Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

7,812 views

Published on

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

No Downloads

Total views

7,812

On SlideShare

0

From Embeds

0

Number of Embeds

11

Shares

0

Downloads

8

Comments

5

Likes

18

No embeds

No notes for slide

- 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.

No public clipboards found for this slide

Login to see the comments