INVERTERS – IN SEARCH OF POWER

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.

Fig.(A):- Single Phase Half Bridge Voltage Source Inverter

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

**Fig.(B) :- Quadrant Operation Of Switches and Diodes**

•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.

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

**Fig.(E) :- Output voltage (Vo) waveform of any load.**

**Fig.(C) :- Waveform of Output Voltage (Vo) and Output Current (Io)**
**In case of purely resistive load**

**Fig.(D) :- Waveform of Output Current (Io)**
**In case of R load, L load and R-L load.**

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