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

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3 phase diode rectifier ppt.It clears your concepts how it works.Hope you will gain knowledge about the nature of diode in different combinations.

### Transcript of 3 phase diode rectifiers/power electronics

POWER POINT PRESENTATION

on Three Phase Diode Rectifier

By – Nitish kumar singh EEE

36

GURU GOBIND SINGH EDUCATIONAL SOCIETY’S TECHNICAL CAMPUS

• Three phase half wave rectifier

• Three phase mid point 6-pulse rectifier

• Three phase bridge rectifier

• Three phase 12-pulse rectifier

CLASSIFICATION OF THREE PHASE DIODE RECTIFIER

Three Phase Half wave Diode Rectifier

VD1

D2

D3

R

Va

VcVbnA

B

C

A

B

C

Common cathode terminal

io

Vo

Fig.1 Three Phase half-wave diode rectifier with common cathode arrangement

Three Phase Half wave Diode Rectifier

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

D3 D1 D2 D3

030⁰ 150⁰ 270⁰

0

Vc Va Vb Vc

Fig.(a)

Fig.(b)

Fig.(c)

Vo

Voltage of neutral ‘n’

Voltage of terminal ‘P’

Conduction of Diodes in proper sequence

Vmp

0.5 Vmp

Fig.

030⁰ 150⁰ 270⁰

icia ib icio

Imp= Vmp /R

90⁰ 210⁰ 330⁰

is= ia i D1

0

is= ia= iD1

Fig.(d) Load current or output current.

Fig.(e) Source current

Fig.

0

D3 D1 D2 D3

150⁰ 390⁰270⁰

-1.5 Vmp

-√3 Vmp

VD1

30⁰

Fig.(f) Voltage across Diode D1

Fig.

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 – VoWhen Diode D1 conduct:Vo = Va Therefore , VD1 = Va – Va = 0When 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

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)

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 VmpForm Factor = Vor/Vo = 1.0165R.M.S value of O/P current (Ir) = 0.84068Vmp/R = 0.84068 Imp

Pdc = Vo Io = (3√3/2∏) Vmp Imp

ᾳ2

ᾳ15∏/6

∏/6

5∏/6

∏/6

22 1/2

2 2

2

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 ImpTransformer Utilization Factor(TUF)= (Pdc / Transformer VA Rating) = 0.684VmpImp/1.0295 VmpImp

= 0.6644

2

22 1/2∏/6

5∏/6

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.

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.

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

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

Specification :1. Off state voltage max. =1.6kV2. Rated current =160A3. Max. Forward impulse current =1.2kA 4. Thermal resistance =0.16K/W5. 4000 Vrms isolating voltage