Three phase full wave rectifier

three phase full wave rectifier uncontroled

1. THREE PHASE FULL WAVE rectifier Submittd to submitted by Mr. dharmendra vinay singh Upaddahya 1404620904

2. THREE PHASE BRIDGE RECTIFIER USING 6 DIODES. UPPER DIODES D1,D3,D5 CONSTITUTES + ve GROUP. LOWER DIODES D4,D6,D2 CONSTITUTES – ve GROUP. THREE PHASE T/F FEEDING THE BRIDGE IS CONNECTED IN DELTA-STAR . CONSTRUCTION

3. Positive group of Diodes conduct When these have the most positive anode. Negative group of diodes conduct if these have the most negative anode. WORKING + Ve group - Ve group This group will conduct during + ve half cycle of I/P source. This group will conduct during - ve half cycle of I/P source.

4. Three-Phase, Full-Bridge Rectifier

5. A B C A B C a b c D1 D5 D3 D4 D2 D6 R Va Vc Vb Vo i a i c i b n Fig. Three phase Bridge rectifier using Diodes CIRCUIT DIAGRAM

6. D5 D1 D3 D5 D6 D2 D4 D6 Vo ᾠ t Vcb Vab Vac Vbc Vba Vca Vcb 90⁰ 360⁰ 270⁰ 180⁰ Fig.2(a) Fig.2(c) Fig.2(b) Fig.

7. Vo ᾠ t Vcb Vab Vac Vbc Vba Vca Vcb 90⁰ 360⁰ 270⁰ 180⁰ Fig.2(c) output voltage waveform i a or i s 30⁰ 270⁰ 210⁰ 150⁰ 90⁰ 330⁰ 390⁰ i ab i ac i D1 V ml /R = √3V mp /R = I ml Fig.2(d) Input current waveform Fig.2(e) Diode curent waveform through D1 Fig.

8. 150⁰ 390⁰ 270⁰ -1.5 Vmp -√3 Vmp or V ml VD1 30⁰ D5 D1 D3 D5 D6 D2 D4 D6 Fig .2(f) Voltage variation across Diode D1 . Voltage variation across D1 can be obtained in a similar manner as in the case of 3-phase half wave diode rectifier. Fig.

9. Average output voltage V =(1/periodicity) ∫ Vm L sin( ᾠ t+30⁰) d( ᾠ t) = (3/∏) ∫ Vm L sin( ᾠ t+30⁰) d( ᾠ t) = (3/ ∏ ) Vm L = (3√ 2/ ∏ ) V L = (3√6/ ∏ ) V p Where, VmL = maximum value of line voltage VL = rms value of line voltage Vp = rms value of phase voltage R.M.S value of output voltage( Vor ) = [ 3/∏ ∫ Vm L sin ᾠ t d( ᾠ t) ] = 0.9558 V m L Ripple Voltage ( Vr ) = √ ( Vrms – Vavg .) = 0.0408 V mL Voltage ripple factor ( VRF ) = Vr /Vo = 0.0408 V mL /(3/ ∏ ) V mL = 0.0427 or 4.27% Form Factor = V or /V o = 1.0009 R.M.S value of O/P current ( I or ) = 0.9558Vml/R = 0.9558 I mL ᾳ 2 ᾳ 1 ∏/2 ∏/6 ∏/3 2∏/3 2 2 1/2 2 2

10. P dc = Vo Io = (3/ ∏ ) V mL I mL Pac = Vr Ir = 0.9558 V mL I mL Rectifier efficiency = Pdc /Pac = 0.9982 % Rectifier efficiency = 0.9982 ×100 = 99.82% Rms value of source voltage(Vs) = Vmp /√2 = VmL /√6 (Since, V mL = √3 V mp ) Rms value of line current(Is) = rms value of T/F secondary current = [ 2 /∏ ∫ I mL sin ᾠ t d( ᾠ t ) ] = 0.7804 I mL VA rating of transformer = 3Vs Is = 3 ( VmL /√6) × 0.7804 I mL = 0.955791 V mL I mL Transformer Utilization Factor(TUF) = ( Pdc / Transformer VA Rating ) = (3/∏)^2 /0.955791 = 0.9541 2 2 2 1/2 ∏ /3 2 ∏ /3 2

11. Working of 3 phase bridge rectifier

12. Summary Line-frequency diode rectifiers converts line-frequency ac into dc in an uncontrolled manner Various diodes rectifier circuits have been discussed Three-phase rectifiers are preferable in most respects over the single-phase ones Rectifiers inject large amounts of harmonic currents into the utility systems – remedies would have to be implemented