The document describes a three-phase, full-wave rectifier circuit using 6 diodes arranged in a bridge configuration. The upper diodes (D1, D3, D5) form the positive group and conduct during the positive half cycles of the input voltage. The lower diodes (D2, D4, D6) form the negative group and conduct during the negative half cycles. Calculations are provided for the output voltage, current, power, ripple, efficiency and transformer utilization factor of the three-phase full-wave rectifier.

1. THREEPHASE FULLWAVE rectifier
Submittdto submittedbyMr.dharmendra
vinaysingh
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
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 D5D3
D4 D2D6
R
Va
Vc Vb
Vo
ia
ic
ib
n
Fig. Three phase Bridge rectifier using Diodes
CIRCUIT DIAGRAM

6. D5 D1 D3 D5
D6 D2 D4 D6
Vo
ᾠt0
Vcb Vab Vac Vbc Vba Vca Vcb
90⁰ 360⁰270⁰180⁰
Fig.2(a)
Fig.2(c)
Fig.2(b)
Fig.

7. Vo
ᾠt0
Vcb Vab Vac Vbc Vba Vca Vcb
90⁰ 360⁰270⁰180⁰
Fig.2(c) output voltage waveform
ia or is
0
30⁰
270⁰210⁰
150⁰90⁰
330⁰
390⁰
iab iac
0
iD1
Fig.2(d) Input current waveform
Fig.2(e) Diode curent waveform through D1
Fig.

8. 0
150⁰ 390⁰270⁰
-1.5 Vmp
-√3 Vmp
or Vml
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 V0 =(1/periodicity) ∫VmL sin(ᾠt+30⁰) d(ᾠt)
=(3/∏) ∫VmL sin(ᾠt+30⁰) d(ᾠt)
= (3/∏)VmL = (3√2/ ∏)VL = (3√6/∏)Vp
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/∏ ∫VmL sinᾠt d(ᾠt)]
= 0.9558 VmL
Ripple Voltage (Vr) = √(Vrms – Vavg.) = 0.0408 VmL
Voltage ripple factor (VRF) = Vr/Vo = 0.0408 VmL/(3/∏)VmL = 0.0427 or 4.27%
Form Factor = Vor/Vo = 1.0009
R.M.S value of O/P current (Ior) = 0.9558Vml/R = 0.9558 ImL
ᾳ2
ᾳ1
∏/2
∏/6
∏/3
2∏/3
22 1/2
2 2

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

12. Working of 3 phase bridge rectifier

13. 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