# What is a Voltage Source Inverter : Circuit & Its Working

The term inverter in power electronics refers to a device called a converter, which converts direct current (DC) power at a particular frequency to alternating current at another frequency using solid-state electronics. There are 2 traditional approaches for converting a static ac frequency, like cyclo converter and rectifier inverter approaches. A cyclo converter converts directly DC power at a particular frequency to AC power at another frequency, whereas a rectifier inverter converts AC power firstly to a DC power and converts DC power to AC power at variable frequency. A rectifier inverter consists of a rectifier and an inverter. These inverters can be constructed in any of 2 techniques like external commutation and self-commutation. The external commutation inverters, acquire sources externally from motors or power supply and the self-commutated inverters control the circuit with the help of capacitor function. Self-commutated inverters are classified as current source inverters and voltage source inverters. This article gives an overview of a voltage source inverter.

## What is Voltage Source Inverter?

**Definition:** A voltage source inverter or VSI is a device that converts unidirectional voltage waveform into a bidirectional voltage waveform, in other words, it is a converter that converts its voltage from DC form to AC form. An ideal voltage source inverter keeps the voltage constant through-out the process.

### Construction

A VSI usually consists of a DC voltage source, voltage source, a transistor for switching purposes, and one large DC link capacitor. A DC voltage source can be a battery or a dynamo, or a solar cell, a transistor used maybe an IGBT, BJT, MOSFET, GTO. VSI can be represented in 2 topologies, are single-phase and a 3-phase inverter, where each phase can be further classified into a Half-bridge inverter and full-bridge inverter.

#### Single Phase Half Bridge Voltage Source Inverter

It consists of 1 DC voltage source, 4 transistors S1, S2, S3, S4, and 4 anti-parallel diodes D1, D2, D3, D4 for switching purpose and one large DC link capacitor “C” as shown below

#### 3 Phase Full Bridge Voltage Source Inverter

It consists of 6 transistors with T1, T2, T3, T4, T5, T6, 6 anti-parallel diodes like D1, D2, D3, D4, D5, D6, 3 load terminals, one DC source, and one large DC linked capacitor, and a thyristor is connected along with commutation circuit. The three outputs like “ABC”, where “A” is connected to T1 & T4, “B” is connected to T3 and T6, and “C” is connected to T5 and T2. These ABCs are in-turn connected to a 3 phase balanced load.

A balanced load consists of 2 main components a source and a load, where a balanced source implies phase and magnitude are equal and are phase-shifted by 120 degrees. According to the KCL principle, the balanced load implies all the load impedances in all the 3 phases are equal in magnitude and phase. The thyristors T1, T3, T5 supply current to the load or act as forwarding path, whereas thyristors T6, T4, T2 carry the current back to the source and acts like return path, as shown below.

### Working of Single-Phase Voltage Source Inverter

A voltage source inverter can operate in any of 2 conduction mood, i.e,

- 180 degree and
- 120degree conduction mood.

Let us consider the scenario of 180-degree conduction mode in a three-phase inverter. The three-phase inverter is represented in 180-degree conduction mode because both switches S1 and S2 conduct at 180 degrees. Whereas in a full-bridge voltage source inverter all the 4 switches S1, S2, S3, S4 conducts at 180 degrees. Considering the circuit diagram shown below, the switch T1 and T4 are connected to a phase, out of which T1 and T4 conduct at 180 degrees each, where the total duration is given as 1800 + 1800 = 3600. If both the switches conduct together it may lead to a short circuit. From the graph shown below

Where X-axis is wt and Y-axis is amplitude, from the graph we can observe that

- T1 conducts from 0 to 1800
- T4 conducts from 1800 to 3600
- For a balanced output, A phase and B phase must have a phase shift of 1200 That is when angle A is 00, angle B should be 1200 and angle C should be -2400.
- Therefore, T3 will start conducting from 1200 to 3000. After this phase, T6 starts conducting from 3000 to 3600. T6 also conducting from 00 to 1200
- T5 starts conducting after 2400 to 3600 and also conducts from 00 to 600
- T2 conducts from 600 to 2400.
- From the above graph, we can conclude that at a time 3SCR switches are conducting.

**Case1:** From 0 to 600 T1, T6, and T5 conduct, out of which T1 and T5 carry the current into the load, and T6 carries the current out of the load.

**V _{AN} = IZ ……..1**

**V _{CN} = IZ …….. 2**

**V _{BN} = – 2 IZ ……3**

**V _{AB} = 3 IZ = V_{CB} = Vdc ………4**

**IZ = V _{dc} / 3……5**

Substituting the value of IZ in 1,2,3 we get

**V _{AN} = V_{dc} / 3 ……..6**

**V _{CN} = V_{dc} / 3 …….. 7**

**V _{BN} = – 2 V_{dc} / 3 ….8**

**Case2:** From 600 to 1200, the devices T1, T6, T2 conducts, out of which T1 carries the current into the load, and T2 and T6 carry the current out of the load.

**V _{AN} = 2 IZ ……..9**

**V _{BN} = – IZ …….. 10**

**V _{CN} = – IZ ……11**

**V _{AB} = 3 IZ = V_{CB} = Vdc ………12**

**IZ = V _{dc} / 3……13**

Substituting the value of IZ in 9, 10, 11 we get

**V _{AN} = 2 V_{dc} / 3 ……..14**

**V _{CN} = – V_{dc} / 3 …….. 15**

**V _{BN} = – V_{dc} / 3 ….16**

**Case3:** From 1200 to 1800, the devices T1, T3, T2 conducts, out of which T1and T3 carry the current into the load, and T2 carries the current out of the load.

**V _{AN} = IZ ……..17**

**V _{BN} = IZ …….. 18**

**V _{CN} = – 2IZ ……19**

**V _{AB} = 3 IZ = V_{CB} = Vdc ………20**

**IZ = V _{dc} / 3……21**

Substituting the value of IZ in 17, 18, 19 we get

**V _{AN} = V_{dc} / 3 ……..22**

**V _{CN} = – 2 V_{dc} / 3 …….. 23**

**V _{BN} = V_{dc} / 3 ….24**

**Case4:** From 1800 to 2400, the devices T4, T3, T2 conducts, out of which T3 carries the current into the load and T4 and T2 carries the current out of the load.

*V _{AN} = – IZ ……..25*

*V _{BN} = – IZ …….. 26*

*V _{CN} = – 2IZ ……27*

*V _{AB} = 3 IZ = V_{CB} = Vdc ………28*

*IZ = V _{dc} / 3……29*

Substituting the value of IZ in 17, 18, 19 we get

**V _{AN} = – V_{dc} / 3 ……..30**

**V _{CN} = – V_{dc} / 3 …….. 31**

**V _{BN} = -2 V_{dc} / 3 ….32.**

### Waveforms

The following are the waveforms obtained from the above equations

- The waveform for the A-phase
- Waveform for VB
- Waveform of VCN

Line phase voltages waveforms are given as

- The waveform of VAB = VAN – VBN
- Waveform of VBC
- Waveform of VCA

#### Formulas

**Phase voltage b _{n} = 2 / 2π [ _{0}∫^{π/3} V_{dc} /3 Sin w_{0} t d (w_{0} t) +_{π/3}∫^{2π/3} 2 V_{dc} /3 Sin w_{0} t d (w_{0} t) + _{2} _{π/3} ∫^{π} V_{dc} /3 Sin w_{0} t d (w_{0} t) – _{π}∫^{4π/3} V_{dc} /3 Sin w_{0} t d (w_{0} t) – _{4π/12}∫^{5π/3} 2V_{dc} /3 Sin w_{0} t d (w_{0} t) — _{5π/3}∫^{2π} V_{dc} /3 Sin w_{0} t d (w_{0} t)]**

on solving the above equation, we obtain^{ }

**b _{n} = 4 V_{dc }/ 3nπ [ 1 + sin nπ/6 + sin nπ/2]**

**V _{AB} = ∑_{n=6k (+/-) 1} 4V_{dc }/nπ Cos (nπ/6) Sin ( nw_{0}t + nπ/6)**

**V _{BC} = ∑_{n=6k(+/-)1} 4V_{dc }/nπ Cos (nπ/6) Sin ( nw_{0}t – nπ/2)**

**V _{CA} = ∑_{n=6k(+/-)1} 4V_{dc }/nπ Cos (nπ/6) Sin ( nw_{0}t – 7nπ/6)**

#### Differences Between Current Source and Voltage Source Inverter

The differences between the current source and the voltage source inverter are represented in the table below

Current Source Inverter |
Voltage Source Inverter |

A stiff current source is provided along with the inverter | A stiff voltage source is provided along with the inverter |

The high internal impedance of a DC source | The low internal impedance of a DC source |

Absence of feedback diodes |
Requires feedback diodes |

Output voltage varies due to a change in load |
Output voltage varies slightly due to the capacitor. |

Simple in construction | Complicated in construction |

Short circuit protection is possible | Short circuit protection is not possible |

High output impedance | Low output impedance |

Examples: Capacitor commutated current source inverter and auto sequential commutated inverter (ASCI). |
Examples: Half-bridge, Full bridge, square wave, and pulse width modulated inverters |

#### Advantages

The following are the advantages of voltage source inverter

- Occupies less area
- The output voltage is independent of the load that is used
- Uses simple logic
- More than one motors can be operated with a single voltage source inverter
- Design range up to 500 Hz

#### Disadvantages

The following are the disadvantages of voltage source inverter

- Not safe, when a short circuit occurs
- Less speed
- The input power factor is less I,e 1s.

#### Applications

The following are the applications of voltage source inverter

- Uninterruptable power supply
- AC speed drivers
- Filters
- Electronic frequency changer circuits.

Thus, an inverter is a device that converts DC to AC. Self-commutated inverters are classified as current source inverters and voltage source inverters. A voltage source inverter is a device that converts its voltage from DC form to AC form. It can be represented in a single phase or in 3 phases. The following article explains about 3 phase VSI and its working. Where each phase experiences 180 degrees and 120degree conduction mood. The main advantage is that the output voltage is independent of the load that is used.