Detail Explanation of 6-phase star rectifier along with performance parameters.

1. 6-Winding Rectifier (Multi-Phase Star Rectifier)
Presented By
FA13-R09-005 Muqadsa Iftikhar
FA13-R09-013 Zunaib Ali
FA13-R09-024 Madiha Naeem

2. Diagram of 6 Phase Rectifier
D2
D3
D1
D5
D4
V1
V2
V3
V4
V5
D6
V6
N
R
The circuit for 6-phase
start rectifier is shown in
figure
Six phase can be
achieved by using
multiphase windings at
transformer secondary.
The circuit can be
considered as 6-single
phase half-wave
rectifier.

3. 3 Transformer Configurations
complex 3 6 pulse: delta/wye/wye (DUU)
primary secondary
Only one secondary winding carries current at a particular time and as
result the primary must be connected in delta to eliminate the dc
component in the input side of the transformer. This minimizes the
harmonics of primary line current
Diagram cont…

4. Diagram cont…

5. This circuit consists essentially of two three-phase star rectifiers with their neutral
points interconnected through an interphase transformer or reactor.
The polarities of the corresponding secondary windings in the two
interconnected systems are reversed with respect to each other, so that the
rectifier output voltage of one three-phase unit is at a minimum when the
rectifier output voltage of the other unit is at a maximum
Diagram cont…

6. The interphase transformer causes the output voltage vL to be the average of the
rectified voltages v1 and v2
In addition, the ripple frequency of the output voltage is now six times that of the
mains and, therefore, the component size of the filter (if there is any) becomes
smaller.
In a balanced circuit, the output currents of two three-phase units flowing in opposite
directions in the interphase transformer winding will produce no dc magnetization
current. Similarly, the dc magnetization currents in the secondary windings of two
three-phase units cancel each other out
By virtue of the symmetry of the secondary circuits, the three primary currents add up
to zero at all times. Therefore, a star primary winding with no neutral connection
would be equally permissible.
Diagram cont…

7. So we have six sinusoidal voltages at secondary (V1 V2 V3 V4 V5 V6)
having same frequency and magnitude i.e.
Vm1 = Vm2 = Vm3 = Vm4 = Vm5 = Vm6 = Vm
f1 = f2 = f3 = f4 = f5 = f6 = f
Let
V1 = Vm Sin (wt)
V2 = Vm Sin (wt + π/3)
V3 = Vm Sin (wt + 2π/3)
V4 = Vm Sin (wt + π)
V5 = Vm Sin (wt + 4π/3)
V6 = Vm Sin (wt + 5π/3)

8. When Diode Conducts
D2
D3
D1
D5
D4
D6
N
R
V1
V2
V3
V4
V5
V6
As shown in figure, let
V1= 2V, V1= 3V,
V1= 4V, V1= 5V,
V1= 6V, V1= 7V,
Now the diode with
highest voltage level will
conduct.
Hence it is concluded that
diode with phase voltage
higher than others will
conduct.

9. Graphs for output Voltage and input Current
The conduction period of each diode is (2π/6 or π/3 or 60o)
The 6 phase input voltage waveform and the corresponding output is
shown in figure
ωt
i1(ωt)
π/3 2π/3 π 4π/3 5π/3 2π
V2 V3V4V5V6V1
D2 D2D3D4D5D6D1
6 Phase
Input to
Rectifier
Vo(ωt)
π/3 2π/3 2π
ωt

10. Output DC Level
𝑉𝑑𝑐 =
1
2𝜋
6
𝑉𝑚 sin 𝜔𝑡 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝑉𝑑𝑐 =
3𝑉𝑚
𝜋
sin 𝜔𝑡 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝑉𝑑𝑐 =
3𝑉𝑚
𝜋
−cos 𝜔𝑡
2𝜋
3
𝜋
3
𝑉𝑑𝑐 =
−3𝑉𝑚
𝜋
cos 2𝜋
3 − cos 𝜋
3
𝑉𝑑𝑐 =
3𝑉𝑚
𝜋
𝑉𝑑𝑐 = 0.954929𝑉𝑚
The output dc voltage of rectifier can be calculated from Fig.. The relation is:

11. Output DC Level
𝑉𝑑𝑐 =
1
2𝜋
6
𝑉𝑚 cos 𝜔𝑡 𝑑 𝜔𝑡
𝜋
6
− 𝜋
6
𝑉𝑑𝑐 =
3𝑉𝑚
𝜋
cos 𝜔𝑡 𝑑 𝜔𝑡
𝜋
6
− 𝜋
6
𝑉𝑑𝑐 =
3𝑉𝑚
𝜋
sin 𝜔𝑡
𝜋
6
− 𝜋
6
𝑉𝑑𝑐 =
3𝑉𝑚
𝜋
sin 𝜋
6 − sin − 𝜋
6
𝑉𝑑𝑐 =
3𝑉𝑚
𝜋
So the value of 𝐼𝑑𝑐 =
𝑉 𝑑𝑐
𝑅
The output dc voltage of rectifier can be calculated from Fig.. The relation In terms
of Cosine. Assuming a cosine wave from -π/6 to π/6 :

12. Output Voltage rms
𝑉0𝑟𝑚𝑠 =
1
2𝜋
6
𝑉𝑚 sin 𝜔𝑡 2 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝑉0𝑟𝑚𝑠 =
3𝑉𝑚
2
𝜋
sin2 𝜔𝑡 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝑉0𝑟𝑚𝑠 =
3𝑉𝑚
2
2𝜋
1 − cos⁡(2𝜔𝑡 )𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝑉0𝑟𝑚𝑠 =
3𝑉𝑚
2
2𝜋
𝜔𝑡 −
sin⁡(2𝜔𝑡)
2
2𝜋
3
𝜋
3
𝑉0𝑟𝑚𝑠 =
3𝑉𝑚
2
2𝜋
2𝜋
3 − 𝜋
3 −
sin 4𝜋
3 − sin 2𝜋
3
2
𝑉0𝑟𝑚𝑠 =
3𝑉𝑚
2
2𝜋
𝜋
3 + 0.8666
𝑉0𝑟𝑚𝑠 = 0.913967𝑉𝑚
2 = 0.9560162𝑉𝑚
The rms value of output voltage of rectifier can be calculated from Fig.. The
relation is:

13. Output DC Level
𝐼0𝑟𝑚𝑠 =
1
2𝜋
6
𝐼 𝑚 sin 𝜔𝑡 2 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝐼0𝑟𝑚𝑠 =
3𝑉𝑚
2
𝜋𝑅2 sin2 𝜔𝑡 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝐼0𝑟𝑚𝑠 =
𝑉0𝑟𝑚𝑠
𝑅
𝐼0𝑟𝑚𝑠 =
0.9560162𝑉 𝑚
𝑅
The rms value of output current of rectifier can be calculated from Fig.. The
relation is:

14. Input Current rms
𝐼𝑖𝑟𝑚𝑠 =
1
2𝜋
𝐼 𝑚 sin 𝜔𝑡 2 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝐼𝑖𝑟𝑚𝑠 =
𝑉 𝑚
2
2𝜋𝑅2 sin2
𝜔𝑡 𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝐼𝑖𝑟𝑚𝑠 =
𝑉 𝑚
2
4𝜋𝑅2 1 − cos⁡(2𝜔𝑡 )𝑑 𝜔𝑡
2𝜋
3
𝜋
3
𝐼𝑖𝑟𝑚𝑠 =
𝑉 𝑚
2
4𝜋𝑅2 𝜔𝑡 −
sin⁡(2𝜔𝑡)
2
2𝜋
3
𝜋
3
𝐼𝑖𝑟𝑚𝑠 =
𝑉𝑚
2
4𝜋 𝑅2
2𝜋
3 − 𝜋
3 −
sin 4𝜋
3 − sin 2𝜋
3
2
𝐼𝑖𝑟𝑚𝑠 =
𝑉𝑚
2
4𝜋 𝑅2
𝜋
3 + 0.8666
𝐼𝑖𝑟𝑚𝑠 = 0.390294
𝑉𝑚
𝑅
For resistive load the peak current through diode is Im = Vm/R. The rms value of a diode
current (or transformer secondary current) can be calculated as follow

15. Performance Parameters
Efficiency:
ŋ =
𝑉𝑑𝑐 × 𝐼𝑑𝑐
𝑉0𝑟𝑚𝑠 × 𝐼0𝑟𝑚𝑠
ŋ =
3𝑉𝑚
𝜋
×
3𝑉𝑚
𝜋𝑅
0.9569162𝑉𝑚 × 0.9569162
𝑉𝑚
𝑅
ŋ = 99.874%
Form Factor
𝐹𝐹 =
𝑉0𝑟𝑚𝑠
𝑉𝑑𝑐
𝐹𝐹 =
0.9569162𝑉𝑚
3𝑉𝑚
𝜋
𝐹𝐹 = 1.00157

16. Ripple Factor:
𝑅𝐹 =
𝑉𝑎𝑐
𝑉𝑑𝑐
= 𝐹𝐹2 − 1
𝑅𝐹 = 0.056057
Transfer Utilization Factor
𝑇𝑈𝐹 =
𝑉𝑑𝑐 𝐼𝑑𝑐
6𝑉𝑖𝑟𝑚𝑠 𝐼𝑖𝑟𝑚𝑠
𝑇𝑈𝐹 =
3𝑉𝑚
𝜋
×
3𝑉𝑚
𝜋𝑅
6 ×
𝑉𝑚
2
× 0.390294
𝑉𝑚
𝑅
𝑇𝑈𝐹 = 0.5503
The best Transformer Utilization Factor (TUF) that can be achieved with a single-way connection is
TUF = 0.79 while with a bridge configuration it is possible to reach higher values, up to TUF = 0.955
Performance Parameters

17. Power Factor:
𝑃. 𝐹 =
𝑃𝑎𝑣 (𝐷𝑒𝑙𝑖𝑒𝑣𝑒𝑟𝑑)
6 𝑉𝑖𝑟𝑚𝑠 𝐼𝑖𝑟𝑚𝑠
𝑃. 𝐹 =
𝑉0𝑟𝑚𝑠 𝐼0𝑟𝑚𝑠
6 𝑉𝑖𝑟𝑚𝑠 𝐼𝑖𝑟𝑚𝑠
𝑃. 𝐹 =
0.9569162𝑉𝑚 × 0.9569162
𝑉𝑚
𝑅
6 ×
𝑉𝑚
2
× 0.390294
𝑉𝑚
𝑅
𝑃. 𝐹 = 0.552993
As resistive load, no displacement factor
So, Displacement factor =1
Distortion factor =0.552993 ( distortion factor is due to disturbance in input current wavform)
Performance Parameters

18. Table: Performance Parameters for Rectifier
Performance Parameters

19. S.No Performance Parameters Calculated Values Ideal Values
1 Efficieny, Rectification ratio (RR) 99.874% 100%
2 Form factor (FF) 1.00157 1
3 Transfer Utilization Factor (TUF) 0.5503 1
4 Ripple Factor (RF) 0.056057 0
1 Power Fcator (PF) 0.552993 1
Table # 2: calculated values of RR, FF, TUF & RF PF.
Performance Parameters
Table

20. Simulations
Continuous
pow ergui
v
+
-
Voltage Measurement2
v
+
-
Voltage Measurement
V6V5V4V3
V2V1
In1
Specturm1
Scope9
Scope8
Scope7
Scope6
Scope5
Scope4
Scope3
Scope2
Scope11
Scope10
Scope1
Scope
signalrms
RMS4
signalrms
RMS3signalrms
RMS2
signalrms
RMS1
signalrms
RMS
R
0.5519
Power Factor
In Mean
Mean Value3In Mean
Mean Value2
In Mean
Mean Value1
In Mean
Mean Value
signal
magnitude
angle
Fourier
Irms
I1rms
Theta
DSTF
DSPF
PF
zunaib
Embedded
MATLAB Function
0.5519
Distortion Factor
1
Displacement Factor
D6
D5
D4
D3
D2
D1
i
+
-
Current M1
i
+
-
Current M
Clock

21. Output Voltage Fourier Series
The frequency of output is 6𝑓
𝑏 𝑛 =
2
2𝜋
6
𝑉𝑚 cos 𝜔𝑡 sin 𝑛𝜔𝑡 𝑑 𝜔𝑡
𝜋
6
− 𝜋
6
𝑏 𝑛 = 0
𝑎 𝑛 =
2
2𝜋
6
𝑉𝑚 cos 𝜔𝑡 cos 𝑛𝜔𝑡 𝑑 𝜔𝑡
𝜋
6
− 𝜋
6
Solving 𝑎 𝑛 𝑤𝑒 𝑔𝑒𝑡
𝑎 𝑛 =
−12𝑉𝑚
𝜋(𝑛2 − 1)
cos
𝑛𝜋
6
sin
𝑛𝜋
6
Where 𝑛 = 6, 2 × 6, 3 × 6 …
The dc component already calculated
𝑉𝑑𝑐 =
1
2𝜋
6
𝑉𝑚 cos 𝜔𝑡 𝑑 𝜔𝑡
𝜋
6
− 𝜋
6
𝑉𝑑𝑐 = 0.954929𝑉𝑚
So the Fourier series is:
𝑉0 𝑡 = 𝑉𝑑𝑐 + 𝑎 𝑛
∝
𝑛=6,12,18
cos 𝑛𝜔𝑡
Substituting we get
𝑉0 𝑡 = 0.954929𝑉𝑚 (1 +
2
35
cos 6𝜔𝑡 +
2
143
cos 12𝜔𝑡 + ⋯ )

22. Output Voltage Spectrum
0 360 720 1080 1440 1800 2160 2520 2880 3240 3600 3960
0
10
20
30
40
50
60
70
80
90
100
Multi pHase Rectifier Output Voltage
Frequency Hz
NormalizedHarmonicMagnitude

23. Output Current Spectrum
0 360 720 1080 1440 1800 2160 2520 2880 3240 3600 3960
0
10
20
30
40
50
60
70
80
90
100
Multi pHase Rectifier Output Current Specturm
Frequency Hz
NormalizedHarmonicMagnitude

24. Parameters using Matlab
clc
Iom=Iomean(:,2); % output mean current
Ior=Iorms(:,2); % output rms current
Iim=Iiimean(:,2); % input mean current
Iir=Iiirms(:,2); % input rms current
VinR=inVrms(:,2); % input rms voltage
VinM=inVmean(:,2); % input mean voltage
VoM=oVmean(:,2); % output mean voltage
VoR=oVrms(:,2); % output rms voltage
Iom=mean(Iom);
Ior=mean(Ior);
Iim=mean(Iim);
Iir=mean(Iir);
VinR=mean(VinR);
VinM=mean(VinM);
VoM=mean(VoM);
VoR=mean(VoR);
efficiency= (VoM * Iom)/(VoR * Ior) % Rectification ratio
Form_Factor= VoR/VoM % From factor
Ripple_Factor= sqrt(FF^2 -1) % Ripple factor
TUF=(VoM* Iom)/(6*VinR* Iir) % Transformer Utilization factor

25. Power Factor Using Matlab
function [DSTF,DSPF,PF] = zunaib(Irms,I1rms,Theta)
DSTF=((I1rms/(sqrt(2))/Irms));
DSPF=cosd(Theta);
PF=DSTF*DSPF;

26. Input Current Specturm
0 60 120 180 240 300 360 420 480 540 600 660
0
10
20
30
40
50
60
70
80
90
100
Input Current Specturm
Frequency Hz
NormalizedHarmonicMagnitude

27. S.No Performance Parameters 3-Pulse 6-Pulse
1 Efficieny, Rectification ratio
(RR)
96.77% 99.874%
2 Form factor (FF) 1.0165 1.00157
3 Transfer Utilization Factor
(TUF)
0.6643 0.5503
4 Ripple Factor (RF) 0.1824 0.056057
5 Power Factor 0.6844 0.552993
6 Output Ripple Frequency 3f 6f
7 Diode PIV
Table # 2: calculated values of RR, FF, TUF & RF .
Comparison of Performance Parameters of 3pulse
and 6pulse star rectifier
3𝑉𝑚 6𝑉𝑚

28. Conclusion
A Multiphase rectifier increases the amount of dc component and lowers the amount of the
harmonic components. The output voltage of p-phase rectifier contains harmonics whose
frequencies are multiple of 𝑝 ( 𝑝 times the supply frequency), 𝑝𝑓
In practice, for single-way connections, the maximum number of pulses is normally 12.