Gandhinagar Institute of Technology

ACTIVE LEARNING ASSIGNMENTS ON

THREE PHASE UNCONTROLLED RECTIFIER

Prepared by:-

Sodha Manthan (140120109057)

Gundigara Naitik (150123109003)

Ranpura Rutul (150123109013)

SEM-5TH

BATCH -B3

POWER ELECTRONICS-1

ELECTRICAL DEPT.(2016)

Guided by:-

Prof. Supraja Giddaluru

Necessity of 3-phase full wave rectifierMost industrial power supplies for motor drives and welding applications use three-phase ac voltage. This means that the rectifier for these circuits must use a three-phase bridge, which has six diodes to provide full-wave rectification (two diodes for each line of the three phases).The three-phase full-wave bridge rectifier is used where the required amount of dc power is high and the transformer efficiency must be high. Since the output waveforms of the half-waves overlap, they provide a low ripple percentage. In this circuit, the output ripple is six times the input frequency. Since the ripple percentage is low, the output dc voltage is usable without much filtering. 

A

B

C

A

B

C

a

bc

D1 D5D3

D4 D2D6

R

Va

Vc Vb

Vo

ia

ic

ibn

Fig. Three phase full wave rectifier using Diodes

CIRCUIT DIAGRAM

This diagram also shows the waveforms for the three-phase sine waves that supply power to the bridge, and for the six half-waves of the output pulsing dc voltage. Notice that since the six half-waves overlap, the dc voltage does not have a chance to get to the zero voltage point; thus, the average dc output voltage is very high.Fig. Electronic schematic of the three-phase full wave rectifier that is connected to the secondary winding of a three-phase transformer. a,b,c Three-phase input sine waves. D1,D2,D3,D4,D5,D6 Six half-waves for the dc output.

THREE PHASE FULL WAVE RECTIFIER

USING 6 DIODES. UPPER DIODES D1,D3,D5 CONSTITUTES +ve GROUP. LOWER DIODES D4,D6,D2 CONSTITUTES –ve GROUP.

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.

D5 D1 D3 D5

D6 D2 D4 D6

Vo

ᾠt0

Vcb Vab Vac Vbc Vba Vca Vcb

90⁰ 360⁰270⁰180⁰

Fig.(a)

Fig.(c)

Fig.(b)

Vo

ᾠt0

Vcb Vab Vac Vbc Vba Vca Vcb

90⁰ 360⁰270⁰180⁰

Fig.(c) output voltage waveform

ia or is

030⁰

270⁰210⁰

150⁰90⁰330⁰

390⁰

iab iac

0

iD1

Vml/R = √3Vmp/R = Iml

Fig.(d) Input current waveform

Fig.(e) Diode current waveform through D1

0150⁰ 390⁰270⁰

-1.5 Vmp

-√3 Vmp or Vml

VD1

30⁰

D5 D1 D3 D5

D6 D2 D4 D6

Fig .(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.

= .d(ωt) It is seen from fig© that the valueof at ωt=0 is and its periodicity is 60 or rad.

= = = Where = maximum value of line voltage = rms value of phase voltage = rms value of phase voltage.

Average value of output voltage can also be obtained as under

(1) Take any sinusoidal wave and integrate it from 60 to 120. it is because the voltage pule area require extends from ωt = 60 to ωt=120 for the function.

(2) For a cosine function , voltage pulse of 60 duration extends 30 to the left of its peak and 30 to the right of its peak.

=

Rms value of o/p voltage , = 0.9558

Ripple voltage, = 0.0408 Voltage ripple factor, VRF= = 0.0427 or 4.27% FF = = 1.0009 Rms vlue of output current , = = 09558 = = = = Rectifier efficiency = 99.82%

Rms value of source voltage ()= Rms value of line current ()= = 0.7804 VA rating of transformer = 3 = 3 × × 0.7804 TUF = = 0.9541 Average value of diode current = = Rms value of diode current, = = 0.582

A 3-phase bridge rectifier, using diodes, delivers power to a load of R=10 Ὠ at a dc voltage of 400 V . Determine the rating of the diodes and of the three phase delta-star transformer.

It is the given that dc output voltage = = 400 V Maximum value of line voltage = 418.88 V Rms value of phase voltage for star secondary, = = 171 V Maximum value of load current, = = 41.89 A

rms value of phase current

= 0.7804 = 32691 A

VA rating of transformer = 3 = 3(171)(32.961)

Rms value of diode current,

=

=0.5518

Average value of diode current,

= =13.334 A

Peak diode current= =41.89 A, PIV = = 418.88 V

= = (400)*(400/10)=16000w

Transformer rating

TUF = = 16000/0.9541 = 16769..73 VA

Thank You Reference by:-

P S Bhimrah