUCTION
Applications:
Induction motor drives, traction, standby power
supplies, and uninterruptible ac power supplies (UPS).
10. The total RMS value of the load output voltage,
The instantaneous output voltage is: (refer to slide 30/pp:88)
Single-phase Half-bridge
Square-wave Inverter
2
2
2 2
/
0
2
DC
T
DC
O
V
dt
V
T
V =
= ∫
The fundamental rms output voltage (n=1)is
2,4,....
n
for
0
sin
2
,...
5
,
3
,
1
=
=
= ∑
=
n
DC
O t
n
n
V
v ω
π
DC
DC
O V
V
V 45
.
0
2
1
2
1 =
=
π
11. In the case of RL load, the instantaneous load current io ,
where
Single-phase Half-bridge
Square-wave Inverter
∑
∞
=
−
+
=
,..
5
,
3
,
1
2
2
)
sin(
)
(
2
n
n
DC
o t
n
L
n
R
n
V
i θ
ω
ω
π
)
/
(
tan 1
R
L
nω
θ −
=
where
The fundamental output power is
)
/
(
tan 1
R
L
n
n ω
θ −
=
+
=
=
=
2
2
2
1
1
1
1
1
)
(
2
2
cos
L
R
V
R
I
I
V
P
DC
o
o
o
o
ω
π
θ
12. The total harmonic distortion (THD),
Single-phase Half-bridge
Square-wave Inverter
= ∑
∞
2
1
V
THD
= ∑
= ,....
7
,
5
,
3
2
1
1
n
n
O
V
V
THD
( )
2
1
2
1
1
o
o
O
V
V
V
THD −
=
13. Example 3.1
The single-phase half-bridge inverter has a resistive load of R =
2.4Ω and the DC input voltage is 48V. Determine:
(a) the rms output voltage at the fundamental frequency
(b) the output power
Single-phase Half-bridge
Square-wave Inverter
(c) the average and peak current of each transistor.
(d) the THD
14. Solution
VDC = 48V and R = 2.4Ω
(a) The fundamental rms output voltage,
Vo1 = 0.45VDC = 0.45x48 = 21.6V
Single-phase Half-bridge
Square-wave Inverter
(b) For single-phase half-bridge inverter, the output voltage
Vo = VDC/2
Thus, the output power,
W
R
V
P o
o
240
4
.
2
)
2
/
48
(
/
2
2
=
=
=
15. Solution
(c) The transistor current Ip = 24/2.4 = 10 A
Because each of the transistor conducts for a 50% duty cycle,
the average current of each transistor is IQ = 10/2 = 5 A.
Single-phase Half-bridge
Square-wave Inverter
(d)
2
1
2
1
1
o
o
O
V
V
V
THD −
=
( )
%
34
.
48
45
.
0
2
45
.
0
1 2
2
=
−
= DC
DC
DC
V
V
xV
16. The switching in the second leg is delayed by 180
degrees from the first leg.
The maximum output voltage of this inverter is twice
that of half-bridge inverter.
Single-phase Full-bridge
Square-wave Inverter
17. The output RMS voltage
And the instantaneous output voltage in a Fourier series is
Single-phase Full-bridge
Square-wave Inverter
DC
DC
O V
dt
V
T
V =
= ∫
2
2
The fundamental RMS output voltage
In the case of RL load, the instantaneous load current
∑
=
=
,...
5
,
3
,
1
sin
4
n
DC
O t
n
n
V
v ω
π
DC
DC
V
V
V 9
.
0
2
4
1 =
=
π
( )
( )
n
n
DC
o t
n
L
n
R
n
V
i θ
ω
ω
π
−
+
= ∑
∞
=
sin
4
,...
5
,
3
,
1
2
2
18. Example 3.2
A single-phase full-bridge inverter with VDC = 230 and
consist of RLC in series. If R = 1.2Ω, ωL = 8 Ω and
1/ωC = 7 Ω, find:
(a) The amplitude of fundamental rms output current,
Single-phase Full-bridge
Square-wave Inverter
(a) The amplitude of fundamental rms output current,
io1
(b) The fundamental component of output current in
function of time.
(c) The power delivered to the load due to the
fundamental component.
19. Example 3.3
A single-phase full-bridge inverter has an RLC load with R
= 10Ω, L = 31.5mH and C = 112µF. The inverter frequency
is 60Hz and the DC input voltage is 220V. Determine:
(a) Express the instantaneous load current in Fourier series.
(b) Calculate the rms load current at the fundamental
Single-phase Full-bridge
Square-wave Inverter
(b) Calculate the rms load current at the fundamental
frequency.
(c) the THD of load current
(d) Power absorbed by the load and fundamental power.
(e) The average DC supply current and
(f) the rms and peak supply current of each transistor
21. Viewed as extensions of the single-phase bridge circuit.
The switching signals for each switches of an inverter leg are
displaced or delayed by 120o.
With 120o conduction, the switching pattern is T6T1 – T1T2 –
T2T3 – T3T4 – T4T5 – T5T6 – T6T1 for the positive A-B-C
sequence.
When an upper switch in an inverter leg connected with the
Three-Phase Inverter
When an upper switch in an inverter leg connected with the
positive DC rail is turned ON, the output terminal of the leg
(phase voltage) goes to potential +VDC/2.
When a lower switch in an inverter leg connected with the
negative DC rail is turned ON, the output terminal of that leg
(phase voltage) goes to potential -VDC/2.
22.
23. The line-to-neutral voltage can be expressed in Fourier series
Three-Phase Inverter
∑
∑
∞
=
∞
=
−
=
+
=
,..
5
,
3
,
1
,..
5
,
3
,
1
2
sin
3
sin
2
6
sin
3
sin
2
n
dc
bn
n
dc
an
t
n
n
n
V
v
t
n
n
n
V
v
π
ω
π
π
π
ω
π
π
The line voltage is vab = √3van with phase advance of 30o
∑
∞
=
−
=
,..
5
,
3
,
1 6
7
sin
3
sin
2
n
dc
cn t
n
n
n
V
v
π
ω
π
π
( )
∑
∑
∑
∞
=
∞
=
∞
=
−
=
−
=
+
=
,..
5
,
3
,
1
,..
5
,
3
,
1
,..
5
,
3
,
1
sin
3
sin
3
2
3
sin
3
sin
3
2
3
sin
3
sin
3
2
n
dc
ca
n
dc
bc
n
dc
ab
t
n
n
n
V
v
t
n
n
n
V
v
t
n
n
n
V
v
π
ω
π
π
π
ω
π
π
π
ω
π
π
24. Fourier series is a tool to analyze the wave shapes of the
output voltage and current in terms of Fourier series.
Fourier Series and
Harmonics Analysis
( )
∫
=
π
θ
π
2
0
1
d
v
f
ao
Inverse Fourier
( ) ( )
∑
∞
+
+
= sin
cos
1
n
bn
n
a
a
v
f θ
θ
π 0
( ) ( )
∫
=
π
θ
θ
π
2
0
cos
1
d
n
v
f
an
( ) ( )
∫
=
π
θ
θ
π
2
0
sin
1
d
n
v
f
bn
( ) ( )
∑
=
+
+
=
1
0 sin
cos
2
1
n
n n
bn
n
a
a
v
f θ
θ
Where
t
ω
θ =
25. If no DC component in the output, the output voltage and
current are
Fourier Series and
Harmonics Analysis
( )
∑
∞
=
+
=
1
sin
)
(
n
n
n
o t
n
V
t
v θ
ω
( )
∑
∞
+
= sin
)
( t
n
I
t
i φ
ω
The rms current of the load can be determined by
( )
∑
=
+
=
1
sin
)
(
n
n
n
o t
n
I
t
i φ
ω
2
1
1
2
,
2
∑
∑
∞
=
∞
=
=
=
n
n
n
rms
n
rms
I
I
I
Where
n
n
n
Z
V
I =
26. The total power absorbed in the load resistor can be
determined by
Fourier Series and
Harmonics Analysis
∑
∑
∞
=
∞
=
=
=
1
2
,
1 n
rms
n
n
n R
I
P
P
=
= 1
1 n
n
27. Since the objective of the inverter is to use a
DC voltage source to supply a load that
requiring AC voltage, hence the quality of
the non-sinusoidal AC output voltage or
current can be expressed in terms of THD.
Total Harmonics
Distortion
current can be expressed in terms of THD.
The harmonics is considered to ensure that
the quality of the waveform must match to
the utility supply which means of power
quality issues.
This is due to the harmonics may cause
degradation of the equipments and needs to
be de-rated.
29. The THD of the load voltage is expressed as,
Total Harmonics
Distortion
rms
rms
rms
rms
n rms
n
v
V
V
V
V
V
THD
,
1
2
,
1
2
,
1
2
2
, )
( −
=
=
∑
∞
=
The current THD can be obtained by replacing the
harmonic voltage with harmonic current,
rms
n rms
n
i
I
I
THD
,
1
2
2
, )
(
∑
∞
=
=
30. Harmonics of Square-
wave Waveform
DC
V
0
1
0
2
0 =
−
+
= ∫ ∫
π π
π
θ
π
DC
DC V
d
V
a
0
)
cos(
)
cos(
2
=
−
= ∫ ∫
π π
θ
θ
θ
θ
π
d
n
d
n
V
a DC
n
π
2
π
t
ω
θ =
DC
V
−
0
)
cos(
)
cos(
0
=
−
= ∫ ∫
π
θ
θ
θ
θ
π
d
n
d
n
an
[ ]
[ ]
( )
[ ]
π
π
π
π
π
π
π
θ
θ
π
θ
θ
θ
θ
π
π
π
π
π π
π
n
n
n
n
n
n
V
n
n
n
V
d
n
d
n
V
b
DC
DC
DC
n
cos
1
2
)
cos
2
(cos
)
cos
0
(cos
)
cos(
)
cos(
)
sin(
)
sin(
2
0
0
2
−
=
−
+
−
=
+
−
=
−
= ∫ ∫
31. When the harmonics number, n of a waveform is
even number, the resultant of
Therefore,
Harmonics of Square-wave
Waveform
1
cos =
π
n
0
=
b
Therefore,
When n is odd number,
Hence,
0
=
n
b
1
cos −
=
π
n
π
n
V
b DC
n
4
=
32. Spectrum of Square-
wave
• Harmonic decreases with
a factor of (1/n).
• Even harmonics are
absent
• Nearest harmonics is the
(0.33)
(0.2)
(0.14)
(0.11)
(0.09)
• Nearest harmonics is the
3rd. If fundamental is
50Hz, then nearest
harmonic is 150Hz.
• Due to the small
separation between the
fundamental and 3rd
harmonics, output low-
pass filter design can be
very difficult.
35. Example 3.4
The full-bridge inverter with DC input voltage of 100V,
load resistor and inductor of 10Ω and 25mH
respectively and operated at 60 Hz frequency.
Determine:
Determine:
(a) The amplitude of the Fourier series terms for the
square-wave load voltage.
(b) The amplitude of the Fourier series terms for load
current.
(c) Power absorbed by the load.
(d) The THD of the load voltage and load current for
square-wave inverter.
36. Amplitude and
Harmonics Control
π π
2
t
ω
α α α α
The output voltage of the full-
bridge inverter can be controlled
by adjusting the interval of on
each side of the pulse as zero .
37. The rms value of the voltage waveform is
The Fourier series of the waveform is expressed as
Amplitude and
Harmonics Control
π
α
ω
π
α
π
α
2
1
)
(
1 2
−
=
= ∫
−
DC
DC
rms V
t
d
V
V
The amplitude of half-wave symmetry is
∑
=
odd
n
n
O t
n
V
t
v
,
)
sin(
)
( ω
)
cos(
4
)
(
)
sin(
2
α
π
ω
ω
π
α
π
α
n
n
V
t
d
t
n
V
V DC
DC
n
=
= ∫
−
38. The amplitude of the fundamental frequency is controllable by
adjusting the angle of α.
Amplitude and
Harmonics Control
α
π
cos
4
1
= DC
V
V
The nth harmonic can be eliminated by proper choice of
displacement angle α if
α
π
cos
1
=
V
0
cos =
α
n
OR
n
o
90
=
α
39. Pulse-width modulation provides a way to decrease
the total harmonics distortion (THD).
Types of PWM scheme
Natural or sinusoidal sampling
Pulse-Width Modulation
(PWM)
Natural or sinusoidal sampling
Regular sampling
Optimize PWM
Harmonic elimination/minimization PWM
SVM
40. Several definition in PWM
(i) Amplitude Modulation, Ma
Pulse-Width Modulation
(PWM)
tri
,
sin
,
carrier
,
,
m
e
m
m
reference
m
a
V
V
V
V
M =
=
If Ma ≤ 1, the amplitude of the fundamental
frequency of the output voltage, V1 is linearly
proportional to Ma.
(ii) Frequency Modulation, Mf
tri
,
carrier
, m
m
e
tri
carrier
f
f
f
f
f
M
sin
reference
=
=
41. Bipolar Switching of PWM
Pulse-Width Modulation
(PWM)
vtri (carrier)
vsine (reference)
VDC
-VDC
(a)
(b)
S1 and S2 ON when Vsine Vtri
S3 and S4 ON when Vsine Vtri
43. Unipolar Switching
of PWM
Pulse-Width Modulation
(PWM)
vtri (carrier)
vsine (reference)
(a)
(b)
(c)
S4
S2
Vo
Vdc
0
Vdc
0
S1 is ON when Vsine Vtri
S2 is ON when –Vsine Vtri
S3 is ON when –Vsine Vtri
S4 is ON when Vsine Vtri
45. ■ Advantages of PWM switching
- provides a way to decrease the THD of load current.
- the amplitude of the o/p voltage can be controlled with the
modulating waveform.
- reduced filter requirements to decrease harmonics.
Pulse-Width Modulation
(PWM)
- reduced filter requirements to decrease harmonics.
■ Disadvantages of PWM switching
- complex control circuit for the switches
- increase losses due to more frequent switching.
46. Harmonics of Bipolar PWM
Assuming the PWM output is symmetry, the
harmonics of each kth PWM pulse can be
expressed
PWM Harmonics
Finally, the resultant of the integration is
+
=
=
∫ ∫
∫
+
+
+
k
k
k k
k
t
d
t
n
-V
t
d
t
n
V
t
d
t
n
t
v
V
DC
DC
T
nk
δ
α
α
α
δ
α
ω
ω
ω
ω
π
ω
ω
π
1
k
)
(
)
sin(
)
(
)
(
)
sin(
2
)
(
)
sin(
)
(
2
0
[ ]
)
(
cos
2
cos
cos
2
1 k
k
k
k
DC
nk n
n
n
n
V
V δ
α
α
α
π
+
−
+
= +
48. Harmonics of Bipolar PWM
The Fourier coefficient for the PWM waveform is the
sum of Vnk for the p pulses over one period.
PWM Harmonics
∑
=
p
V
V
The normalized frequency spectrum for bipolar
switching for ma = 1 is shown below
∑
=
=
k
nk
n V
V
1
ma = 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
n=1 1.00 0.90 0.80 0.70 0.60 0.50 o.40 0.30 0.20 0.10
n = mf 0.60 0.71 0.82 0.92 1.01 1.08 1.15 1.20 1.24 1.27
n=mf +2 0.32 0.27 0.22 0.17 0.13 0.09 0.06 0.03 0.02 0.00
Normalized Fourier Coefficients Vn/Vdc for Bipolar PWM
49. Example 3.5
The inverter has a resistive load of 10Ω and
inductive load of 25mH connected in series with
the fundamental frequency current amplitude of
9.27A. The THD of the inverter is not more than
9.27A. The THD of the inverter is not more than
10%. If at the beginning of designing the inverter,
the THD of the current is 16.7% which is does
not meet the specification, find the voltage
amplitude at the fundamental frequency, the
required DC input supply and the new THD of the
current.
50. Example 3.6
The single-phase full-bridge inverter is used to produce
a 60Hz voltage across a series R-L load using bipolar
PWM. The DC input to the bridge is 100V, the
amplitude modulation ratio is 0.8, and the frequency
amplitude modulation ratio is 0.8, and the frequency
modulation ratio is 21. The load has resistance of R =
10Ω and inductance L = 20mH. Determine:
(a) The amplitude of the 60Hz component of the output
voltage and load current.
(b) The power absorbed by the load resistor
(c) The THD of the load current
