Operation with Mode Diagram

1. 3 Phase Inverter
Prepared by Malarselvam

2. Circuit Diagram

4. Three Phase Inverter- 180 Degree
Conduction Mode
• let’s start switching sequence by closing the switch S1 in the first segment
of the ideal circuit and let’s name the start as 00. Since the selected time
of conduction is 1800 the switch S1 will be closed from 0º to 1800.

5. • But after 1200of the first phase, the second phase will also have a positive
cycle as seen in the three-phase voltage graph, so switch S3 will be closed
after S1. This S3 will also be kept closed for another 1800. So S3 will be
closed from 1200 to 3000 and it will be open only after 3000.

8. Step1: (for 0-60) S1, S4&S5 are closed while the remaining three switches are
open. In such a case, the simplified circuit can be as shown below.
So for 0 to 60: Vao = Vco= Vs/3 ; Vbo = -2Vs/3
By using these we can derive the line voltages as:
Vab = Vao – V bo = Vs
Vbc = Vbo – Vco = -Vs
Vca = Vco – Vao = 0

9. Step2: (for 60 to 120) S1, S4&S6 are closed while the remaining three
switches are open. In such a case, the simplified circuit can be as shown
below.
So for 60 to 120: Vbo = Vco= -Vs/3 ; Vao = 2Vs/3
By using these we can derive the line voltages as:
Vab = Vao – Vbo = Vs
Vbc = Vbo – Vco = 0
Vca = Vco – Vao = -Vs

10. Step3: (for 120 to 180) S1, S3&S6 are closed while the remaining three
switches are open. In such a case, the simplified circuit can be drawn as
below.
So for 120 to 180: Vao = Vbo= Vs/3 ; Vco = -2Vs/3
By using these we can derive the line voltages as:
Vab = Vao – Vbo = 0
Vbc = Vbo – Vco = Vs
Vca = Vco – Vao = -Vs

11. Similarly, we can derive the phase voltages and line voltages for
the next steps in the sequence. And it can be shown as the figure
given below:
Time
Period
Switches
in On
State
Vao Vbo Vco Vab Vbc Vca
0 – 60 S1, S4, S5 Vs/3 -2Vs/3 Vs/3 Vs -Vs 0
60 – 120 S1, S4, S6 2Vs/3 -Vs/3 -Vs/3 Vs 0 -Vs
120 – 180 S1,S3,S6 Vs/3 Vs/3 -2Vs/3 0 Vs -Vs
180 – 240 S2,S3,S6 -Vs/3 2Vs/3 -Vs/3 -Vs Vs 0
240 – 300 S2, S3,S5 -2Vs/3 Vs/3 Vs/3 -Vs 0 Vs
300 – 360 S2,S4,S5 -Vs/3 -Vs/3 2Vs/3 0 -Vs Vs

14. Three Phase Inverter- 120 Degree
Conduction Mode

15. The 120º mode is similar to 180º at all aspects except the closing time of each switch is reduced
to 120, which were 180 before. As usual, let’s start switching sequence by closing the switch S1 in
the first segment and be the start number to 0º. Since the selected time of conduction is 120º
the switch S1 will be opened after 120º, so the S1 was closed from 0º to 120º.

16. Now after 120º of the first phase, the second phase will also have a positive
cycle as mentioned before, so switch S3 will be closed after S1. This S3 will
also be kept closed for another 120º. So S3 will be closed from 120º to 240º.

17. Similarly, the third phase also has a positive cycle after 120º of the second phase
positive cycle so the switch S5 will be closed after 120º of S3 closing. Once the switch
is closed, it will be kept closed for coming 120º before being opened and with that,
the switch S5 will be closed from 240º to 360º

18. This cycle of symmetrical switching will be continued for achieving the desired three-
phase voltage. If we fill in the beginning and ending switching sequence in the above
table we will have a complete switching pattern for 120º conduction mode as below.

19. From the above table we can understand that:
From 0-60: S1&S4 are closed while remaining switches are opened.
From 60-120: S1 &S6 are closed while remaining switches are opened.
From 120-180: S3&S6 is closed while remaining switches are opened.
From 180-240: S2&S3 are closed while remaining switches are opened
From 240-300: S2&S5 are closed while remaining switches are opened
From 300-360: S4&S5 are closed while remaining switches are opened
And this sequence of steps goes on like that. Now let us draw the
simplified circuit for each step to better understand the current flow
and voltage parameters of the 3 Phase Inverter circuit.

20. Step1: (for 0-60) S1, S4 are closed while the remaining four switches are
open. In such a case, the simplified circuit can be shown as below.
So for 0 to 60: Vao = Vs/2, Vco= 0 ; Vbo = -Vs/2
By using these we can derive the line voltages as:
Vab = Vao – Vbo = Vs
Vbc = Vbo – Vco = -Vs/2
Vca = Vco – Vao = -Vs/2

21. Step2: (for 60 to 120) S1 &S6 are closed while the remaining
switches are open. In such a case, the simplified circuit can be
shown as below.
So for 60 to 120: Vbo =0, Vco= -Vs/2 & Vao = Vs/2
By using these we can derive the line voltages as:
Vab = Vao – Vbo = Vs/2
Vbc = Vbo – Vco = Vs/2
Vca = Vco – Vao = -Vs

22. Step3: (for 120 to 180) S3&S6 are closed while the remaining switches are
open. In such a case, the simplified circuit can be shown as below.
So for 120 to 180: Vao =0, Vbo= Vs/2 & Vco = -Vs/2
By using these we can derive the line voltages as:
Vab = Vao – V bo = -Vs/2
Vbc = Vbo – Vco = Vs
Vca = Vco – Vao = -Vs/2

23. Similarly, we can derive the phase voltages and line voltages for
the next steps in the sequence. And it can be shown as the figure
given below:
Time
Period
Switches
in On
State
Va0 Vb0 Vc0 Vab Vbc Vca
0 – 60 S1, S4, S5 Vs/2 -Vs/2 0 Vs -Vs/2 -Vs/2
60 – 120 S1, S4, S6 Vs/2 0 -Vs/2 Vs/2 Vs/2 -Vs
120 – 180 S1,S3,S6 0 Vs/2 -Vs/2 -Vs/2 Vs -Vs/2
180 – 240 S2,S3,S6 -Vs/2 Vs/2 0 -Vs Vs/2 Vs/2
240 – 300 S2, S3,S5 -Vs/2 0 Vs/2 -Vs/2 -Vs/2 Vs
300 – 360 S2,S4,S5 0 -Vs/2 Vs/2 Vs/2 -Vs Vs/2

24. Phase Voltage

25. Line Voltage

26. Conclusion
• It can be seen in the output graphs of both 180o and
120o switching cases that we have achieved an
alternating three-phase voltage at the three output
terminals.
• Although the output waveform is not a pure sine wave,
it did resemble the three-phase voltage waveform.
• This is a simple ideal circuit and approximated
waveform for understanding 3 phase inverter working.
• You can design a working model based on this theory
using thyristors, switching, control, and protection
circuitry.