Chapter
3
INVERTER
(Dc to AC Converter)
SCHOOL OF ELECTRICAL ENGINEERING
COLLEGE OF ENGINEERING
UNIVERSITI TEKNOLOGI MARA
Learning Objectives
◎ To study the operation and characteristic of Inverters.
◎ To learn switching technique for dc-ac converters
◎ To understand the performance parameters of
inverters.
◎ To study the effect of load impedance on the load
current.
◎ To learn the techniques for simulating the inverter
circuits.
3.1
Introduction
3.1 Introduction to Inverter
◎
◎
Converts DC to AC power by switching the
DC input voltage (or current)
in
.
a pre determined sequence to generate AC voltage (or current) output
Inverters are used in a wide range of applications, from small switching
power supplies in computers, to large electric utility applications that
transport bulk power.
General block diagram
3.1 Introduction to Inverter


In Malaysia power is generated at 50 Hz.


dc to ac power conversion makes use of static switches such as transistors or SCRs.



What if there need for a power source at some other frequency such as 400 Hz or 20 kHz.

The best way to obtain the power at a frequency other than that frequency is to first ac
power to dc and then reconvert to ac at a desired frequency.
MOSFETs and IGBTs are good candidates for the switches in dc-to-ac conversion
techniques.
The main drawback of dc-to-ac conversion is that the output voltage is not sinusoid
because of rich with harmonic components.
The harmonic components must be filtered out using high-frequency filters on the output
side.
There are many technique can be used to controlling the output voltage such as PWM,
SPWM, MPWM
3.1 Introduction to Inverter
TYPICAL APPLICATIONS:
1)
DC Power Utilization
2)
Uninterruptible power supply (UPS)
3)
Induction heating
4)
High voltage direct current (HVDC) power transmission
5)
Variable-speed ac motor drives
6)
Electric vehicle drives
2 Commonly used circuits for dc-ac conversion
•
Half-Bridge Inverter – satisfactory for low-power
application
•
Full-Bridge Inverter – allows the use of PWM
technique to control its output
2 Switching scheme can be considered:1)
Square-wave switching scheme (SW)
2)
Quasi-square wave switching scheme (QSW)
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TYPICAL INVERTER APPLICATIONS: DC POWER UTILIZATION
◎ An inverter allows the 12 or 24 volt
(battery) DC power available in an
automobile or from solar panels to
supply AC power to operate equipment
that is normally supplied from a main
power source.
◎ Inverters are also used to provide a
source of AC power from photovoltaic
solar cells and fuel cell power supplies.
Back
TYPICAL INVERTER APPLICATIONS: UNINTERUPTABLE POWER
SUPPLY (UPS)
One type of uninterruptible power supply
uses batteries to store power and an
inverter to supply AC power from the
batteries when main power is not
available. When main power is restored,
a rectifier is used to supply DC power to
recharge the batteries
Back
TYPICAL INVERTER APPLICATIONS: INDUCTION HEATING
Invertersare used to convert low
frequency main AC power to a
higher frequency for use in
induction heating. To do this, AC
power is first rectified to provide
DC power. The inverter then
changes the DC power to high
frequency ACpower.
Back
TYPICAL INVERTER APPLICATIONS: HVDC SYSTEM
Gurun HVDC Converter Station (Source from HVDC
Transmission, Gurun)
One-line diagram of HVDC
interconnection between Malaysia
and Thailand (Source from HVDC
Transmission, Gurun)
Back
TYPICAL INVERTER APPLICATIONS: VIARABLE-SPEED AC
MOTOR DRIVE
Back
TYPICAL INVERTER APPLICATIONS: VIARABLE-SPEED AC
MOTOR DRIVE
Back
3.1 Introduction to Inverter
INVERTER (DC  AC)
a. Forced commutation or standalone inverter
Single-phase Inverter:Half-bridge inverter
Full-bridge inverter
Output voltage: square, quasi-square & PWM
Three-phase Inverter
b. Grid Inverter (require synchronization process)
Line commutated Inverter
Commercial Inverter
3.1 Introduction to Inverter :Standalone Inverter
Voltage
Current
Time
AC Load
Inverter
Magnitude
Magnitude
DC Source
Voltage
Current
Time
3.1 Introduction to Inverter: Grid Inverter
DC Source
Inverter
TNB Power distribution
system
synchronization
3.1 Introduction to Inverter: Single-Phase Inverter
DC Source
Inverter
L
N
Single-phase output
voltage
3.1 Introduction to Inverter: Three-Phase Inverter
DC Source
Inverter
R
Y
B
Three-phase output
voltage
3.1 Introduction to Inverter: Multi-Stage Power Conversion
AC Input
AC Output
Rectifier
Inverter
3.1.1 Introduction to Inverter: Inverter Output Waveform
•
•
•
•
•
•
Inverter produces a square voltage waveform as opposed to the sinusoidal
waveform that is the usual waveform of an AC power supply.
Using Fourier analysis, periodic waveforms are represented as the sum of an
infinite series of sine waves.
The sine wave that has the same frequency as the original waveform is
called the fundamental component.
The other sine waves, called harmonics, that are included in the series have
frequencies that are integral multiples of the fundamental frequency.
The quality of the inverter output waveform can be expressed by using the
Fourier analysis data to calculate the
total harmonic distortion (THD).
The total harmonic distortion is the square root
of the sum of the squares of the harmonic
voltages divided by the fundamental voltage:
3.1.1 Introduction to Inverter: Inverter Output Waveform
3.1.1 Introduction to Inverter: Inverter Output Waveform
Half-Bridge
Inverter Circuit
Single-Supply
Square-wave
Dual-Supply
Square-wave
Square-wave
Full-Bridge
Quasi-Squarewave
PWM
3.2
Square-Wave
Switching Scheme
3.2.1 Switching Scheme topology : Square-wave (Single-Supply)
• In case there is only one available dc source, we can split its voltage
equally into two by using the circuit shown below.
• Once the two large electrolytic capacitors are fully charged, they
behave as the voltage sources.
• Two equal resistors in parallel with the capacitors not only ensure
that the voltages on the two capacitor are the same but also provide
the paths for the capacitors to discharge once the half-bridge
inverter is switched off.
3.2.1 Switching Scheme topology : Square-wave (Single-Supply)
Which switches are turn ON and OFF to produce +ve & -ve output voltage
?
3.2.1 Switching Scheme topology : Square-wave (Dual-Supply)
• It has two identical dc voltage sources connected in series, two
static switches, and two diodes.
• The diodes are needed to protect the switches especially when
these switches have to carry currents in the reverse directions.
• Since most single-phase loads are inductive in nature, we have
drawn the circuit with an inductive load.
3.2.1 Switching Scheme topology : Square-wave (Dual-Supply)
Which switches are turn ON and OFF to produce +ve & -ve output voltage
?
3.2.2 Single-Phase Half-Bridge Inverter (R load)
?
?
Square-wave
Single-supply
Dual-supply
1.
2.
How to determine the magnitude of the output voltage waveform ???
How to set output frequency???
3.2.2 Single-Phase Half-Bridge Inverter (R load)
Single-supply
for Single - supply;
- Supply voltage = Vs
- output voltage = stored voltage in capacitor (C1 @ C2 )
Vs
2
V
- Voltage stored in C2 = s
2
V 
Vo _ rms =  s  V
2
Vo _ ave = 0 V
- Voltage stored in C1 =
Dual-supply
for Dual - supply;
- output voltage = supply voltage (V1 @ V2 )
- V1 = voltage supply for + ve half cycle
- V2 = voltage supply for − ve half cycle
Vo _ rms = [V1 ] V
Vo _ ave = 0 V
Try to derive the average and
RMS output voltage
3.2.3 Single-Phase Half-Bridge Inverter (R-L load)
Square-wave
Single-supply
Dual-supply
3.2.3 Single-Phase Half-Bridge Inverter (R-L load) - Analysis
t1=first
zero
crossing
t2=second
zero
crossing
What is the differences of output current
waveform in previous slide with this slide ???
3.2.3 Single-Phase Half-Bridge Inverter (R-L load) - Analysis
for the analysis,
Vs = Vo (load voltage)
∴ better use 'Vo ' in writing/calculation
3.2.3 Single-Phase Half-Bridge Inverter (R-L load) - Analysis
for the analysis,
Vs = Vo (load voltage)
∴ better use 'Vo ' in writing/calculation
3.2.3 Single-Phase Half-Bridge Inverter (R-L load) - Analysis
for the analysis,
Vs = Vo (load voltage)
∴ better use 'Vo ' in writing/calculation
Example 3.1
• The half-bridge inverter is operating from a dual 48-V dc
source and supplies power to a single-phase motor whose
equivalent resistance and inductance are 2 Ω and 5 mH,
respectively. If the operating frequency is 2 kHz, determine
a) the minimum & maximum currents in the motor,
I max = 1.2 A
I min = − I max = −1.2 A
b) the expression of current for first half-cycle,
i (t ) = 24 − 25.2 ⋅ e −400t A
c) the time at which the current is zero during each half period,
t1 = 0.122 ms
d) the expression for the current during negative half-cycle.
i (t ) = −24 + 25.2 ⋅ e −400(t −0.25 ms ) A
3.2.4 Single-Phase Full-Bridge Inverter (R-L load) – Square-Wave
3.2.4 Single-Phase Full-Bridge Inverter (R-L load) – Square-Wave
[1] When switch Q1 and Q4 operated
3.2.4 Single-Phase Full-Bridge Inverter (R-L load) – Square-Wave
[2] When diode D2 and D3 conducted,
3.2.4 Single-Phase Full-Bridge Inverter (R-L load) – Square-Wave
[3] When switch Q2 and Q3 operated,
3.2.4 Single-Phase Full-Bridge Inverter (R-L load) – Square-Wave
[4] When diode D1 and D4 conducted,
3.2.4 Single-Phase Full-Bridge Inverter (R-L load) – Square-Wave
Output waveform - During steady State condition
4
1
2
3
4
3.3
Quasi -Square-Wave
Switching Scheme
3.3.1 & 3.3.2 Single-Phase Full-Bridge Inverter (R-L load)
Quasi-Square-Wave
Dead time
Delay angle/time
Vo _ rms =
Vs
π
Vo _ ave = 0 V
2
[π − 2α ]
V
Try to derive the average and
RMS output voltage
3.3.1 & 3.3.2 Single-Phase Full-Bridge Inverter (R-L load)
Quasi-Square-Wave
[1] When switch Q3 and diode D4 conducted,
3.3.1 & 3.3.2 Single-Phase Full-Bridge Inverter (R-L load)
Quasi-Square-Wave
[2] When diode D1 and D4 conducted,
3.3.1 & 3.3.2 Single-Phase Full-Bridge Inverter (R-L load)
Quasi-Square-Wave
[3] When switch Q1 and Q4 conducted,
3.3.1 & 3.3.2 Single-Phase Full-Bridge Inverter (R-L load)
Quasi-Square-Wave
[4] When switch Q1 and diode D2 conducted,
3.3.1 & 3.3.2 Single-Phase Full-Bridge Inverter (R-L load)
Quasi-Square-Wave
[5] When diode D2 and D3 conducted,
3.3.1 & 3.3.2 Single-Phase Full-Bridge Inverter (R-L load)
Quasi-Square-Wave
[6] When switch Q2 and Q3 conducted,
Inverter output : Harmonic Analysis
-
The output voltage as
shown will have harmonics
components in addition to
the fundamental
components.
-
The square wave output
voltage having a significant
low-order of harmonic
component as shown in
the frequency spectrum.
-
The harmonics
components can be filtered
out by placing a filter on
the output side.
Inverter output : Harmonic Analysis
The instantaneous output voltage can be expressed in Fourier Series,
ao ∞
Vo (ωt ) = + ∑ (an cos(nωt ) + bn sin( nωt )
2 n =1
•
For square (rectangular) waveform/quarter-wave symmetry along X-axis,
ao = 0
an = 0
∞
∴Vo (ωt ) = ∑ (bn sin( nωt )
n =1
Inverter output : Harmonic Analysis (continued…)
bn can be obtained from,
bn =
1
π
0
∫π − V
o
sin( nωt ) d (ωt ) +
−
0
1
π
π
∫V
o
sin( nωt ) d (ωt )
0
π
V  cos(nωt ) 
Vo  − cos(nωt ) 
+
bn = o 


π 
π 
n
n
−π
0
V
bn = o [(1) − (−1)] + [(1) − (−1)]
nπ
4V
bn = o
nπ
4Vo
πn
4V
bn = o (cos(nα ))
πn
bn =
Vo=load voltage
(Square - wave)
(Quasi - Square - wave)
Inverter output : Harmonic Analysis (continued…)
RMS output voltage of nth harmonic order,
bn
Von =
2
Total Harmonic Distortion (THD) of output voltage
n
THD =
2
V
(
)
∑ on
n =3
Vo1
×100%
3.4
Pulse Width Modulation(PWM)
Switching Scheme
3.4.1 Introduction of PWM
Provides a way to decrease the THD of load current.

Advantages:
i. Amplitude of the output voltage w/f can be easily
controlled by varying the modulating w/f of PWM
ii. low order harmonics minimized
iii. High order harmonics filtered by load inductance
iv. Economical control

Control of the switches for sinusoidal PWM output requires
i. a reference signal/ modulating/ control signal – a sinusoid
ii. a carrier signal – a triangular (controls the switching freq)

PWM Switching scheme
i. Bipolar – output is alternates between +ve and –ve
ii. Unipolar – output is switched from either high to zero, or low to zero

Disadvantage
i) More complex control circuits are required for the switches
ii) Increase losses due to frequency switching.
3.4.1 Introduction of PWM
Triangulation method (Natural sampling)
- Amplitudes of the triangular wavecarrier)
(
and
sine wave modulating)
(
are compared to obtain
PWM waveform. Simple analogue comparator
can be used.
Two type of switching scheme
- Bipolar
- Unipolar
3.4.1 Introduction of PWM
Modulation index, mi
mi =
Vcontrol
Vtriangle
Frequency Modulation, m f
mf =
f triangle
f carrier
Fundamental output magnitude, Vo1
Vo1 = mi × Vd
Fundamental output magnitude, Vo1
f output = f control
Design Equation
3.4.1 Introduction of PWM
3.4.1 Introduction of PWM
3.4.1 Introduction of PWM
3.4.1 Introduction of PWM
3.4.1 Introduction of PWM
3.4.1 Introduction of PWM
3.4.1 Introduction of PWM
3.4.2 Unipolar PWM Switching Scheme
•
•
•
•
Two comparison (VC1-VT) & (VC2-VT)
Each leg of H-bridge driven
independently.
3-level output
Less harmonic distortion than
bipolar PWM
3.4.2 Unipolar PWM Switching Scheme
Vm intersect Vcr at ‘V’
produce Va=+VAN
Vm- intersect Vcr at ‘V’
produce Vb=+VBN
Output= VAN - VBN
Thank You