Transfers power from DC Circuit to another AC Circuit at required voltage and required frequency.

**APPLICATIONS OF INVERTER**

The applications areas of inverters such as:-

•Adjustable-speed ac motor drives

•Uninterrupted power supplies (UPS)

•Running appliances of ac used in an automobile battery.

•power transmission industry such as reactive power controllers and adaptive power filters

**TYPES OF INVERTER**

#### CURRENT SOURCE INVERTER (VSI)

- Unidirectional and bipolar
- Good Short Circuit Protection
- Load current depend on source current
- Load Voltage depends on load

**VOLTAGE SOURCE INVERTER (CSI)**

- Unipolar and Bidirectional
- Short Circuit Damages
- Load Voltage Depends on Source Voltage
- Load current depends on load

**SINGLE PHASE HALF BRIDGE VOLTAGE SOURCE INVERTER**

**OPERATION**

- Switch S1 is ON for the time duration of 0<t<T/2
- Switch S2 is ON for the time duration of T/2<t<T
- When Switch S1 is turned ON the instantaneous voltage across the load Vo = Vdc/2.
- When Switch S2 is turned ON the instantaneous voltage across the load Vo = -Vdc/2
- Antiparallel diode does not change voltage polarity but reverses current direction.

**QUADRANT OPERATION OF SWITCH AND DIODE OF SINGLE PHASE HALF BRIDGE VSI**

**QUADRANT OPERATION OF SWITCH AND DIODE OF SINGLE PHASE HALF BRIDGE VSI**

•Switch (S1 & S2) Provides Operation in 1^{st} and 3^{rd} quad

•Diode (D1 & D2) Provides Operation in 2^{nd} and 4^{th} quad

- Ig1 and Ig2 is called a gate pulse which is used for turned on the switch S1 and S2
- In case of Resistive load during the period of 0<t<T/2 the output voltage across the resistive load is Vo = Vdc/2 and during the period of T/2<t<T the Vo = -Vdc/2.
- In case of Resistive load during the period of 0<t<T/2 the current flowing through the resistive load is Io = Vdc/2R and during the period of T/2<t<T the Io = -Vdc/2R.
- In case of purely (inductance load) L load current Io symmetric about t-axis so that dc component = 0 and current is linearly from minimum peak current (-Ip) to maximum peak current (+Ip).In this case diode D1 is conduct in 0<t<T/4, switch S1 is conduct in T/4<t<T/2, Diode D2 is conduct in T/2<t<3T/4 and switch S2 is conduct in 3T/4<t<T.

•L load (0<t<T/2)

V=Vdc=Ldi/dt

L[Ip-(-Ip)]/T/2=Vdc/2

Ip=Vdc/8fL

In case of R-L load exponentially rise from (-Ip to +Ip). In this case diode D1 is conduct in 0<t<T/4, switch S1 is conduct in T/4<t<T/2, Diode D2 is conduct in T/2<t<3T/4 and switch S2 is conduct in 3T/4<t<T.

- For any type of Load,
**Output Voltage waveform will remain same**but Current waveform depends on the nature of the load.

Or

Output Voltage waveform is Half Wave Symmetric hence **all even harmonics are absent**.