controlled rectifiers

33
SEMINAR ON AC – DC CONVERTERS BY: ANKUR MAHAJAN M.E. I&C REGULAR -2011 ROLL NO. 112505

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Controlled Rectifiers in power Electronics. Ankur Mahajan NITTTR Chandigarh

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Page 1: controlled rectifiers

SEMINAR ON AC – DC CONVERTERS

BY:ANKUR MAHAJAN

M.E. I&C REGULAR -2011ROLL NO. 112505

Page 2: controlled rectifiers

GOAL OF THE SEMINAR SINGLE PHASE HALF WAVE & FULL WAVE

CONVERTERS.

SEMICONDUCTOR DEVICES – THE

HEART OF INDUSTRIAL

ELECTRONICS

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MY OWN QUOTES : “THE DEVELOPMENT OF EVERY NATION DEPEND UPON

ITS POWER SOURCES, & POWER ELECTRONICS IS A MAGIC WAND WHICH GENERATE POWER SOURCES”

“POWER ELECTRONICS IS A LIFE BLOOD FOR MODERN COMPUTERS, COMMUNICATION & ELECTRONIC SYSTEMS , THUS WITHOUT IT OUR LIFE COMES TO HALT”

WHAT IS THE MOST IMPORTANT REASON FOR RAPID DEVELOPMENT IN THE FIELD OF POWER

ELECTRONIC DEVICES ?

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PIE CHART OF WORL’S ENERGY PRODUCTION LEVEL

FOSSIL FUELSNUCLEAR

RENEWABLES

IN INDIA 70% OF ELECTRICAL ENERGY COMES FROM COAL

ACCORDING TO IEEE JOURNAL ON

POWER DELIVERY: WE HAVE NO

FOSSIL & NUCLEAR FUEL BY THE

END OF 22ND CENTURY.

87%

6%

7%

(A)USE EE EFFICIENTLY.(B) IMPROVE CONVERSION EFFICIENCY.

(C) USE RENEWABLES

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INTRODUCTION

WHAT IS RECTIFICATION ? RECTIFICATION REFERS TO PROCESS OF

CONVERTING AN A.C. VOLTAGE OR CURRENT TO D.C. VOLTAGE OR CURRENT.

RECTIFIERS – RECTIFIER REFER TO POWER ELECTRONIC CONVERTERS WHERE THE ELECTRIC POWER FLOWS FROM A.C. SIDE TO THE D.C. SIDE.

NOTE: IN MANY SITUATIONS THE SAME CONVERTER CARRIES ELECTRICAL POWER FROM D.C. SIDE TO A.C. SIDE - INVERTERS

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RECTIFIERS CLASSIFICATION

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BECAUSE ON & OFF STATE OF DIODE IS DETERMINED BY THE STATE OF THE CIRCUIT.

BECAUSE THYRISTOR CAN BE TURNED ON BY CONTROL TERMINAL. i.e. GATE

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ASSUMPTIONS :

ALL THE DEVICES & CIRCUIT COMPONENTS ARE IDEAL i.e. Ton =Toff =zero

INPUT IS PURE SINE WAVE

CHARACTERISTICS OF AN IDEAL SWITCH

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TERMINOLOGIES

LET ‘f’ BE THE INSTANTANEOUS VALUE OF ANY VOLTAGE OR CURRENT ASSOCIATED WITH RECTIFIER CIRCUIT, THEN

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TERMINOLOGIES (CONT.)

Angle atWhich SCR starts

conductingAngle at which SCR stops conducting

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SINGLE PHASE FULLY CONTROLLED H/W RECTIFIER

CAN BE TURNED ON BY

APPLYING GATE SIGNAL

WHEN SCR IS FORWARD

BIASED.

OPERATION IS ALMOST

SAME AS THAT OF DIODE

CIRCUIT

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FOR PURELY RESISTIVE LOAD

CURRENT FLOWS FROM TO THROUGH LOAD BECAUSE SCR IS IN FPRWARD CONDUCTION MODE.

π-α

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CALCULATIONS

Average value of output voltage

On solving

ttdvv iav sin22

1

)cos1(2

iav

vv

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CALCULATION OF RMS VALUE OF VOLTAGE

tdvv orms

2

2

1

tdtvv irms

22 sin2

2

1

tdtv

v irms

)2cos1(

2

2

2

1

])2

2sin(

1[2

tt

vv irms

)2cos1(2

12 ttSin

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CALCULATION OF RMS VALUE OF VOLTAGE (CONT.)

On solving

Form factor

2

1

)2

2sin1(

2

irms

vv

)cos1(2

iav

vv

cos1

)22sin

1( 2

1

av

rms

v

vformfactor

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RESISTIVE – INDUCTIVE LOAD

α to π π to β

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RESISTIVE – INDUCTIVE LOAD

Thyristor turns ON at ωt= α & voltage appears across Load &

current starts building up. Inductor does not changes

current instantaneously, so load current does not become zero at

ωt = π It continue to flow but the

negative supply voltage decreases its magnitude

At ωt= β current becomes zero and thyristor starts blocking until

again turned ON.

Since thyristor does not conduct over the entire input so it is called

DISCONTINUOUS CONDUCTION MODE

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CALCULATIONS t

tvtvvv imio sin2sin

otherwisev ,00

Average value of output voltage

ttdvv iav sin22

1

)cos(cos2

,

iav

vvonsolving

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RMS VALUE OF OUTPUT VOLTAGE

2

122 )sin2

2

1( ttdvv irms

21

)2cos1(1

2tdt

vv irms

21

2

2sin

2

2sin(

1

2

irms

vv

21

2

2sin2sin(

2

irms

vv

)2cos1(2

12 ttSin

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SINGLE PHASE HALF CONTROLLED BRIDGE

Assume, load is highly inductive Because 90% of loads are inductiveCASE I

T1

T4

D3

D2

LOA

D A B

DURING +VE HALF CYCLE T1 &

D2 ARE FORWARD BIASED.

AT ωt = α, T1 IS TRIGGERED AND ITS

STARTS CONDUCTING.

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SINGLE PHASE HALF CONTROLLED BRIDGE

At ωt =π, D3 starts conducting and -ve voltage appear across T1 & it

starts blocking voltage. THIS PROCESS IS CALLED LINE

COMMUTATION

T1

D2

A B

α to π

SINCE LOAD IS HIGHLY INDUCTIVE FREEWHEELING CURRENT FLOWS

THROUGH D2-D3FROM ωt =π To π+α

D3

D2

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SINGLE PHASE HALF CONTROLLED BRIDGE

At ωt = π, T4 is in forward blocking mode at ωt = π +α T4 is triggered & +ve voltage appears across D2 and it stops conducting.

At ωt = 2π, D2 starts conducting and T4 is OFF, this is called LINE COMMUTATION

D3

T4

A

B

From ωt = 2π to 2π+α, only D2-D3 conducts,

because of load

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SINGLE PHASE HALF CONTROLLED BRIDGE

2π+α,α

π

π+α

Since thyristor does not conduct over the entire input so it is called

DISCONTINUOUS CONDUCTION MODE

Avg. current rating of thyristor is < that of diode, which is not desirable

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CALCULATIONS

cos1 mav

vV

Vav is +ve and Iav is also +ve so source is supplying power to the load or we can say that load is PASSIVE.

Vav

α

------

----

----

---

1

0.5

π/2

π

Vo

Io

Operation in 1st Quadrant

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SINGLE PHASE HALF CONTROLLED BRIDGE

CASE II: CONTINUOUS CONDUCTION MODE

T1 T3

D4 D2

LOAD

DURING α to π OPRATION OS SAME AS THAT OF CASE I.

FROM π to π+α D4 STARTS CONDUCTING & T3 IS IN FORWARD BLOKING MODE. THUS NO CONDUCTION.T1 & D4 CONDUCTS.

A B

α - π

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SINGLE PHASE HALF CONTROLLED BRIDGE

FROM π to π+α FROM π+α to 2π

T1

D4

At ωt = π+α, T3 IS TRIGGERED & T1

COMMUTATES ( LINE COMMUTATION)

T3

A B

D4

FROM 2π, D2 STARTS CONDUCTING & T1 IS IN FORWARD BLOKING MODE. THUS NO CONDUCTION.T3 & D2 CONDUCTS.

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SINGLE PHASE HALF CONTROLLED BRIDGE

2π+α,α

2π π

π+α

Since thyristor conducts over the entire input so it is called

CONTINUOUS CONDUCTION MODE

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SINGLE PHASE HALF CONTROLLED BRIDGECONTINUOUS CONDUCTION MODE

DISCONTINUOUS CONDUCTION MODE

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FULLY CONTROLLED BRIDGE

ASSUMPTIONS:a) LOAD CURRENT IS CONSTT. & RIPPLE FREE.

b) Io IS CONTINUOUS.IN THE +VE HALF, T1 T2 ARE FORWARD

BIASED.

IN THE NEGATIVE HALF T3 T4 ARE FORWARD

BIASED.

TI & T2 CONTINUE TO CONDUCT TILL T3 & T4

ARE TRIGGERED.

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FULLY CONTROLLED BRIDGE

SINCE THERE ARE TWO PULSESPER CYCLE SO IT ISCALLED :TWO PULSE CONVERTER

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AVERAGE VALUE OF OUTPUT VOLTAGE

ttdvV mav

sin1

cos

2 mav

vV

onsolving

Thus, Vav is +ve from o ˂α ˂ π/2 Vav is –ve from π/2 ˂ α ˂ π

Vo

Io

-Vo

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REFERENCES

M.H.RASHID,POWER ELECTRONICS: CIRCUITS, DEVICES & APPLICATIONS, PENTICE HALL OF INDIA,(IIIRD Ed.),2004

CYRIL LANDER, POWER ELECTRONICS, MCGRAW HILLS CO. ,(IIIRD Ed.), 1993

B.K. BOSE, MODERN POWER ELECTRONICS & A.C. DRIVES, PEARSON EDUCATION INC.,2002

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