THREE PHASE FULL WAVE rectifier

Submittd to submitted by Mr. dharmendra

vinay singh Upaddahya

1404620904

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

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.

Three-Phase, Full-Bridge Rectifier

A

B

C

A

B

C

a

bc

D1 D5D3

D4 D2D6

R

Va

Vc Vb

Vo

ia

ic

ibn

Fig. Three phase Bridge rectifier using Diodes

CIRCUIT DIAGRAM

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.

Vo

ᾠt0

Vcb Vab Vac Vbc Vba Vca Vcb

90⁰ 360⁰270⁰180⁰

Fig.2(c) output voltage waveform

ia or is

030⁰

270⁰210⁰

150⁰90⁰330⁰

390⁰

iab iac

0

iD1

Vml/R = √3Vmp/R = Iml

Fig.2(d) Input current waveform

Fig.2(e) Diode curent waveform through D1

Fig.

0150⁰ 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.

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∏/322 1/2

2 2

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 21/2

∏/3

2∏/3

2

Working of 3 phase bridge rectifier

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