Download Three-Phase Inverters

Three-Phase Inverters
Consider three single-phase inverters in
parallel, driven 120° apart.
Three-Phase Inverter (continued)
Three single-phase full bridge inverters
12 transistors, 12 diodes, 3 transformers
Could it be simpler?
Alternative (Preferred) Configuration
6 transistors, 6 diodes
conduction for 120° or 180°
180° Conduction
• Three transistors ON at a time
Summary Table
Waveforms for 180 Conduction
Phase Voltages for 180 Conduction
Waveforms for 180 Conduction
Summary Table
Mode 1 Operation
Mode 1 Operation

0  t 
3
R 3R
Req  R  
2
2
Vs 2Vs
i1 

Req 3R
Q1, Q5, Q6 conduct
i1 R Vs
van  vcn 

2
3
2Vs
vbn  i1 R 
3
Summary Table
Mode 2 Operation
Mode 2 Operation

2
 t 
3
3
R 3R
Req  R  
2
2
Vs 2Vs
i2 

Req 3R
Q1, Q2, Q6 conduct
2Vs
van  i2 R 
3
i2 R Vs
vbn  vcn 

2
3
Summary Table
Mode 3 Operation
Mode 3 Operation
2
 t  
3
R 3R
Req  R  
2
2
Vs
2Vs
i3 

Req 3R
Q1, Q2, Q3 conduct
i3
van  vbn 
2
2Vs
vcn  i3 R 
3
Phase Voltages for 180 Conduction
Fourier Series for Line-to-Line Voltages
ao 
vab    (an cos(nt )  bn sin(nt ))
2 n 1
5
 56

6
1

bn    Vs d (t )   Vs d (t ) 
 5

 6

6
4Vs
n
n
bn 
sin( ) sin( )
n
2
3

4Vs
n

vab  
sin
sin n(t  )
3
6
n 1,3,5,... n
For the other Line-to-Line Voltages
4Vs
n

vbc  
sin
sin n(t  )
3
2
n 1,3,5,... n

4Vs
n
7
vca  
sin
sin n(t  )
3
6
n 1,3,5,... n

Line-to-Line rms Voltage

 2
VL  
2

2
3


2
0 Vs d (t ) 

2
VL  Vs  0.8165Vs
3
1
2
rms value of the nth Component
4Vs
n
VLn 
sin
3
2n
n  1 Fundamental Component
4Vs sin 60
VL1 
 0.7797Vs
2
Line-to-Neutral Voltages
2Vs
VL
Vp 

 0.4714Vs
3
3
Phase Voltages (Y-connected load)
vaN 
vbN 
vcN 


n 1,3,5,..


n 1,3,5,..


n 1,3,5,..
4Vs
n
sin( )sin(nt )
3
3n
4Vs
2
n
sin( )sin n(t  )
3
3
3n
4Vs
4
n
sin( )sin n(t  )
3
3
3n
Line Current for an RL load


4Vs
n 

ia  
sin
sin(nt   n )


3
 n R 2  (n L) 2 
n 1,3,5,...
3
 


1  n L 
 n  tan 

 R 

DC Supply Current
vs is  vab (t )ia (t )  vbc (t )ib (t )  vca (t )ic (t )
....
Vo1
Is  3
I o cos(1 )
Vs
Vo1
Is  3
I L cos(1 )
Vs
IL=√3Io is the rms load line current
Vo1= fundamental rms output line voltage
Io is the rms load phase current
Θ1 = the load impedance angle at the fundamental
frequency
Three-Phase Inverter with RL Load