C2 c3 rectification all student

POWER ELECTRONICS

Ministry of Education and TrainingHCMC University of Technology and Education

➔Convert an AC power supply source voltage to

a controlled DC load voltage

➔Diode and SCR

AC – DC CONVERTER

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AC – DC CONVERTER

Input

- AC

- U1,f1= constant

Output

- DC

- U2= DC

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AC – DC CONVERTER

Input

- AC

- U1,f1= constant

Output

- DC

- U2= DC

▪ …….

IG> 0

VKVA

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DC MotorAC – DC – Motor Speed

▪ Adapter▪ Battery Charger▪ DC Voltage source▪ AC – DC - Device

Generator ExciterAC – DC – Exciter

10

I INTRODUCTION

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UPSAC – DC - ACWind energy

AC – DC - AC

VFDAC – DC - AC

11

I INTRODUCTION

Diode

RuIN(q )

Positive half wave

Uin= UPN+Um

-Um+Um

u0(V)

uin(V)

q (rad)

q (rad)u0(V)

2

][sin2

sin

)()

(

Ut

U

vv

m

PNi

q

qq

==

=

II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load

Diode

RuIN(q )

Uin= UPN+Um

-Um+Um

u0(V)

uin(V)

q (rad)

q (rad)u0(V)

2

][sin2

sin

)()

(

Ut

U

vv

m

PNi

q

qq

==

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The average load voltage

The average load current

RUId=d

2m

0m

2

00d

U45,0U

dsinU

21

dv2

1U

==

=qq

=

=q

=





II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load

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II Single-Phase Half-Wave Rectifiers

II.1 Operationwith R Load

u0(q )P

N

Diode

RuIN(q )

UNCONTROLLED

Diode

RuIN(q )

][sin2

sin

)()

(

Ut

U

vv

m

PNi

q

qq

==

=

II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load

UNCONTROLLED

U2 = 100 V, R = 10 Ω

➔ Ud = 0,45U2 = 45V➔ Id= Idiode= = Ud/R = 45/10 = 4,5 A

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II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load

Ut

sinU

)(v

)(

CONTROLLED

sinU

)(v

)(

sinU

)(v

)(

SCR

RuIN(q )

2

IG

]V[sin2

Ut

sinU

)(v

)(v

2m

PNi

q=

=

q=

SCR

RuIN(q )

Positive half wave

2

IG

]V[sin2

Ut

sinU

)(v

)(v

2m

PNi

q=

=

q=

SCR

RuIN(q )

2

IG

]V[sin2

Ut

sinU

)(v

)(v

2m

PNi

q=

=

q=

q

II Single-Phase Half-WaveRectifiers

II.1 Operation with R Load

Negative half wave

CONTROLLED

Open

The average load voltage

The average load current

2U

dsinU

21U

m

md

+

=

qq

CONTROLLED

CONTROLLED

CONTROLLED

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II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load = /2

vi

vGv0i0

vSCR

CONTROLLED

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II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load = 2/3

vi

vGv0

i0

vSCR

CONTROLLED

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II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load = 5/6

vi

vGv0

i0

vSCR

CONTROLLED

)()

(

Ut

U

vv

m

PNi

q

qq

==

=

II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load

CONTROLLED

u0(q )P

N

SCR

RuIN(q )

➔ Ud = 𝟏𝟎𝟎 𝟐

𝟐 𝐜𝐨𝐬𝟔𝟎 + 𝟏 = 𝟑𝟑, 𝟕𝟔 𝐕

➔ Id= Idiode= = Ud/R = 33,76/10 = 3,376 A

)V)(1(cos

2U

dsinU

21U

m

md

+

=

qq

=



)()

(

Ut

U

vv

m

PNi

q

qq

==

=

II Single-Phase Half-Wave Rectifiers

II.1 Operation with R Load

dsinU

21U

m

md

+

=

qq

=



II.2 Operationwith R + L Load

II Single-Phase Half-Wave Rectifiers

U

tsin

U

)(v

)(v

2m

PNi

q=

=

q=

q

UNCONTROLLED

absorbing power (Charging)

vL> 0-

When Inductor is supplying power (Discharging) ➔ iLis descreasing, vL< 0

iL> 0-

vL> 0+

II Single-Phase Half-Wave Rectifiers

]V[sin2

U

tsin

U

)(v

)(v

2m

PNi

q=

=

q=

q

β

II Single-Phase Half-Wave unControlled Rectifiers

]V[sin2

Ut

sinU

)(v

)(

UNCONTROLLED

)(

);(

UE

m

qq

II Single-Phase Half-Wave unControlled Rectifiers

]V[sin2

Ut

sinU

)(v

)(

UNCONTROLLED

)(

);(

UE

m

qq

The average load voltage

)()sin

(2

12

1

VdEU

EUd=+qm−

q

qq



)(

);(

UE

m

qq

][sin2

sin)

()

(

U

tU

v

q

q

q

=

==

The average load voltage

)()sin

(2

12

1

VdEU

EUd=+qm−

q

qq



)(

);(

UE

m

qq

][sin2

sin)

()

(

U

tU

v

q

q

q

=

==

𝟏𝟎𝟎 𝟐𝒔𝒊𝒏q − 𝟏𝟎𝟎 𝒅q

The average load current

dDIODEd

REU

I=−()=

Peak Inverse Voltage

EU

UPIV=m+

][sin2

sin)

()

(

U

tU

v

q

q

q

=

==

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u0

RD3

The load voltage

][sin2

sin)

()

(

U

tU

v

q

q

q

=

==

The average load voltage

20

9,0sin

1

Ud

UUd=mqq=



][sin2

sin)

()

(

U

tU

v

q

q

q

=

==

N

u0

RD3

D2

D4

D1

The average load current

)

( A

RUId

d=

Peak Inverse Voltage

mPIVUU=

)(2A

IIdDIODE=d

The average Diode current

N

u0

RD3

D2

D4

D1

điện áp tạiAnode caonhấtdẫn.

Các linhkiện (Diode, SCR) mắcchung Anode (a) được gọi là nhómchuyển mạchÂm Tại một thời điểm, chỉ có 1 linh kiện mà có điện áptạiCathode thấp nhấtdẫn.

)V)(1(cos

U

dsinU

1U

m

md

+

=

qq

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The full-wave controlled bridge rectifier has an AC input of

450 a) Sketch the input and output voltage waveforms?b) Determine the average current and voltage in the load?

u0

R

ED3

D2

D4

D1

III.2 Operation with R + E Load

III Single-Phase full-Wave Rectifiers

)()sin

(12

1

VdEU

EUd=+qm−

q

qq



)(

);(

UE

m

qq

][sin2

sin)

()

(

U

tU

v

q

q

q

=

==

The average load voltage

UNCONTROLLED

u0

R

ED3

D2

D4

D1

III.2 Operation with R + E Load

III Single-Phase full-Wave Rectifiers

The average load current

)

( A

REU

Id=d−

Peak Inverse Voltage

EU

UPIV=m+

)(2A

IId

dDIODE=

The average Diode current

UNCONTROLLED

)V(d)Esin

U(1EU

2m



q−

q

+=

)(

);(

UE

m

qq

CONTROLLED

u0

R

LD3

D2

D4

D1

III.3 Operationwith R +L Load

III Single-Phase full-Wave Rectifiers

The average load voltage

20

9,0sin

1

Ud

UUd=mqq=



][sin2

sin)

()

(

U

tU

v

q

q

q

=

==

UNCONTROLLED

u0

R

LD3

D2

D4

D1

III.3 Operationwith R +L Load

III Single-Phase full-Wave Rectifiers

The average load current

)

( A

RUId

d=

Peak Inverse Voltage

mPIVUU=

)(2A

IIdDIODE=d

The average Diode current

UNCONTROLLED

24

➔ j > 300

v0(V)

Load Voltage

Discont load current

2π 2π

t (rad)Fig a

Voltage

Cont load current

Discontinuous load current

Continuous load current

III Single-Phase Full-Wave Rectifiers

III.2 Operation with RL Load (Full Control)

RLarctan=

j

RLarctan=

j>

CONTROLLED

Critical load Current

2π

t (rad)Fig a

Fig b

t (rad)

Critical Load Current

RLarctan=

j=

CONTROLLED

III.2 Operation with RL Load (Full Control)

The average load voltage

)V(cosU

9,0d

sinU

1Udmqq=2

=+



+j



qq

1Usind

RLarctan=

j

RLarctan=

j>

RLarctan=

j=Ud1Umsinqdq=0,9U2cos(V)

=+



1 Continuous Load Current = Continuous Conduction Mode CCM

2 Discontinuous Load Current = Discontinuous Conduction Mode DCM

3.Critical Load Current

CONTROLLED

a) Is the circuit working in CCM or DCM? Determine the average load voltage?b) Which factors do the CCM or DCM of the circuit depend on? Explain?

9,0d

sinU

1Udmqq=2

=+



= 0,9*220*cos60 = 99 V

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)V)(1(cos

U

dsinU

1U

m

md

+

=

qq

=



CONTROLLED

u0

R

LD3

III Single-Phase full-Wave Rectifiers

UNCONTROLLED

IV Three-Phase Half-Wave Rectifiers

The average load voltage

)V(U17,1d

sinU

32

1

65

6m

=





])[34sin(

2)

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

A0

v0

RD3

D2D1

B

C0

IV Three-Phase Half-Wave Rectifiers

The average load current

)

( A

RUId=d

Peak Inverse Voltage

)(63UU2VUPIV=m=

)(3A

IIdDIODE=d

The average Diode current

A0

v0

RD3

D2D1BC0

i0

vD1

UNCONTROLLED

RD3

D2D1

B

C0

0

RD3

D2D1

B

C0

0

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

=

A0

v0

R3

21

B

C0

0

CONTROLLED

VA0VB0VC0

IV.1 Operation with R Load

The average load voltage

)V(cosU

17,1d

sinU

32

1

65

6m

+

+0

30

0

30>

0

30=

1 Continuous Load Current

2 Discontinuous Load Current

3.Critical Load Current

)]6cos(

1[31.U17,1

dsinU

32

1U

2

6md

++

=

qq

=

+

)V(cosU

17,1d

sinU

32

1

65

6m

+

+

CONTROLLED

Cho bộ chỉnh lưu mạch hình tia 3 pha điều khiển mắc vào tải chứa R = 10 Ω Điện áp pha của nguồn xoay chiều 3 pha có trị hiệu dụng U = 220 V Tính trị trung bình của điện áp chỉnh lưu và dòng chỉnh lưu khi:

a) Góc kích  = 00b) Góc kích  = 200c) Góc kích  = 900

R3

21

B

C0

0

A three-phase half-wave controlled rectifier shown in figure 4 has an AC input line-to-neutral voltage of 220 V (rms) at 50 Hz

with a load R = 10Ω Calculate the output voltage and current in

cases of the firing angles of 00, 200and 600, respectively.

IV Three-Phase Half-Wave Rectifiers

The average load voltage

)V(U17,1d

sinU

32

1

65

6m



]

)[34sin(

2)

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

q(rad)

A0

v0

R

LD3

D2D1

BC0

0

v0i0

UNCONTROLLED

IV Three-Phase Half-Wave Rectifiers

The average load current

)

( A

RUId=d

Peak Inverse Voltage

)(63UU2VUPIV=m=

)(3A

IIdDIODE=d

The average Diode current

A0

v0

RLD3

D2D1

BC0

i0

vD1

UNCONTROLLED

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

=

A0

v0

R3

21

BC0

0

L

CONTROLLED

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

=

A0

v0

R3

21

BC0

0

L

CONTROLLED

P

sinU

32

1

65

6m

+

+



+=>

L30or

30

00

1 Continuous Load Current

2 Discontinuous Load Current

l+

qq

=

6m

32

1U

IV Three-Phase Half-Wave Rectifiers

IV.2 Operation with RL Load



=>

L

30or

0

CONTROLLED

IV Three-Phase Half-WaveRectifiers

The average load voltage



q

q

q−

q

+

1

d)Esin

U(32

1E

mmEU2

q(rad)

A0

v0

RD3

D2

D1

BC0

);(

UE

m

qq

UNCONTROLLED

IV Three-Phase Half-WaveRectifiers

mmEU2

U



VA0VB0VC0

2π

4ππ/65π/6

UEU

2

The average load current

)A(R

EU

Id=d−

Peak Inverse Voltage

)(63UU2VUPIV=m=

)(3A

IId

dDIODE=

The average Diode current

UNCONTROLLED

IV Three-Phase Half-WaveRectifiers

2UE

q(rad)

A0

v0

RD3

D2

D1

BC0

sinU

32

1

65

6m





UNCONTROLLED

IV Three-Phase Half-WaveRectifiers

VA0VB0VC0

2π

4ππ/65π/6

EU

Id=d−

Peak Inverse Voltage

)(63UU2VUPIV=m=

)(3A

IId

dDIODE=

The average Diode current

2UE

C2

UNCONTROLLED

IV Three-Phase Half-WaveRectifiers

m

U



UEU

2

The average load current

)A(R

EU

Id=d−

Peak Inverse Voltage

)(63UU2VUPIV=m=

)(3A

IId

dDIODE=

The average Diode current

UNCONTROLLED

q

q

q−

q

+

1

d)Esin

U(32

1E

b) E = 50 V

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V.1 Operation with R Load

])[34sin(

2)

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

=

A0

v0

R1

B0

35

462

V Three-Phase Full-WaveRectifiers

])[34sin(

2)

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

=

A0

v0

R1

B0

UNCONTROLLED

π

V Three-Phase Full-WaveRectifiers

The average load voltage

)(34,2)

(31

22

6

vUd=A−BO=





q

The average load current

)

( A

RUId=d

Peak Inverse Voltage

)(63UU2VUPIV=m=

)(3A

IId

vD1

UNCONTROLLED

•Tính điện áp trung trên trải, dòng điện trung bình qua tải và qua mỗi diode;•Tính điện áp ngược lớn nhất trên mỗi diode;

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V Three-Phase Full-Wave Rectifiers

V.1 Operation with R Load = 300

V.1 Operation with R Load = 600

CONTROLLED

VA0VB0VC0

VABVACVBCVBAVCAVCB

V.1 Operation with R Load = 900

CONTROLLED

0

60

0

60>

1 Continuous Load Current

2 Discontinuous Load Current

V Three-Phase Full-Wave Rectifiers

V.1 Operation with R Load

Ud= 2,34U1 cos()

3



3.Critical Load Current

)V(cosU

34,2d

)vv

(3

A

+

+

CONTROLLED

60(vv)d2,34Ucos(V)

31

2

6

0B0

A

+

+

0

60>

d= 2,34U1 cos()

3



V Three-Phase Full-WaveUncontrolled Rectifiers

])[34sin(

2)

(

])[32sin(

2)

(

][sin2)

(

20

20

20

VU

v

VU

v

VU

v

CBA

q

q

q

q

qq

−=

−=

=

A0

v0

R1

B0

π

V Three-Phase Full-WaveUncontrolled Rectifiers

The average load voltage

)(34,2)

(31

22

6

vUd=A−BO=





q

The average load current

)

( A

RUId=d

Peak Inverse Voltage

)(63UU2VUPIV=m=

)(3A

IIdDIODE=d

The average Diode current

RD3

D2

D1

BC

0

0

D4

VI.1 Operationwith R Load

VI Six-Phase Half-WaveUncontrolled Rectifiers

The average load voltage

)V(U35,1d

sinU

31

32

3md=qq=





)(v]

V)[34sin(2U)(v

)(v]

V)[32sin(2U)(v

)(v]

V[sin2U)(v

0'C2

0C

0'B2

0B

0'A2

0A

q−

=

−q=

q

q−

=

−q=

q

q−

=q

=q

VA0 VC’0 VB0 VA’0 VC0 VB’0

D3

D2

D1

BC

0

0

D4

VI.1 Operationwith R Load

VI Six-Phase Half-WaveUncontrolled Rectifiers

Peak Inverse Voltage

)V(U22U

2UPIV=m=2

)A(6IIdDIODE=d

The average Diode current

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