Page 1
INVERTERS
Introduction to Inverters
The word „inverter? in the context of power-electronics denotes a class of power conversion (or power
conditioning) circuits that operates from a dc voltage source or a dc current source and converts it into ac
voltage or current. The inverter does reverse of what ac-to-dc converter does (refer to ac to dc converters).
Even though input to an inverter circuit is a dc source, it is not uncommon to have this dc derived from an
ac source such as utility ac supply. Thus, for example, the primary source of input power may be utility ac
voltage supply that is converted to dc by an ac to dc converter and then „inverted? back to ac using an
inverter. Here, the final ac output may be of a different frequency and magnitude than the input ac of the
utility supply
A single phase Half Bridge DC-AC inverter is shown in Figure below
Figure: 5.1 Single phase Half Bridge DC-AC inverter with R load
The analysis of the DC-AC inverters is done taking into accounts the following assumptions and
conventions.
1) The current entering node a is considered to be positive.
2) The switches S1 and S2 are unidirectional, i.e. they conduct current in one direction.
3) The current through S1 is denoted as i1 and the current through S2 is i2.
The switching sequence is so design is shown in Figure below. Here, switch S1 is on for the time
duration 0 = t = T1 and the switch S2 is on for the time duration T1 = t = T2. When switch S1 is turned
on, the instantaneous voltage across the load is ? o = Vin/ 2
When the switch S2 is only turned on, the voltage across the load is
? o = ? Vin/ 2.
Page 2
INVERTERS
Introduction to Inverters
The word „inverter? in the context of power-electronics denotes a class of power conversion (or power
conditioning) circuits that operates from a dc voltage source or a dc current source and converts it into ac
voltage or current. The inverter does reverse of what ac-to-dc converter does (refer to ac to dc converters).
Even though input to an inverter circuit is a dc source, it is not uncommon to have this dc derived from an
ac source such as utility ac supply. Thus, for example, the primary source of input power may be utility ac
voltage supply that is converted to dc by an ac to dc converter and then „inverted? back to ac using an
inverter. Here, the final ac output may be of a different frequency and magnitude than the input ac of the
utility supply
A single phase Half Bridge DC-AC inverter is shown in Figure below
Figure: 5.1 Single phase Half Bridge DC-AC inverter with R load
The analysis of the DC-AC inverters is done taking into accounts the following assumptions and
conventions.
1) The current entering node a is considered to be positive.
2) The switches S1 and S2 are unidirectional, i.e. they conduct current in one direction.
3) The current through S1 is denoted as i1 and the current through S2 is i2.
The switching sequence is so design is shown in Figure below. Here, switch S1 is on for the time
duration 0 = t = T1 and the switch S2 is on for the time duration T1 = t = T2. When switch S1 is turned
on, the instantaneous voltage across the load is ? o = Vin/ 2
When the switch S2 is only turned on, the voltage across the load is
? o = ? Vin/ 2.
Figure: 5.2 Single phase Half Bridge DC-AC inverter output waveforms
The r.m.s value of output voltage ? o is given by,
The instantaneous output voltage ? o is rectangular in shape. The instantaneous value of ? o can be
expressed in Fourier series as,
Due to the quarter wave symmetry along the time axis , the values of a0 and an are zero. The value of bn
is given by,
Substituting the value of bn from above equation , we get
Page 3
INVERTERS
Introduction to Inverters
The word „inverter? in the context of power-electronics denotes a class of power conversion (or power
conditioning) circuits that operates from a dc voltage source or a dc current source and converts it into ac
voltage or current. The inverter does reverse of what ac-to-dc converter does (refer to ac to dc converters).
Even though input to an inverter circuit is a dc source, it is not uncommon to have this dc derived from an
ac source such as utility ac supply. Thus, for example, the primary source of input power may be utility ac
voltage supply that is converted to dc by an ac to dc converter and then „inverted? back to ac using an
inverter. Here, the final ac output may be of a different frequency and magnitude than the input ac of the
utility supply
A single phase Half Bridge DC-AC inverter is shown in Figure below
Figure: 5.1 Single phase Half Bridge DC-AC inverter with R load
The analysis of the DC-AC inverters is done taking into accounts the following assumptions and
conventions.
1) The current entering node a is considered to be positive.
2) The switches S1 and S2 are unidirectional, i.e. they conduct current in one direction.
3) The current through S1 is denoted as i1 and the current through S2 is i2.
The switching sequence is so design is shown in Figure below. Here, switch S1 is on for the time
duration 0 = t = T1 and the switch S2 is on for the time duration T1 = t = T2. When switch S1 is turned
on, the instantaneous voltage across the load is ? o = Vin/ 2
When the switch S2 is only turned on, the voltage across the load is
? o = ? Vin/ 2.
Figure: 5.2 Single phase Half Bridge DC-AC inverter output waveforms
The r.m.s value of output voltage ? o is given by,
The instantaneous output voltage ? o is rectangular in shape. The instantaneous value of ? o can be
expressed in Fourier series as,
Due to the quarter wave symmetry along the time axis , the values of a0 and an are zero. The value of bn
is given by,
Substituting the value of bn from above equation , we get
The current through the resistor ( iL ) is given by,
Half Bridge DC-AC Inverter with L Load and R-L Load
The DC-AC converter with inductive load is shown in Figure below. For an inductive load, the load
current cannot change immediately with the output voltage.
Figure: 5.3 Single phase Half Bridge DC-AC inverter with RL load
The working of the DC-AC inverter with inductive load is as follow is:
Case 1: In the time interval 0<=t<= T1 the switch S1 is on and the current flows through the inductor
from points a to b. When the switch S1 is turned off (case 1) at t-T1, the load current would continue to
flow through the capacitor C2 and diode D2 until the current falls to zero, as shown in Figure below.
Figure: 5.4 Single phase Half Bridge DC-AC inverter with L load
Page 4
INVERTERS
Introduction to Inverters
The word „inverter? in the context of power-electronics denotes a class of power conversion (or power
conditioning) circuits that operates from a dc voltage source or a dc current source and converts it into ac
voltage or current. The inverter does reverse of what ac-to-dc converter does (refer to ac to dc converters).
Even though input to an inverter circuit is a dc source, it is not uncommon to have this dc derived from an
ac source such as utility ac supply. Thus, for example, the primary source of input power may be utility ac
voltage supply that is converted to dc by an ac to dc converter and then „inverted? back to ac using an
inverter. Here, the final ac output may be of a different frequency and magnitude than the input ac of the
utility supply
A single phase Half Bridge DC-AC inverter is shown in Figure below
Figure: 5.1 Single phase Half Bridge DC-AC inverter with R load
The analysis of the DC-AC inverters is done taking into accounts the following assumptions and
conventions.
1) The current entering node a is considered to be positive.
2) The switches S1 and S2 are unidirectional, i.e. they conduct current in one direction.
3) The current through S1 is denoted as i1 and the current through S2 is i2.
The switching sequence is so design is shown in Figure below. Here, switch S1 is on for the time
duration 0 = t = T1 and the switch S2 is on for the time duration T1 = t = T2. When switch S1 is turned
on, the instantaneous voltage across the load is ? o = Vin/ 2
When the switch S2 is only turned on, the voltage across the load is
? o = ? Vin/ 2.
Figure: 5.2 Single phase Half Bridge DC-AC inverter output waveforms
The r.m.s value of output voltage ? o is given by,
The instantaneous output voltage ? o is rectangular in shape. The instantaneous value of ? o can be
expressed in Fourier series as,
Due to the quarter wave symmetry along the time axis , the values of a0 and an are zero. The value of bn
is given by,
Substituting the value of bn from above equation , we get
The current through the resistor ( iL ) is given by,
Half Bridge DC-AC Inverter with L Load and R-L Load
The DC-AC converter with inductive load is shown in Figure below. For an inductive load, the load
current cannot change immediately with the output voltage.
Figure: 5.3 Single phase Half Bridge DC-AC inverter with RL load
The working of the DC-AC inverter with inductive load is as follow is:
Case 1: In the time interval 0<=t<= T1 the switch S1 is on and the current flows through the inductor
from points a to b. When the switch S1 is turned off (case 1) at t-T1, the load current would continue to
flow through the capacitor C2 and diode D2 until the current falls to zero, as shown in Figure below.
Figure: 5.4 Single phase Half Bridge DC-AC inverter with L load
Case 2: Similarly, when S2 is turned off at t = T1 , the load current flows through the diode D1 and
capacitor C1until the current falls to zero, as shown in Figure below.
Figure: 5.5 Single phase Half Bridge DC-AC inverter with L load
When the diodes D1 and D2 conduct, energy is feedback to the dc source and these diodes are known as
feedback diodes. These diodes are also known as freewheeling diodes. The current for purely inductive
load is given by,
Similarly, for the R – L load. The instantaneous load current is obtained as,
Where,
Operation of single phase full bridge inverter
A single phase bridge DC-AC inverter is shown in Figure below. The analysis of the single phase DC-AC
inverters is done taking into account following assumptions and conventions.
1) The current entering node a in Figure 8 is considered to be positive.
2) The switches S1, S2, S3 and S4 are unidirectional, i.e. they conduct current in one direction.
Page 5
INVERTERS
Introduction to Inverters
The word „inverter? in the context of power-electronics denotes a class of power conversion (or power
conditioning) circuits that operates from a dc voltage source or a dc current source and converts it into ac
voltage or current. The inverter does reverse of what ac-to-dc converter does (refer to ac to dc converters).
Even though input to an inverter circuit is a dc source, it is not uncommon to have this dc derived from an
ac source such as utility ac supply. Thus, for example, the primary source of input power may be utility ac
voltage supply that is converted to dc by an ac to dc converter and then „inverted? back to ac using an
inverter. Here, the final ac output may be of a different frequency and magnitude than the input ac of the
utility supply
A single phase Half Bridge DC-AC inverter is shown in Figure below
Figure: 5.1 Single phase Half Bridge DC-AC inverter with R load
The analysis of the DC-AC inverters is done taking into accounts the following assumptions and
conventions.
1) The current entering node a is considered to be positive.
2) The switches S1 and S2 are unidirectional, i.e. they conduct current in one direction.
3) The current through S1 is denoted as i1 and the current through S2 is i2.
The switching sequence is so design is shown in Figure below. Here, switch S1 is on for the time
duration 0 = t = T1 and the switch S2 is on for the time duration T1 = t = T2. When switch S1 is turned
on, the instantaneous voltage across the load is ? o = Vin/ 2
When the switch S2 is only turned on, the voltage across the load is
? o = ? Vin/ 2.
Figure: 5.2 Single phase Half Bridge DC-AC inverter output waveforms
The r.m.s value of output voltage ? o is given by,
The instantaneous output voltage ? o is rectangular in shape. The instantaneous value of ? o can be
expressed in Fourier series as,
Due to the quarter wave symmetry along the time axis , the values of a0 and an are zero. The value of bn
is given by,
Substituting the value of bn from above equation , we get
The current through the resistor ( iL ) is given by,
Half Bridge DC-AC Inverter with L Load and R-L Load
The DC-AC converter with inductive load is shown in Figure below. For an inductive load, the load
current cannot change immediately with the output voltage.
Figure: 5.3 Single phase Half Bridge DC-AC inverter with RL load
The working of the DC-AC inverter with inductive load is as follow is:
Case 1: In the time interval 0<=t<= T1 the switch S1 is on and the current flows through the inductor
from points a to b. When the switch S1 is turned off (case 1) at t-T1, the load current would continue to
flow through the capacitor C2 and diode D2 until the current falls to zero, as shown in Figure below.
Figure: 5.4 Single phase Half Bridge DC-AC inverter with L load
Case 2: Similarly, when S2 is turned off at t = T1 , the load current flows through the diode D1 and
capacitor C1until the current falls to zero, as shown in Figure below.
Figure: 5.5 Single phase Half Bridge DC-AC inverter with L load
When the diodes D1 and D2 conduct, energy is feedback to the dc source and these diodes are known as
feedback diodes. These diodes are also known as freewheeling diodes. The current for purely inductive
load is given by,
Similarly, for the R – L load. The instantaneous load current is obtained as,
Where,
Operation of single phase full bridge inverter
A single phase bridge DC-AC inverter is shown in Figure below. The analysis of the single phase DC-AC
inverters is done taking into account following assumptions and conventions.
1) The current entering node a in Figure 8 is considered to be positive.
2) The switches S1, S2, S3 and S4 are unidirectional, i.e. they conduct current in one direction.
Figure: 5.6 Single phase Full Bridge DC-AC inverter with R load
When the switches S1 and S2 are turned on simultaneously for a duration 0 = t = T1 , the the input
voltage Vin appears across the load and the current flows from point a to b.
Q1 – Q2 ON, Q3 – Q4 OFF ==> ? o = Vs
Figure: 5.7 Single phase Full Bridge DC-AC inverter with R load
If the switches S3 and S4 turned on duration T1 = t = T2, the voltage across the load the load is reversed
and the current through the load flows from point b to a.
Q1 – Q2 OFF, Q3 – Q4 ON ==> ? o = -Vs
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