Unit 2

UNIT II

PHASE CONTROLLED CONVERTERS

EE2301- POWERELECTRONICS

Phase-Control ConvertersPhase-Control Converters

Single-PhaseSingle-Phase

SemiconverterSemiconverter

Three-PhaseThree-Phase

Full converterFull converter

Dual converterDual converter

SemiconverterSemiconverter

Full converterFull converter

Dual converterDual converter


Semiconverter..is a one-quadrant converter and it has one polarity

Semiconverter..is a one-quadrant converter and it has one polarity

Full converter..is a two-quadrant converter and the polarity of its

output can be either positive or negative.However

the output current of full converter has one polarity only

Full converter..is a two-quadrant converter and the polarity of its

output can be either positive or negative.However

the output current of full converter has one polarity only

Dual converter..can operate in four quadrants ; both the output

voltage and current can be either positive or negative

Dual converter..can operate in four quadrants ; both the output

voltage and current can be either positive or negative



cos12

sin2

1 mmdc

VttdVV

cos12

sin2

1 mmdc

VttdVV

Average Output VoltageAverage Output Voltage

m

dm

VV

m

dm

VV Maximum

Output Voltage

MaximumOutput Voltage

cos15.0 dm

dcn V

VV cos15.0

dm

dcn V

VVNormalizing

Output Voltage

NormalizingOutput Voltage

2

2sin1

2sin

2

1 22

mmrms

VttdVV

2

2sin1

2sin

2

1 22

mmrms

VttdVV

RMS Output VoltageRMS Output Voltage


If the converter has a purely resistive load of R and the delay angle is , determine

(a) the rectification efficiency(b) the form factor FF(c) the ripple factor RF

and (d) the peak inverse voltage PIV of thyristor T1




2/


%27.203536.0

1592.0

3536.02

22

sin2

1

2

1592.0

2cos1

2sin

2

1

2

2

2

2

2

m

m

rms

dc

mm

rms

mdc

mmdc

V

V

V

V

VV

V

VV

VttdVV

%27.203536.0

1592.0

3536.02

22

sin2

1

2

1592.0

2cos1

2sin

2

1

2

2

2

2

2

m

m

rms

dc

mm

rms

mdc

mmdc

V

V

V

V

VV

V

VV

VttdVV








2/

221.21592.0

3536.0

m

m

dc

rms

V

V

V

VFF 221.2

1592.0

3536.0

m

m

dc

rms

V

V

V

VFF








2/

983.11221.21 22 FFRF 983.11221.21 22 FFRF

mVPIV mVPIV

EE2301- POWERELECTRONICSSemiconverter


Single-Phase SemiconverterSingle-Phase Semiconverter

ttdVV

VttdVV

mrms

mmdc

22 sin2

2

cos1sin2

2

ttdVV

VttdVV

mrms

mmdc

22 sin2

2

cos1sin2

2


Single-Phase Semiconverter (RL-load)Single-Phase Semiconverter (RL-load)

L

R

L

R

LLL

LL

eR

EeItiI

ERidt

diL

1

0

011

11

L

R

L

R

LLL

LL

eR

EeItiI

ERidt

diL

1

0

011

11

tL

RS

LS

L

SLL

eZ

V

R

EI

R

Et

Z

VI

tVERidt

diL

sin2

sin2

sin2

12

22

tL

RS

LS

L

SLL

eZ

V

R

EI

R

Et

Z

VI

tVERidt

diL

sin2

sin2

sin2

12

22

Mode 1Mode 1 t0

Mode 2Mode 2 t

R

L 1tan R

L 1tan 22 LRZ 22 LRZ


Single-Phase Semiconverter (RL-load)Single-Phase Semiconverter (RL-load)

RMS Currentfor Thyristor


tdiI LR

222

1



tdiI LA 22

1

RMS OutputCurrent

RMS OutputCurrent

tditdiI LLrms

220

21 2

1

2

1

AVG OutputCurrent

AVG OutputCurrent

tditdiIdc 20 1 2

1

2

1


The single-phase semiconverter has an RL load of L = 6.5mH, R = 2.5 Ohm, and E = 10 V. The input v

oltage is VS = 120 V(rms) at 60 Hz. Determine(a) the load current IL0 at , and the load curren

t IL1 at ,(b) the average thyristor current IA

(c) the rms thyristor current IR

(d) the rms output current Irms

and (e) the average output current Idc

The single-phase semiconverter has an RL load of L = 6.5mH, R = 2.5 Ohm, and E = 10 V. The input v

oltage is VS = 120 V(rms) at 60 Hz. Determine(a) the load current IL0 at , and the load curren

t IL1 at ,(b) the average thyristor current IA

(c) the rms thyristor current IR

(d) the rms output current Irms

and (e) the average output current Idc

0t 60t


Single-PhaseFull Converter

RectificationMode

RectificationMode

InversionMode

InversionMode


Single-Phase Full ConverterSingle-Phase Full Converter

2

sin2

2

cos2

sin2

2

22 mmrms

mmdc

VtdtVV

VtdtVV

2

sin2

2

cos2

sin2

2

22 mmrms

mmdc

VtdtVV

VtdtVV


Single-Phase Full Converter (RL-load)Single-Phase Full Converter (RL-load)

tL

RS

LS

L eZ

V

R

EI

R

Et

Z

VI

sin2

sin2

0 tL

RS

LS

L eZ

V

R

EI

R

Et

Z

VI

sin2

sin2

0

Mode 1 = Mode 2Mode 1 = Mode 2R

L 1tan R

L 1tan 22 LRZ 22 LRZ


Single-Phase Full Converter (RL-load)Single-Phase Full Converter (RL-load)



tdiI LR

2

2

1



tdiI LA 2

1

RMS OutputCurrent

RMS OutputCurrent RRRrms IIII 222

AVG OutputCurrent

AVG OutputCurrent AAAdc IIII 2


DualConverter


Single-Phase Dual ConverterSingle-Phase Dual Converter

High-Power Variable-Speed DrivesHigh-Power Variable-Speed Drives

21

22

11

cos2

cos2

dcdc

mdc

mdc

VV

VV

VV

21

22

11

cos2

cos2

dcdc

mdc

mdc

VV

VV

VV


Three-PhaseSemiconverter


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.


3-Phase Half Wave Converter(3-Pulse Converter)

with RL Load

Continuous & ConstantLoad Current Operation



Vector Diagram of 3 Phase Supply Voltages

V A N

V C N

V B N

1 2 00

1 2 00

1 2 00 RN AN

YN BN

BN CN

v v

v v

v v


3 Phase Supply Voltage Equations

We deifine three line to neutral voltages

(3 phase voltages) as follows


0

0

0

sin ;

Max. Phase Voltage

2sin

3

sin 120

2sin

3

sin 120

sin 240

RN an m

m

YN bn m

m

BN cn m

m

m

v v V t

V

v v V t

V t

v v V t

V t

V t


van vbn vcn van


io=Ia

Constant Load Current

Ia

Ia

Each thyristor conducts for 2/3 (1200)


To Derive an Expression for the

Average Output Voltage of a 3-Phase Half Wave Converter

with RL Loadfor Continuous Load Current


01

02

03

0

306

5 150

6

7 270

6

2Each thytistor conducts for 120 or radians

3

T is triggered at t

T is triggered at t

T is triggered at t


5

6

6

5

6

6

3sin .

2

3cos

2

3 5cos cos

2 6 6

mdc

mdc

mdc

VV t d t

VV t

VV


0 0

0

Note from the trigonometric relationship

cos cos .cos sin .sin

5 5cos cos sin sin

6 63

2co

cos 150 cos sin 150 sin3

2 cos 30

s .cos sin sin6 6

.cosm

dc

mdc

A

VV

B A B A B

VV

0sin 30 sin


0 0

0 0 0 0

0 0

0 0

0

0

0

0

0 0

Note: cos 1

cos 180 30 cos sin 180 30 sin3

2 cos 30 .cos sin 30 sin

cos 30 cos sin 30 sin3

2 cos 30 .cos sin 30 s

80 30 cos 30

sin 180 30 sin 30

in

mdc

mdc

VV

VV


032cos 30 cos

2

3 32 cos

2 2

3 3 33 cos cos

2 23

cos2

Where 3 Max. line to line supply voltage

mdc

mdc

m mdc

Lmdc

Lm m

VV

VV

V VV

VV

V V


max

The maximum average or dc output voltage is

obtained at a delay angle 0 and is given by

3 3

2Where is the peak phase voltage.

And the normalized average output voltage is

mdmdc

m

ddcn n

VV V

V

VV V

cosc

dmV


15 26

2 2

6

1

2

The rms value of output voltage is found by

using the equation

3sin .

2

and we obtain

1 33 cos 2

6 8

mO RMS

mO RMS

V V t d t

V V


3 Phase Half Wave Controlled Rectifier Output Voltage Waveforms For RL

Load at

Different Trigger Angles


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

V an

V 0

V 0

V an

= 300

= 600

V bn

V bn

V cn

V cn

t

t

=300

=600


030

060

090

0120

0150

0180

0210

0240

0270

0300

0330

0360

0390

0420

0

V 0

V an

= 900

V bn V cn

t

=900


3 Phase Half Wave Controlled Rectifier With

R Load and

RL Load with FWD


a a

b b

c c

RV 0

L

R V 0

+

T 1

T 2

T 3

n n

T 1

T 2

T 3


3 Phase Half Wave Controlled Rectifier Output

Voltage Waveforms For R Load or RL Load with FWD

atDifferent Trigger Angles


0

0

3 00

3 00

6 00

6 00

9 00

9 00

1 2 00

1 2 00

1 5 00

1 5 00

1 8 00

1 8 00

2 1 00

2 1 00

2 4 00

2 4 00

2 7 00

2 7 00

3 0 00

3 0 00

3 3 00

3 3 00

3 6 00

3 6 00

3 9 00

3 9 00

4 2 00

4 2 00

V s

V 0

V a n

= 0

= 1 5 0

V b n V c n

t

V a n V b n V c n

t

=00

=150


0

0

3 00

3 00

6 00

6 00

9 00

9 00

1 200

1 200

1 500

1 500

1 800

1 800

2 100

2 100

2 400

2 400

2 700

2 700

3 000

3 000

3 300

3 300

3 600

3 600

3 900

3 900

4 200

4 200

V 0

= 3 0 0

V a nV b n V c n

t

V 0

= 6 0 0

V a nV b n V c n

t

=300

=600


To Derive An Expression For The Average Or

Dc Output Voltage Of A 3 Phase Half Wave Converter

With Resistive Load Or

RL Load With FWD


01

0 01

02

0 02

0

306

30 180 ;

sin

5 150

6

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


03

0 03

0

0

7 270

6

270 420 ;

sin 240

sin 120

O cn m

m

T is triggered at t

T conducts from to

v v V t

V t


0

0

0

0

0

0

180

30

0 0

180

30

180

30

3.

2

sin ; for 30 to 180

3sin .

2

3sin .

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


0

0

180

30

0 0

0

0

3cos

2

3cos180 cos 30

2

cos180 1, we get

31 cos 30

2

mdc

mdc

mdc

VV t

VV

VV


Three Phase Semiconverters

• 3 Phase semiconverters are used in Industrial dc drive applications upto 120kW power output.

• Single quadrant operation is possible.• Power factor decreases as the delay angle

increases.• Power factor is better than that of 3 phase half

wave converter.


3 Phase Half Controlled Bridge Converter

(Semi Converter)with Highly Inductive Load & Continuous Ripple free Load

Current



Wave forms of 3 Phase Semiconverter for

> 600




0 0

1

3 phase semiconverter output ripple frequency of

output voltage is 3

The delay angle can be varied from 0 to

During the period

30 210

7, thyristor T is forward biased

6 6

Sf

t

t


1

1 1

If thyristor is triggered at ,6

& conduct together and the line to line voltage

appears across the load.

7At , becomes negative & FWD conducts.

6The load current contin

ac

ac m

T t

T D

v

t v D

1 1

ues to flow through FWD ;

and are turned off.mD

T D


1

2

1 2

If FWD is not used the would continue to

conduct until the thyristor is triggered at

5, and Free wheeling action would

6

be accomplished through & .

If the delay angle , e3

mD T

T

t

T D

ach thyristor conducts

2for and the FWD does not conduct.

3 mD


0

0

0

We deifine three line neutral voltages


sin ; Max. Phase Voltage

2sin sin 120

3

2sin sin 120

3

sin 240

RN an m m

YN bn m m

BN cn m m

m

v v V t V

v v V t V t

v v V t V t

V t

V

is the peak phase voltage of a wye-connected source.m


3 sin6

53 sin

6

3 sin2

3 sin6

RB ac an cn m

YR ba bn an m

BY cb cn bn m

RY ab an bn m

v v v v V t

v v v v V t

v v v v V t

v v v v V t


Wave forms of 3 Phase Semiconverter for

600





To derive an Expression for the

Average Output Voltage of 3 Phase Semiconverter

for > / 3 and Discontinuous Output Voltage


76

6

76

6

For and discontinuous output voltage:3

the Average output voltage is found from

3.

2

33 sin

2 6

dc ac

dc m

V v d t

V V t d t


max

3 31 cos

23

1 cos2

3 Max. value of line-to-line supply voltage

The maximum average output voltage that occurs at

a delay angle of 0 is

3 3

mdc

mLdc

mL m

mdmdc

VV

VV

V V

VV V


17 26

2

6

The normalized average output voltage is

0.5 1 cos

The rms output voltage is found from

3.

2

dcn

dm

acO rms

VV

V

V v d t


• Resistive Load

)3/4sin(2

)3/2sin(2

)sin(2

tVv

tVv

tVv

PHc

PHb

PHa

)6/sin(2 tVvvv LLbaab

Supply Voltages:

ai

bi

ci

1D 3D 5D

4D 6D 2D

dv

av

bv

cv

di

dR

a

b

c

Six-pulse Diode Rectifier


• Waveforms

dvabv acv

1A

bcv cav cbvdv

bav

doV

I II

bv cv avv

0

0

0

ai

61, DD 21, DDON

2t

t

t2

abv

6

2

av

ON

)(6/sin23/

13/

area 2/6/ tdtVV LLdo

1A

LLLL VV 35.123



• Capacitive Load

dCdV

(a)

dV

(b)

R

ai

bi

ci

1D 3D 5D

4D 6D 2D

di

a

b

c

sLav

bv

cv

sL

sL

Assumption:

constant dd VC



• Waveforms

t

t

t

t

t

abv acv bcv bav cav cbv abv acv

2ai

bi

di

61, DD 2

ci

ON21, DD

ON

v

0

0

0

0

0

dV



• Discontinuous Current Operation

3/

1 2 3

di

v abv acv

t

t

dV

3/

4

0

0

pI

5

LL

d

V

V

2sin 1

1

12




3/

1 2 3

di

v abv acv

t

t

dV

3/

4

0

0

pI

5

Differential Equation: dabd

s Vvdt

idL 2 31 t

11 coscos22

1

sin22

11

dLLs

dLLs

d

VVL

tdVtVL

iSolution:

dV

ai

bi

ci

1D 3D 5D

4D 6D 2D

di

a

b

c

sLav

bv

cv

sL

sL




3/

1 2 3

di

v abv acv

t

t

dV

3/

4

0

0

pI

5

Peak dc current:

Average dc current:

2121 coscos22

1

dLLs

p VVL

I

Voltage – theta relation:31

31 coscos

2

LL

d

V

V

diI dd 3

13/

1



• Continuous Current Operation

di

ici

D2 , D3D1 , D2

D1D2D3

I IIIII

On-stateDiode

2

pIbi

t

t

ai

pI

aibi ci

cdV

ai

bi

ci

1D 3D 5D

4D 6D 2D

di

a

b

c

sLav

bv

cv

sL

sL

Note: - With the increase of the load current, the rectifier will enter into continuous current operation. - During commutation interval, three diodes are on.



• Definition of Total Harmonic Distortion (THD)

Phase voltage (pure sine):

Line current (distorted):

RMS line current:

Line current THD:

nnann

a tIi

)(sin2,...3,2,1

2/1

,...3,2,1

22/1

2

0

2 )(2

1

nanaa ItdiI

1

21

2

a

aa

I

IITHD

tVv aa 1sin2



• Definition of Power Factor (PF)

Per-phase average (real) power:

Per-phase apparent power:

Total power factor (PF):

Distortion factor (DF) :

Displacement power factor (DPF) :

tdivP aa

2

02

111 cosaa IV

aa IVS

DPFDFI

I

IV

IV

S

PPF

a

a

aa

aa 1111 cos

cos

aa IIDF /1

1cosDPF

PF = f (THD) :21 THD

DPFPF



ai

1ai 1.41

0 2 3 4

0.282

ai

1ai

1.00

0.50

0.00

-0.50

-1.00

2.00

1.00

0.00

-1.00

-2.00

(a) Ia1 = 0.2 pu

(b) Ia1 = 1 pu

Harmonics n

5 7 11 13 17 19 23 25 THD (%)

1/ aan II (%)

pu2.01aI 63.4 38.7 8.99 8.64 4.22 3.61 2.48 2.02 75.7

1/ aan II (%)

pu11aI 30.4 8.79 6.31 3.40 2.30 1.89 1.04 1.03 32.7

• Typical Waveforms / Harmonic Content



A

B

C

1.0PF

0.6

0.8

0.7

0.9

A

B

C

(b) PF

100THD(%)

20

60

40

80

(a) THD

A: Ls = 0.05B: Ls = 0.10C: Ls = 0.15

A: Ls = 0.05B: Ls = 0.10C: Ls = 0.15

0.2 0.4 0.6 0.8 1aI0 (pu) 0.2 0.4 0.6 0.8 1aI0 (pu)

• THD and PF



Inductive load, quantitative analysis Differential equation

The RMS value of output voltage, output current, and thyristor current can then be

calculated.

0

sin2d

d

o

1oo

ti

tURit

iL

0 20 10060 140 180

20

100

4-3图

60

/(°

)

180

140

/(° )

( 4 - 5 )

S o l u t i o n

tetZ

Ui

t

tg1o )sin()sin(

2

( 4 - 6 ) C o n s i d e r i n g i o = 0 w h e n ω t = α + θ

W e h a v e

tg)sin()sin(

e ( 4 - 7 )


Three Phase Dual Converters

• For four quadrant operation in many industrial variable speed dc drives , 3 phase dual converters are used.

• Used for applications up to 2 mega watt output power level.

• Dual converter consists of two 3 phase full converters which are connected in parallel & in opposite directions across a common load.





Outputs of Converters 1 & 2

• During the interval (/6 + 1) to (/2 + 1), the line to line voltage vab appears across the output of converter 1 and vbc appears across the output of converter 2


0

0

0



sin ;

Max. Phase Voltage

2sin sin 120

3

2sin sin 120

3

sin 240

RN an m

m

YN bn m m

BN cn m m

m

v v V t

V

v v V t V t

v v V t V t

V t


0

0

0



sin ;

Max. Phase Voltage

2sin sin 120

3

2sin sin 120

3

sin 240

RN an m

m

YN bn m m

BN cn m m

m

v v V t

V

v v V t V t

v v V t V t

V t


To obtain an Expression for the Circulating Current

If vO1 and vO2 are the output voltages of converters 1 and 2 respectively, the instantaneous voltage across the current limiting inductor during the interval (/6 + 1) t (/2 + 1) is given by


1 2

3 sin sin6 2

3 cos6

The circulating current can be calculated by

using the equation

r O O ab bc

r m

r m

v v v v v

v V t t

v V t


1

1

6

6

1

max

1.

13 cos .

6

3sin sin

6

3

t

r rr

t

r mr

mr

r

mr

r

i t v d tL

i t V t d tL

Vi t t

L

Vi

L


Four Quadrant Operation


• There are two different modes of operation. Circulating current free

(non circulating) mode of operation Circulating current mode of operation


Non Circulating Current Mode Of Operation

• In this mode of operation only one converter is switched on at a time

• When the converter 1 is switched on,

For 1 < 900 the converter 1 operates in the Rectification mode

Vdc is positive, Idc is positive and hence the average load power Pdc is positive.

• Power flows from ac source to the load


• When the converter 1 is on,

For 1 > 900 the converter 1 operates in the Inversion mode

Vdc is negative, Idc is positive and the average load power Pdc is negative.

• Power flows from load circuit to ac source.



For 2 < 900 the converter 2 operates in the Rectification mode

Vdc is negative, Idc is negative and the average load power Pdc is positive.

• The output load voltage & load current reverse when converter 2 is on.

• Power flows from ac source to the load



For 2 > 900 the converter 2 operates in the Inversion mode

Vdc is positive, Idc is negative and the average load power Pdc is negative.

• Power flows from load to the ac source.• Energy is supplied from the load circuit to the ac

supply.


• Both the converters are switched on at the same time.

• One converter operates in the rectification mode while the other operates in the inversion mode.

• Trigger angles 1 & 2 are adjusted such that (1 + 2) = 1800

Circulating Current Mode Of Operation


When 1 < 900, converter 1 operates as a controlled rectifier. 2 is made greater than 900 and converter 2 operates as an Inverter.

• Vdc is positive & Idc is positive and Pdc is positive.


• When 2 < 900, converter 2 operates as a controlled rectifier. 1 is made greater than 900 and converter 1 operates as an Inverter.

• Vdc is negative & Idc is negative and Pdc is positive.

Unit 2

Documents

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