Power Electronics 1-2015bjr

06/11/2015

by ZEN-15 1

Presented by:Moh. Zaenal Efendi (ZEN)

[email protected] JOSS

Industrial Electrical EngineeringElectronics Engineering Polytechnic of Surabaya

POWER ELECTRONICS

1

PART-1

INTRODUCTION TO POWER ELECTRONICS SINGLE PHASE UNCONTROLLED RECTIFIER THREE PHASE UNCONTROLLED RECTIFIER SINGLE PHASE CONTROLLED RECTIFIER THREE PHASE CONTROLLED RECTIFIER SINGLE PHASE AC-AC VOLTAGE

CONTROLLER

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Generally: Electronics: Solid State Electronics Devices and

their Driving Circuits. Power: Static and Dynamic Requirements for

Generation, Conversion and Transmission of Power.

Control: The Steady State and Dynamic Stability of the Closed Loop system.

POWER ELECTRONICS may be defined as the application of Solid State Electronics for the Control and conversion of Power.

3

Definition of Power Electronics

DEFINITION:To convert, or to process and control the flow of electric power by supplying voltages and currents in a form that is optimally suited for user loads.

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Detailed Block Diagram of Power Electronics System

Filter&

RectifyPE Circuit Load

Filter&

Rectify

ControlCircuit

Mechanical VariableFeedback

Electrical VariableFeedback

Input

Form ofelectricalenergy

ElectricalMechanical

Pre-stage Power proc. stage Post stage Output

Form of elec. ormechan. energy

Switc

hD

rives

Process feedbacksignals and decide

on control

Could generateundesirablewaveforms

Mostly ac linevoltage (singleor three phase)

Mostlyunregulate

d dcvoltage

Interface betweencontrol and power

circuits 5

Applications Static applications involves non-rotating or moving mechanical

components.Examples:

DC Power supply Un-interruptible power supply, Power generation

and transmission (HVDC), Electroplating, Welding, Heating, Cooling, Electronic ballast

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Applications

Drive applicationscontains moving or rotating components such

as motors.Examples:

Electric trains, Electric vehicles, Air-conditioning System, Pumps, Compressor,

Conveyer Belt (Factory automation).

7

Other Applications

Heating, coolingElectroplating, Welding

Photovoltaic Systems.eV (fuel cell, Solar)

Wind-electric systems.8

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Charge Controller 12V Battery

12VDC/220VAC

Power Inverter

12V/100W

Solar Panel

9

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Power Electronics Converters

AC to DC: RECTIFIER

DC to DC: CHOPPER

DC to AC: INVERTER

AC to AC: CYCLOCONVERTER

11

Types Of Power Conversion

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14Superconducting Magnet Energy Storage System

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FUEL CELL DRIVER

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Power Electronics

AC TO DC CONVERSION (RECTIFIER)

Re-edited by : ZEN (ELIN-PENS)20

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Overview Single-phase, half wave

rectifier Uncontrolled R load R-L load R-C load Free wheeling diode Controlled

Single-phase, full wave rectifier R load R-L load, Continuous and

discontinuous current mode

controlled

Three-phase rectifier uncontrolled controlled

21

Rectifiers

DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches.

Basic block diagram

22

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Rectifiers

Input can be single or multi-phase (e.g. 3-phase).

Output can be made fixed or variable

Applications: DC welder, DC motor drive, Battery charger,

DC power supply, HVDC23

Root-Mean-Squares (RMS)

2

0

.21 td

2(.)

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Root Mean Squares of f

2)( fStep 1:

2

0

2)(21 tdfStep 2:

2

0

2)(21 tdfStep 3:

25

Concept of RMS

tv

v2 Average of v2

Square root of the average of v2 Average

of v=026

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Performance parameters

27

)()()( dcodcodco IVP

)()()( rmsormsoaco IVP

)(

)(

aco

dco

PP

)(

)(

dco

rmso

VV

FF

11

;

2

2

)(

)(

2)(

2)(

)(

FFVV

RF

VVVVVRF

dco

rmso

dcormsoacdco

acRipple Factor :

Form Factor :

Effiency:

DC Output Power:

AC Output Power:


28

Power Factor :

Crest Factor:

Transformer Ratio:

Transformer Utilization Factor (TUF):

sekunders

primers

VV

a(

)(

)()(

2)(

)()(

2)(

)()(

)(

//

rmsormss

dco

rmsormss

dco

rmssrmss

dco

VVV

RVVRV

IVP

TUF

)(

)(

rmss

peaks

II

CF

)(

)(

)()(

2)(

)()(

2)(

///

rmss

rmso

rmsormss

rmso

rmssrmss

rmso

VV

RVVRV

IVRV

SPpf

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Single-Phase Uncontrolled Half-Wave Rectifier

Resistor Load

Considering the diode is ideal, the voltage at R-load during forward biased is the positive cycle of voltage source, while for negative biased, the voltage is zero.

29

Resistor Load

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Output Voltage

mm

o

mDCaveo

ss

VVV

tdtVVVV

voltageoutputDCtVtvSourceVoltageGiven

318.0

)sin(21

,""),sin()(,

0

VV

22V

V

1)(1)((2V

V

0)(-Cos-Cos2V

V

tCos2V

V

dt0dttSinV21V

v(t)dtT1VV

m

m

m

m

m

m

o(dc)

o(dc)

o(dc)

o(dc)

0o(dc)

2

0o(dc)

T

0o(dc)avg

31

RMS OUTPUT VOLTAGE

2

4

0sin2102sin

21

4

2sin21

4

)(2cos121

2

)(sin(21

)(

2

)(

)(

0

2

)(

0

2

)(

0

2)(

mrmso

mrmso

mrmso

mrmso

mrmso

mrmso

VV

VV

VV

ttVV

tdVV

tdtVV

32

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Half-wave with R-C load In some applications in which a constant output

is desirable, a series inductor is replaced by a parallel capacitor.

The purpose of capacitor is to reduce the variation in the output voltage, making it more like dc.

The resistance may represent an external load, while the capacitor is a filterof rectifier circuit.

33

Half-wave with R-C load Assume the capacitor is

uncharged, and as source positively increased, diode is forward biased

Capacitor is charged to Vm as input voltage reaches its positive peak at t = /2.

As diode is on, the output voltage is the same as source voltage, and capacitor charges.

As source decreases after t = /2, the capacitor discharges into load resistor. As diode is reversed biased, the load is isolated from source, and the output voltage (capacitive charge) decaying exponentially with time constant RC.

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Half-wave with R-C load

The effectiveness of capacitor filter is determined by the variation in output voltage, or expressed as maximum and minimum output voltage, which is peak-to-peak ripple voltage.

35

Half-wave with R-C load (Ripple Voltage) The ripple can be approximated as:

fRCV

RCVV mmo

2

The ripple factor

The average output voltage:

2)(

omdco

VVV

)(;22:; dcodcoac

dc

ac VVVVwhereVVRF

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Assuming ideal diodes

offdiodeseVonetV

tvRCt

m

m

, sin

onpair diode ,|sin|)(

)/()(0

the angle where the diodes become reverse biased, which is the same as for the half-wave rectifierand is

)RC(Tan)RC(Tan 11 2t

)2sin( sin )/()( mRC

m VeV0sin)(sin )/()2( RCe

= solved numerically for

Peak-to-peak variation (ripple)

)sin1(|)2sin(| mmmo VVVV

In practical circuits where RC

2 , 2

minimal output voltage occurs at 2t

)/(2)/()222(

0 )2(RC

m

RC

m eVeVv

fRCV

RCV

RCV

eVeVVV

mm

m

RCm

RCmmo

2

211

1 )/()/(

f

xxxe

TheoryTaylor

x

2

...321

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Battery Charger To supply a dc source

from an ac source

The diode will remain off as long as the voltage of ac source is less than dc voltage.

Diode starts to conduct at t=. Given by,

39

For Vs>E, diode D conducts. The angle whwn the diode starts conducting can be found from condition:

Vm sin = E

and = arcsin(E/Vm)

Diode will be tirned off when Vs

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The average charging current:

EEVR

I

tdR

EtVI

mdc

mdc

2cos22

1

)(sin21

Eficiency:

RIPIEPPP

PrmsRdcdc

Rdc

dc

2;;

The rms battery current:

2

22

2

22

2

cos42sin2

222

1

)(sin21

rmsrms

mmm

rms

mrms

II

EVVEVR

I

tdR

EtVI

41

FULL WAVE RECTIFIERThe objective of full wave rectifier is to

produce a voltage or current which is purely dc or has some specified dc component.

While the purpose of full wave rectifier is basically the same as that of half-wave rectifier, full wave rectifier have some fundamental advantages.

The output of the full wave rectifier has inherently less ripple than half wave rectifier.

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FULL WAVE RECTIFIER Can be asBridge rectifierCenter-tapped transformer

Center-tapped rectifier requires center-tap transformer. Bridge does not.

Center tap requires only two diodes, compared to four for bridge. Hence, per half-cycle only one diode volt-drop is experienced. Conduction losses is half of bridge.

Bridge Rectifier

Center-tapped Transformer Rectifier

43

44

RV

RV

I

VtdtVV

mdcodco

mmdco

2

2)()(sin1

)()(

0)(

The dc component of the output voltage is the average value, and load current is simply the resistor voltage divided by resistance.

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Bridge waveforms

45

Center-tapped waveforms

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Full-wave with R-C load (Ripple Voltage)

fRCV

RCVV mmo 2

Ripple can be approximated as:

2)(

omdco

VVV

the average output voltage:

)(;22:; dcodcoac

dc

ac VVVVwhereVVRF

The ripple factor:

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Fullwave Rectifier Battery Charger

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Key waveforms

49

The average charging current:

EEVR

I

EEVR

I

tdR

EtVI

mdc

mdc

mdc

2cos21

2cos22

2

)(sin22

Eficiency:

RIPIEPPP

PrmsRdcdc

Rdc

dc

2;;

The rms battery current:

2

22

2

22

22

2

22

2

cos42sin2

22

1

cos42sin2

222

2

)(sin22

rmsrms

mmm

rms

mmm

rms

mrms

II

EVVEVR

I

EVVEVR

I

tdR

EtVI

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THREE PHASE UNCONTROLLED RECTIFIER

Edited by : M. Zaenal Efendi

51

THREE PHASE UNCONTROLLED RECTIFIER

Three phase rectifiers are commonly used in industry to produce a dc voltage and current for large loads high power levels output power in excess of 4 kW for a 220/380 AC system. They have several advantages to offer compared with the single-phase diode rectifiers.

These include: Lower output voltage ripple No triple harmonics in a three-wire configuration As already mentioned higher power handling capability

for a given supply-side voltage and current.

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Review on three phase voltage and current

Star connection

phLline

phNLLLline

IIIVVVV

33

53

Review on three phase voltage and current

Delta connection

phLline

phNLLLline

III

VVVV

3

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Delta-Star Transformer

55

Delta-Delta Transformer

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POWER IN THREE PHASE SYSTEM

ACTIVE/TRUE POWER in LOAD

RV

P

RVP

CosIVP

CosIVP

ph

NL

LLL

LNL

2

2

3

3

3

3

57


REACTIVE POWER in LOAD

XV

Q

XVQ

SinIVQ

SinIVQ

ph

NL

LLL

LNL

2

2

3

33

3

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APPARENT POWER in LOAD

ZV

S

ZVS

IVS

IVS

ph

NL

LLL

LNL

2

2

3

33

3

59

3 Phase Half-wave Rectifier

This rectifier may be considered as 3 single phase halfwave rectifiers and can be considered as a half-wave type. The k-th diode will conduct during the period when the voltage of k-th phase is higher than other phases. The conduction period of each diode is 2/3.

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Input and output voltage of 3 phase Half-wave Rectifier

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The average output voltage of 3 phase halfwave rectifier

3

3sin

3sin2

3/2

3sin

3sin

3/2

sin3/2

cos3/2

)()(

)()(

)()(

3

3

)()(

3

3

)()(

NLmdco

NLmdco

NLmdco

NLmdco

NLmdco

VV

VV

VV

tV

V

tdtV

V

63

))(()(

)(max)()(

)max(,)(

)(

)(

)()(

17.1

8274.02

33

3/23

3/3

sin

NLrmssdco

NLsdco

NLsdco

NLm

dco

NLmdco

VVVV

VV

VV

VV

The average output voltage of 3 phase halfwave rectifier

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))(()(

)(max)()(

)()(

3/

0

22)(

189.1841.0

32sin

21

323

)(cos3/2

2

NLrmssrmso

NLsrmso

NLmrmso

mrmso

VVVV

VV

tdtVV

The rms output voltage of 3 phase halfwave rectifier

65

The rms input current of 3 phase halfwave rectifier

RV

RV

RVI

I

NLrmss

NLrmss

rmsormsormss

))((

))((

)()()(

687.03

189.133

The rms output current of 3 phase halfwave rectifier

RV

I rmsormso)(

)(

66

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67

)()()( dcodcodco IVP

)()()( rmsormsoaco IVP

)(

)(

aco

dco

PP

)(

)(

dco

rmso

VV

FF

Ripple Factor :

Form Factor :

Effiency:

DC Output Power:

AC Output Power:

11

;

2

2

)(

)(

2)(

2)(

)(

FFVV

RF

VVVVVRF

dco

rmso

dcormsoacdco

ac

Power Factor

)()(,

2)(

)(

3/

rmssNLrmss

rmso

aco

IVRV

PF

SP

PF

where:

2)max(,

)(,NLs

NLrmss

VV

RV

RV

RVI

I NLrmssNLrmssrmsormsormss)(,)(,)()(

)(

687.03

189.133

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Transformer Utilization Factor

;3

/

)()(,

2)(

)(

rmssNLrmss

dco

dco

IVRV

VAP

TUF

where:

2)max(,

)(,NLs

NLrmss

VV

RV

RV

RVI

I NLrmssNLrmssrmsormsormss)(,)(,)()(

)(

687.03

189.133

69

Example

The 3 phase halfwave rectifier is operated from 380 V 50 Hz supply at secondary side and the load resistance is R=20 . If the source inductance is negligible, Determine (a) Rectification efficiency,(b) Form factor(c) Ripple factor(d) Power Factor(e) TUF

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VVVVV NLsNLrmsS 1.3112*220,22063.2193380

)(max,,)(,

max,max,

)( 827.0233

ss

dco VV

V R

VR

VI ssdc

max,max, 827.0233

max,)( 8407.0 srmso VV RV

I srmsomax,

)(

8407.0

%767.96)()(

)()(

)(

)( rmsormso

dcodco

aco

dco

IVIV

PP

%657.101)(

)( dco

rmso

VV

FF

%28.1811 22)(

2)(

)(

2)(

2)(

FFVV

VVV

VVRF

dco

rmso

dco

dcormso

dc

ac

Solution :

71

3-phase Full-wave rectifiers3-phase bridge rectifier is very common in high-power applications. This is a full-wave rectifier. It can operate with or without a transformer and gives 6-pulse ripple on the output voltage.

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Two series diode are always conducting (one feeds into the circuit, the other forms the return path). The diodes are numbered in order of conduction and each diode conducts for 120o. The conduction sequence for diodes is D1D2; D2D3; D3D4; D4D5; D5D6 and D6D1. The pair of diodes which are connected between that pair of supply lines having the highest amount of line-line voltage will conduct.

phlineline VV 3

There are 6 combinations of VL-L. Considering one period of the source to be 360o, a transition of the highest line-line voltage must take place 360o/6=60o.Because of six transition that occur for each period of the source voltage, the circuit is called A SIX-PULSE RECTIFIER

73

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Input and output voltage of 3 phase Full-wave Rectifier

75

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Considers only one of the six segments. Obtain its average over 60 degres

LLm

LLm

LLm

LLmdco

V

V

tV

tdtVV

,

,

3/23/

,

3/2

3/,)(

955.0

3

cos3

sin3/

1

3 phase full-wave average voltage

77

LLm

LLm

NLm

NLm

NLm

wavehalfdcdco

V

V

orV

V

tdtV

VV

,

,

,

,

6/

0,

)()(

955.0

3654.1

33

cos6/2

2

2

3 phase full-wave average voltage

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LLmLLm

NLmNLm

NLmrmso

VV

VV

tdtVV

,,

,,

6/

0

22,)(

95575.04

3923

3

6554.14

3923

cos6/2

2

RV

I rmsormso)(

)(

3 phase full-wave rms output voltage

)(

)(

2)(

)(

2)(

)( ;

aco

dco

rmsoaco

dcodco

PP

R

VP

R

VP

3 phase full-wave rms output currentEfficiency of 3 phase full-wave rectifier

79

Power Factor

rmssLLrmss

rmso

sNLrmss

rmso

aco

IVRV

orIV

RVS

PPF

,)(,

2)(

)(,

2)(

)(

3/

;3

/

RV

I

orRV

Iwhere

II

II

tdtII

LLmm

NLmm

mrmss

mrmss

mrmss

)(

)(

)(

)(

26/

0

2)(

;3

:

;7804.0

22sin

21

62

)(cos28

where:

2

2

)()(,

)()(,

LLmLLrmss

NLmNLrmss

VV

and

VV

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Transformer Utilization Factor

rmssLLrmss

dco

sNLrmss

dco

dco

IVRV

orIV

RVVA

PTUF

,)(,

2)(

)(,

2)(

)(

3/

;3

/

RV

I

orRV

Iwhere

II

II

tdtII

LLmm

NLmm

mrmss

mrmss

mrmss

)(

)(

)(

)(

26/

0

2)(

;3

:

;7804.0

22sin

21

62

)(cos28

where:

2

2

)()(,

)()(,

LLmLLrmss

NLmNLrmss

VV

and

VV

81

Example

The 3 phase full-wave rectifier is operated from 380 V; 50 Hz supply and the load resistance is R=20 . Determine (a) The efficiency, (b) Form factor (c) Ripple factor(d) Power factor(e) TUF

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VVV LLmdco 14.513955.0 ,)(

AR

VI dcodco 66.25

)()(

VVV LLmrmso 544.51395575.0 ,)(

AR

VI rmsormso 68.25

)()(

%83.99)()(

)()(

)(

)( rmsormso

dcodco

aco

dco

IVIV

PP

Solution :

%08.100)(

)( dco

rmso

VV

FF

%411 22)(

2)(

)(

2)(

2)(

FFVV

VVV

VVRF

dco

rmso

dco

dcormso

dc

ac

83

Homework

1. A three phase halfwave rectifier is supplied by 380V-rms, 50Hz. The primary and secondary of input transformer are connected delta-wye and has turns ratio of n=3:1. The load is a 100 ohm. Determine:a. The average load current; b. The rms load current; c. The rms source current; d. The power factor; e. Efficiency; f. Form factor; g. Ripple factor; h. TUF

2. A three phase fullwave rectifier is supplied by 380V-rms, 50Hz. The primary and secondary of input transformer are connected delta-delta and has turns ratio of n=3:1. The load is a 100 ohm. Determine :a. The average load current; b. The rms load current; c. The rms source current; d. The power factor; e. Efficiency; f. Form factor; g. Ripple factor; h. TUF

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Power Electronics

SINGLE PHASE CONTROLLED AC TO DC RECTIFIER

Presented by : M. Zaenal Efendi

85

Type of input: Fixed voltage, fixed frequency ac power supply.

Type of output: Variable dc output voltage

Type of commutation: Natural / AC line commutation

LineCommutated

Converter

+

-

DC OutputV0(dc)

ACInput

Voltage

86

Block Diagram

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Applications of Phase Controlled Rectifiers

DC motor control in steel mills, paper and textile mills employing dc motor drives.

AC fed traction system using dc traction motor.

Electro-chemical and electro-metallurgical processes.

Magnet power supplies.

Portable hand tool drives.

87

Classification of Phase Controlled Rectifiers Single Phase Controlled Rectifiers. Three Phase Controlled Rectifiers.

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Single Phase Half-Wave ThyristorConverter with a Resistive Load

89

The Controlled Half-wave Rectifier

Previously discussed are classified as uncontrolled rectifiers.

Once the source and load parameters are established, the dc level of the output and power transferred to the load are fixed quantities.

A way to control the output is to use SCR instead of diode. Two condition must be met before SCR can conduct: The SCR must be forward biased (VSCR>0) Current must be applied to the gate of SCR

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Controlled, Half-wave R load

A gate signal is applied at t = , where is the delay/firing angle.

91

Average load Voltage (D.C. Voltage)

DC Voltage

max

max

Maximum average (dc) o/pvoltage is obtained when 0 and the maximum dc output voltage

1 cos0 ; cos 0 12

mdmdc

mdmdc

VV V

VV V

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Control Characteristic

VO(dc)

Trigger angle in degrees0 60 120 180

Vdm

0.2 Vdm

0.6Vdm

93

22

0

12

2 2

The RMS output voltage is given by

1 .2

Output voltage sin ; for to

1 sin .2

OO RMS

O m

mO RMS

V v d t

v V t t

V V t d t

Root Mean Square Values

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Root Mean Square Values

RMS Current

For a resistive load VRMS = IRMS R

Note: for = 0 VRMS = 0.5 Vm

2)2sin(1

2

)()]sin([21,

resistor,by absorbedpower Average

0

2,

22

m

mrmso

rms

V

tdtVVwhere

RVRIP rms

95

Performance Parameters

22

2)cos1(

4

RVP mdc

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Example 1 Design a circuit to produce an average voltage of

40V across 100 load resistor from a 120Vrms 50 Hz ac source. Determine the power absorbed by the resistor, efficiency, FF, RF and the power factor.

To obtain Vo=25% of Vdm, determine firing angle, FF, RF, pF and TUF

97

Example 1 (Cont)

Solution

rad

VV

o

so

07.12.61

]cos1[2

212040

]cos1[2

In such that to achieved 40V average voltage, the delay angle must be

If an uncontrolled diode is used, the average voltage would be

That means, some reducing average resistor to the design must be made. A series resistor or inductor could be added to an uncontrolled rectifier, while controlled rectifier has advantage of not altering the load or introducing the losses

V

VV mrmso

6.752

)07.1(2sin07.112

21202

)2sin(12,

WR

VP rms 1.57100

6.75 22

63.0

1006.75)120(

1.57

pf

VVV so 54)120(2

98

06/11/2015

by ZEN-15 50

99

SINGLE PHASE: FULL WAVE BRIDGE CONTROLLED RECTIFIERThere are 2 types of FW Bridge Controlled Rectifiers:

1. Fully Controlled Bridge Converter (Full Converter)

2. Half Controlled Bridge Converter (Semi-Converter)

Single Phase Full Wave Controlled Rectifier With R Load

100

06/11/2015

by ZEN-15 51

101

DC Output Voltage

0

1 .

1 sin .

cos

cos cos ; cos 1

1 cos

dc OO dct

dc mO dc

mdcO dc

mdcO dc

mdcO dc

V V v d t

V V V t d t

VV V t

VV V

VV V

102

06/11/2015

by ZEN-15 52

RMS Output Voltage

2sin212sin

21

2

2sin21

2

)(2cos121

)(sin(1

2

)(

2

)(

2

)(

2)(

mrmso

mrmso

mrmso

mrmso

VV

ttVV

tdVV

tdtVV

103

22sin1

42sin

221

42sin2sin

221

2sin212sin

21

21

)()(

)(

)(

rmssmrmso

mrmso

mrmso

VVV

VV

VV

Power Electronics

THREE PHASE CONTROLLED AC TO DC RECTIFIER

Presented by : M. Zaenal Efendi104

06/11/2015

by ZEN-15 53

105

3 Phase Controlled Rectifiers Operate from 3 phase ac supply voltage.

They provide higher dc output voltage.

Higher dc output power.

Higher output voltage ripple frequency.

Filtering requirements are simplified for smoothing out load voltage and load current.

Extensively used in high power variable speed industrial dc drives.

Three single phase half-wave converters can be connected together to form a three phase half-wave converter.

105

Vector Diagram of 3 Phase Supply Voltages

106

06/11/2015

by ZEN-15 54

3-pulse (Half Wave) Controlled Rectifier With Resistive Load

107

0

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200

Vs

V0

Van

=0

=150

Vbn Vcn

t

Van Vbn Vcn

t

Waveformes

=00

=150

For R- Loads the current waveform is similar

to the voltage waveforms.

The FWD has no effect for small values of .

For >30 the FWD sets the load voltage at 0

leading to higher DC output voltage.

108

06/11/2015

by ZEN-15 55

0

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200

V0

=300

Van Vbn Vcn

t

V0

=600

Van Vbn Vcn

t

=300

=600

Waveforms for different

109

01

0 01

02

0 02

0

306

30 180 ;

sin5 1506

150 300 ;

sin 120

O an m

O bn m

T is triggered at t

T conducts from to

v v V t

T is triggered at t

T conducts from to

v v V t

Average Voltage (DC Voltage) >30

110

06/11/2015

by ZEN-15 56

0

0

0

0

0

0

180

30

0 0

180

30

180

30

3 .2

sin ; for 30 to 180

3 sin .2

3 sin .2

dc O

O an m

dc m

mdc

V v d t

v v V t t

V V t d t

VV t d t

Average Voltage (DC Voltage)

111

112

)30cos(12

3

)30cos(180cos2

32

3

)()(

)()(

180

30

)()( cos

oNLmdco

ooNLmdco

NLmdco

VV

VV

VV t

o

o

DC Output Voltage of 3-pulse (Half Wave) Controlled Rectifier With Resistive Load

06/11/2015

by ZEN-15 57

113

RMS Output Voltage of 3-pulse (HWR) Controlled Rectifier With Resistive Load

260sin81

42453

)()(sin23

)()(

6

22)()(

oNLmrmso

NLmrmso

VV

tdtVV

114

The RMS input current of Half Wave Controlled Rectifier

RVI

I rmsormsormss

33

)()()(

So, we can calculate the power factor dan TUF of Half Wave Controlled Rectifier

)()(,

2)(

)(

3/

rmssNLrmss

rmso

aco

IVRV

PF

SP

PF

;

3/

)()(,

2)(

)(

rmssNLrmss

dco

dco

IVRV

VAP

TUF

06/11/2015

by ZEN-15 58

115

Three Phase Full Converter(Full Wave Controlled Rectifier)

3 Phase Fully Controlled Full Wave Bridge Converter.

Known as a 6-pulse converter.

Used in industrial applications up to 120kW output power.

Two quadrant operation is possible.

115

Circuit Layout

116

06/11/2015

by ZEN-15 59

Circuit of Three Phase Full Converter

117

Waveforms

= 0118

06/11/2015

by ZEN-15 60

Waveforms

= 30 119

120

The thyristors are triggered at an interval of / 3.

The frequency of output ripple voltage is 6fS. T1 is triggered at t = (/6 + ), T6 is already conducting

when T1 is turned ON.

During the interval (/6 + ) to (/2 + ), T1 and T6 conduct together & the output load voltage is equal to vab = (van vbn)

Operation of the circuit

120

06/11/2015

by ZEN-15 61

T2 is triggered at t = (/2 + ), T6 turns off naturally as it is reverse biased as soon as T2 is triggered.

During the interval (/2 + ) to (5/6 + ), T1 and T2conduct together & the output load voltage vO = vac = (van vcn)

Thyristors are numbered in the order in which they are triggered.

The thyristor triggering sequence is 12, 23, 34, 45, 56, 61, 12, 23, 34,

Operation of the circuit

121

Sequence of Gate Pulses of Thyristor

122

06/11/2015

by ZEN-15 62

Table of the thyristor pair in conduction at any instant.

123

The corresponding line-to-line supply voltages are

3 sin6

3 sin2

3 sin2

RY ab an bn m

YB bc bn cn m

BR ca cn an m

v v v v V t

v v v v V t

v v v v V t

Three phase line voltages

124

06/11/2015

by ZEN-15 63

125

2

6

6 . ; 2

3 sin6

dc OO dc

O ab m

V V v d t

v v V t

The output load voltage consists of 6 voltage pulses over a period of 2 radians, Hence the average output voltage is calculated as

DC output voltage

125

The maximum dc voltage is given by

126

)cos(3

)cos(33

)()6

sin(33

)()(

)()(

2

6)()(

LLmdco

NLmdco

NLmdco

VV

or

VV

tdtVV

)()(

max)(

333 LLmNLmdco

VVV

DC Output Voltage of FULL WAVE Controlled Rectifier With Resistive Load

06/11/2015

by ZEN-15 64

127

RMS Output Voltage of Full Wave Controlled Rectifier With Resistive Load

2cos4

3321

2cos4

33213

)()6

(sin33

)()(

)()(

2

6

22)()(

LLmrmso

NLmrmso

NLmrmso

VV

or

VV

tdtVV

127

128

06/11/2015

by ZEN-15 65

129

The RMS input current of Full Wave Controlled Rectifier

)()( 64

rmsormss II

So, we can calculate the power factor dan TUF of Full Wave Controlled Rectifier

rmssLLrmss

dco

sNLrmss

dco

dco

IVRV

orIV

RVVA

PTUF

,)(,

2)(

)(,

2)(

)(

3/

;3

/

rmssLLrmss

rmso

sNLrmss

rmso

aco

IVRV

orIV

RVS

PPF

,)(,

2)(

)(,

2)(

)(

3/

;3

/

130

AC-AC VOLTAGE CONTROLLER

RE-EDITED BY ZENELIN

Power Electronics

06/11/2015

by ZEN-15 66

131131

Prof. M. Madhusudhan Rao, E&C Dept., MSRIT

AC Voltage Controller (RMS Voltage Controllers)

ACVoltage

Controller

V0(RMS)

fS

Variable AC RMS O/P Voltage

ACInput

Voltagefs

Vs

fs

132132


Applications Lighting / Illumination control in ac power circuits. Induction heating. Industrial heating & Domestic heating. Transformer tap changing (on load transformer

tap changing). Speed control of induction motors (single phase

and poly phase ac induction motor control). AC magnet controls.

06/11/2015

by ZEN-15 67

133133


Single Phase Full Wave AC Voltage Controller

Principle of AC Phase Control

134134


Single Phase Full Wave AC Voltage Controller using Triac

06/11/2015

by ZEN-15 68

135135


136

The principle of operation for a single-phase ac voltage controller using phase control is similar to the controlled half-wave rectifier.An analysis identical to that done for the controlled half-wave rectifier can be done on a half cycle for the voltage controller. Then, by symmetry, the result can be extrapolated to describe the operation for the entire period. Some basic observations about cycloconverters are as follows:

The SCRs cannot conduct simultaneously.The load voltage is the same as the source voltage when either SCR is on. The load voltage is zero when both SCRs are off.

06/11/2015

by ZEN-15 69

137137


Input supply voltage

sin 2 sinOutput voltage across the load resistor

sin ;for to & to 2

S m S

O L m

v V t V t

v v V tt t

Equations

138138


1

2

Thyristor is triggered at is triggered at

Output load currentsin sin ;

for to & to 2

O mO m

L L

T tT t

v V ti I tR R

t t

06/11/2015

by ZEN-15 70

139139


To Derive an Expression forthe RMS Value of Output Voltage

2

2 2 2

0

The RMS value of output voltage

12

For a FW ac voltage controller, we can see that two half cycles of o/p voltage w.f. are symmetrical & o/p pulse time period is radians. We

LO RMS L RMSV V v d t

can also calculatethe RMS o/p voltage by using the expression

140140


2 2 2

02

2 2

0

1 sin

1 ;2

sin ;For to & to 2

mL RMS

LL RMS

L O m

V V t d t

V v d t

v v V tt t

06/11/2015

by ZEN-15 71

141141


2

22 2

22 2 2 2

22

Hence

1 sin sin2

1 sin . sin .2

1 cos 2 1 cos 22 2 2

L RMS

m m

m m

m

V

V t d t V t d t

V t d t V t d t

V t td t d t

142142


2 22

2 22

2

2

cos2 . cos2 .2 2

sin2 sin24 2 2

1 1sin2 sin2 sin4 sin24 2 2

1 12 0 sin2 0 sin24 2 2

m

m

m

m

V d t t d t d t t d t

V t tt t

V

V

06/11/2015

by ZEN-15 72

143143


2

2

2

sin 2sin 224 2 2

sin 2 2sin 224 2 2

sin 2 12 sin 2 .cos2 cos2 .sin 24 2 2

sin 2 0 & cos2 1;

m

m

m

V

V

V

144144


22

2

22

Therefore,sin 2 sin 22

4 2 2

2 sin 24

2 2 sin 24

mL RMS

m

mL RMS

VV

V

VV

06/11/2015

by ZEN-15 73

145145


Taking the square root, we get

2 2 sin 22

2 2 sin 22 2

1 2 2 sin 222

mL RMS

mL RMS

mL RMS

VV

VV

VV

146146


1 sin 222 22

1 sin 222

1 sin 22

1 sin 22

mL RMS

mL RMS

L RMS i RMS

SL RMS

VV

VV

V V

V V

06/11/2015

by ZEN-15 74

147147


M aximum RM S voltage will be applied to the load when 0, in that case the full sine wave appears across the load. RM S load voltage will be the same as

the RM S supply voltage .2

W hen is increased th

mV

e RM S loadvoltage decreases.

148148


0

0

0

1 sin 2 0022

1 022

2

mL RMS

mL RMS

mSL RMS i RMS

VV

VV

VV V V

06/11/2015

by ZEN-15 75

149149


Control Characteristic of Single Phase FW AC Voltage Controller With Resistive Load

1 sin 2 ;

2

Where RMS value of 2

input supply voltage

SO RMS

mS

V V

VV

150150


06/11/2015

by ZEN-15 76

151151


VO(RMS)

Trigger angle in degrees

0 60 120 180

VS

0.2 VS

0.6VS

152Prof. M. Madhusudhan Rao, E&C Dept., MSRIT

Need for Isolation

Pulse Transformer

G1

K1G2

K2

GateTriggerPulse

Generator

152

06/11/2015

by ZEN-15 77

153Prof. M. Madhusudhan Rao, E&C Dept., MSRIT

Problems in AC Voltage Controllers

154154


A single phase full wave controller has an input voltage of 120 V (RMS) and a load resistance of 6 ohm. The firing angle of thyristor is 90o. Find RMS output voltage Power output Input power factor If the output power decreases to 20%

of the maximum power, determine : firing angle and Vo

06/11/2015

by ZEN-15 78

155

0

12

12

90 , 120 V , 62

RM S Value of O utput V oltage

1 sin 22

1 sin 1801202 2

84.85 V olts

S

O S

O

O

V R

V V

V

V

156

2

2

RMS Output Current84.85 14.14 A

6Load Power

14.14 6 1200 wattsI/P Current is same as Load Current

14.14 AmpsInput Supply Volt-Amp

120 14.14 1696.8

OO

O O

O

S O

S S

VIRP I R

P

I I

V I VA

06/11/2015

by ZEN-15 79

Question

220 V and 2.2 kW heater is fed by a singlephase AC chopper which is connected 220 VAC grid.

a) Calculate the resistance of the heaterb) For =90, find load voltage and powerc) To obtain the output volatge 15 % of

maximum voltage, determine the delayangle.

06/11/2015

by ZEN-15 80

159

Find the RMS and average current flowing through the heater shown in figure. The delay angle of both the SCRs is 450.

SCR2

SCR1 io+

1 kW, 220Vheater

1-220V

ac

References1. Muh. H. Rashid, Power Electronics, Circuit, Devices and

Applications, Prentice Hall, 19932. Daniel W. Hart, Introduction to Power Electronics, Prentice

Hall, 19973. William Shepherd, Power Converter Circuits, Taylor&Francis,

20044. Mohd Shawal Jadin, Hand-out of Power Electronics5. Yasser. G. Hegazy, Hand-out of Power Electronics6. M. Madhusudhan Rao, , Hand-out of Power Electronics7. Univ of Central Florida, Hand-out of Power Electronics

160

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