The document discusses a three phase diode rectifier presentation. It describes several three phase rectifier circuits including a half wave rectifier using three diodes, a six pulse midpoint rectifier, and a full wave bridge rectifier using six diodes. Equations are provided for the output voltage and current calculations for each circuit. Key specifications of automotive-grade rectifier diodes are also listed.

1. POWER POINT PRESENTATION
on
Three Phase Diode Rectifier
By – Nitish kumar singh
EEE
36
GURU GOBIND SINGH EDUCATIONAL
SOCIETY’S TECHNICAL CAMPUS

2. • Three phase half wave rectifier
• Three phase mid point 6-pulse rectifier
• Three phase bridge rectifier
• Three phase 12-pulse rectifier

3. VD1
D2
D3
R
Va
Vc
VbnA
B
C
A
B
C
io
Vo
Fig.1 Three Phase half-wave diode rectifier with common cathode arrangement

4. Rectifier element is Diode.
Diode can only conduct at the highest +ve instantaneous voltage.
D1 will conduct for ᾠt =30⁰ to ᾠt =150⁰
D2 will conduct for ᾠt =150⁰ to ᾠt =270⁰
D3 will conduct for ᾠt =270⁰ to ᾠt =390⁰
A Diode with the highest positive voltage will begin to conduct at
the cross over point of the 3-PHASE supply.
Working

5. D3 D1 D2 D3
0
30⁰ 150⁰ 270⁰
0
Vc
Va Vb Vc
Fig.(a)
Fig.(b)
Fig.(c)
Vo
Voltage of neutral ‘n’
Voltage of terminal ‘P’
Vmp
0.5 Vmp
Fig.

6. 0
30⁰ 150⁰ 270⁰
ic
ia ib ic
io
90⁰ 210⁰ 330⁰
is= ia
i D1
0
Fig.(d) Load current or output current.
Fig.(e) Source current
Fig.

7. 0
D3 D1 D2 D3
150⁰ 390⁰270⁰
-1.5 Vmp
-√3 Vmp
VD1
30⁰
Fig.(f) Voltage across Diode D1
Fig.

8. Variation of voltage across Diode D1
Voltage variation across diode D1 can be obtained by applying KVL to the loop
consisiting of diode D1, Phase ‘a’ winding and load R.
So, -VD1 -Vo + Va = 0 or VD1 = Va – Vo
When Diode D1 conduct:
Vo = Va
Therefore , VD1 = Va – Va = 0
When diode D2 conduct :
Vo = Vb
Therefore , VD1 = Va – Vb
At ᾠt = 180⁰, Vb=0.866 Vmp , Va = 0 VD1 = - 0.866 Vmp
At ᾠt = 210⁰, Vb= Vmp , Va = -0.5 Vmp VD1 = -1.5 Vmp
At ᾠt = 240⁰, Vb=0.866 Vmp , Va = 0.866 Vmp VD1 = - √3 Vmp
At ᾠt = 270⁰, Vb=0.5 Vmp , Va = - Vmp VD1 = -1.5 Vmp
HOW

9. When Diode D3 conducts :
VD1 = Va - Vc
At ᾠt = 300⁰, Va=-0.866 Vmp , Vc = 0.866Vmp VD1 = - √3 Vmp
At ᾠt = 330⁰, Va=-0.5 Vmp , Vc = Vmp VD1 = - 1.5 Vmp
At ᾠt = 360⁰, Va=0 , Vc = 0.866Vmp VD1 = -0.866 Vmp
At ᾠt = 390⁰, Va= 0.5 Vmp , Vc = 0.5Vmp VD1 = 0
EQUATIONS OF PHASE VOLTAGES FROM WHERE ABOVE
VALUES ARE OBTAINED---
Va = Vmp sin (ᾠt)
Vb = Vmp sin(ᾠt – 120⁰)
Vc = Vmp sin(ᾠt – 240⁰)

10. Average output voltage V0 =(1/periodicity) ∫Vmp sinᾠt d(ᾠt)
=(3/2∏) ∫Vmp sinᾠt d(ᾠt)
= (3√3/2∏)Vmp
R.M.S value of output voltage(Vor) =[3/2∏ ∫Vmp sinᾠt d(ᾠt)]
= 0.84068 Vmp
Ripple Voltage =√(Vrms – Vavg.) = 0.151 Vmp
Form Factor = Vor/Vo = 1.0165
R.M.S value of O/P current (Ir) = 0.84068Vmp/R = 0.84068 Imp
Pdc = Vo Io = (3√3/2∏) Vmp Imp
ᾳ2
ᾳ1
5∏/6
∏/6
5∏/6
∏/6
22
1/2
2 2
2

11. Pac = Vor Ir = 0.84068 Vmp Imp = 0.706743Vmp Imp
Rectifier efficiency = Pdc/Pac = 0.9677
% Rectifier efficiency = 0.9677 ×100 = 96.77
Rms value of source voltage(Vs) = Vmp/√2 = 0.707 Vmp
Rms value of source current(Is) = [1/2∏ ∫Imp sinᾠt d(ᾠt)]
= 0.4854 Imp
VA rating of transformer = 3Vs Is = 0.707Vmp × 0.4854 Imp
=1.0295 Vmp Imp
Transformer Utilization Factor(TUF)= (Pdc / Transformer VA Rating)
= 0.684VmpImp/1.0295 VmpImp
= 0.6644
2
22
1/2∏/6
5∏/6

12. NOTE
• PIV for each diode = √3 Vmp.
• Each Diode conduct for 120⁰.
• There are three pulses of output voltage, or
output current, during one cycle of input
Voltage.Therefore it is called 3-phase three-pulse
diode rectifier.
• Current in the transformer secondary is
unidirectional,therefore,DC exists in the
transformer secondary current. As a result
transformer core gets saturated leading to more
iron losses and reduced efficiency.

13. 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

14. 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.

15. 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

16. 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.

17. 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.

18. 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.

19. 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

20. 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

22. Disassembled automobile alternator , showing the six diodes that
comprise a full-wave three-phase bridge rectifier.

23. Specification :
1. Off state voltage max. =1.6kV
2. Rated current =160A
3. Max. Forward impulse current =1.2kA
4. Thermal resistance =0.16K/W
5. 4000 Vrms isolating voltage
Additional information :
1. Totally lead (Pb) Free
2. Gross weight=0.23kg
3 Phase Bridge rectifier