Diode Circuits or Uncontrolled Rectifier Rectification: The process of converting the alternating...

Diode Circuits or Uncontrolled Rectifier

Rectification: The process of converting the alternating

voltages and currents to direct currents

The main disadvantages of half wave rectifier are :

• High ripple factor,

• Low rectification efficiency,

• Low transformer utilization factor, and,

• DC saturation of transformer secondary winding.

Performance Parameters

22dcrmsac VVV

acdc PP / rectification effeciency

dcrms VVFF /

11 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

form factor

ripple factor

121

2

21

21

2

S

S

S

SSi

I

I

I

IITHD

121

2

21

21

2

S

S

S

SSv

V

V

V

VVTHD

FaactorntDisplacemeFactorDistortion

I

I

IV

IV

IV

PPF

S

S

SS

SS

SS

*

coscos

1111

Single-phase half-wave diode rectifier with resistive load.

0

sin2

1 mmdc

VtdtVV

2sin

2

1

0

22 mmrms

VtdtVV

R

V

R

VI mdcdc

R

V

R

VI mrmsrms 2

the load and diode currents R

VII mDS 2

Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of diode D1.

mm

mdcVV

tdtVV ))0cos(cos(2

)sin(2

1

0 R

V

R

VI mdcdc

2)sin(

2

1

0

2 mmrms

VtVV

R

VI mrms 2

%53.40

2*

2

*

*

*

RVVR

VV

IV

IV

P

P

mm

mm

rmsrms

dcdc

ac

dc

57.12

2

m

m

dc

rms

V

V

V

VFF

211.1157.11 22 FFV

VRF

dc

ac

)d (It is clear from Fig2.2 that the PIV is

.

mV

Half Wave Diode Rectifier With R-L Load

Fig.2.3 Half Wave Diode Rectifier With R-L Load

tansinsin)(

t

m etZ

Vti

0sinsin)( tan

eZ

Vi m

)cos1(*2

sin*2

0

mmdc

Vtdt

VV

)2sin(1(5.0*2

)sin(*2

1

0

2

Vm

dwttVV mrms

Single-Phase Full-Wave Diode Rectifier Center-Tap Diode Rectifier

m

mdcV

tdtVV2

sin1

0

R

VI mdc

2

2

sin1

0

2 mmrms

VtdtVV

R

VI mrms

2

PIV of each diode = mV2

R

VII mDS 2

Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple

factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current.

%05.81

2*

2

2*

2

*

*

R

VVR

VV

IV

IV

P

P

mm

mm

rmsrms

dcdc

ac

dc

11.1222

2

m

m

dc

rmsV

V

V

VFF

483.0111.11 22 FFV

VRF

dc

ac

The PIV is mV2

Single-Phase Full Bridge Diode Rectifier With Resistive Load

Example 4 single-phase diode bridge rectfier has a purely resistive load of R=15 ohms and, VS=300 sin 314 t and unity transformer ratio.

Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor.

VV

tdtVV mmdc 956.190

2sin

1

0

A

R

VI mdc 7324.12

2

VV

tdtVV mmrms 132.212

2sin

12/1

0

2

%06.81rmsrms

dcdc

ac

dc

IV

IV

P

P11.1

dc

rms

V

VFF

482.011 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac The PIV=300V

Input power factor = 1cosRe

SS

SS

IV

IV

PowerApperant

Poweral

Full Bridge Single-phase Diode Rectifier with DC Load Current

.............,5,3,14

cos0cos2

cos2

sin*2

00

nforn

In

n

I

tnn

ItdtnIb

oo

oon

)..........9sin9

17sin

7

15sin

5

13sin

3

1(sin*

4)( ttttt

Iti o

%4615

1

13

1

11

1

9

1

7

1

5

1

3

1))((

2222222

tITHD s

2

41

oS

II

%34.4814

21

2

41))((

2

2

2

1

o

o

S

Ss I

I

I

ItITHD

Example 5 solve Example 4 if the load is 30 A pure DC

From example 4 Vdc= 190.986 V, Vrms=212.132 V AIdc 30 and rmsI = 30 A

%90rmsrms

dcdc

ac

dc

IV

IV

P

P 11.1dc

rms

V

VFF

482.011 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

The PIV=Vm=300V

AI

I oS 01.27

2

30*4

2

41

Input Power factor= PowerApperant

PoweralRe

LagI

I

IV

IV

S

S

SS

SS 9.01*30

01.27cos*cos* 11

Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier .

m

os

V

ILu

21cos 1 oS

oSrd ILf

ILV 4

2

4

osm

rdceinducsourcewithoutdcactualdc IfLV

VVV 42

tan

32

2 2 uII os

2

sin*2

81

u

u

II oS

32

sin2

uu

upf

Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance Ls=5mH supply to feed 200 A pure DC load, find: (i) Average

DC output voltage, (ii) Power factor. And (iii) Determine the THD of the utility line current.

VVm 155562*11000

VVactualdc 9703200*005.0*50*4

15556*2

.285.015556

200*005.0*50**2*21cos

21cos 11 rad

V

ILu

m

os

917.0

3

285.

2285.0

285.0sin*2

32

sin*2

2cos*1

uu

uu

I

Ipf

S

S

AuI

I oS 85.193

3

285.0

2

200*2

32

2 22

Au

u

II oS 46.179

2

285.0sin*

285.0*2

200*8

2sin*

2

81

%84.40146.179

85.1931

22

1

S

Si I

ITHD

Three-Phase Half Wave Rectifier

mm

mdc VV

tdtVV 827.02

33sin

2

36/5

6/

R

V

R

VI mmdc

*827.0

**2

33

mmmrms VVtdtVV 8407.08

3*3

2

1sin

2

36/5

6/

2

R

VI mrms

8407.0

R

V

R

VII mmSr 4854.0

3

08407

ThePIV of the diodes is mLL VV 32

Example 7 The rectifier in Fig.2.21 is operated from 460 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) Peak inverse voltage (PIV) of each diode.

VVVV mS 59.3752*58.265,58.2653

460

mm

dc VV

V 827.02

33

R

V

R

VI mmdc

0827

2

33

mrms VV 8407.0R

VI mrms

8407.0

%767.96rmsrms

dcdc

ac

dc

IV

IV

P

P

%657.101dc

rms

V

VFF

%28.1811 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

The PIV= 3 Vm=650.54V

Three-Phase Half Wave Rectifier With DC Load Current and zero source induct

New

axis

32

13/

3/

0o

oI

tdIa

harmonicstrepleanallfor

nforn

I

nforn

I

tnn

IdwttnIa

o

o

oon

0

17,16,11,10,5,43*

,....14,13,8,7,2,13*

sincos*1 3/

3/

3/

3/

...8sin

8

17sin

7

15sin

5

14sin

4

12sin

2

1sin

3

3)( tttttt

IItI OO

s

%24.1090924.119

*21

2

3

3/1))((

2

2

2

1

O

o

S

Ss

I

I

I

ItITHD

Example 8 Solve example 7 if the load current is 100 A pure DC

VVV

V mm

dc 613.310827.02

33

AIdc 100

VVV mrms 759.3158407.0 AIrms 100

%37.98100*759.315

100*613.310

rmsrms

dcdc

ac

dc

IV

IV

P

P

%657.101dc

rms

V

VFF

%28.1811 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

The PIV= 3 Vm=650.54V

Three-Phase Full Wave Rectifier With Resistive Load

LLmLLm

mdc VVVV

tdtVV 3505.1654.12333

sin33

3/2

3/

R

V

R

V

R

V

R

VI LLLLmmdc

3505.123654.133

LLmmmrms VVVtdtVV 3516.16554.14

3*9

2

3sin3

33/2

3/

2

R

VI mrms

6554.1

R

V

R

VI mmr 9667.0

3

6554.1

R

VI mS 29667.0

Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load resistance is R=20ohms. If

the source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse

voltage (PIV) of each diode .

VVV

V mm

dc 226.621654.133

AR

V

R

VI mmdc 0613.31

654.133

VVVV mmrms 752.6216554.14

3*9

2

3

AR

VI mrms 0876.31

6554.1

%83.99rmsrms

dcdc

ac

dc

IV

IV

P

P

%08.100dc

rms

V

VFF

%411 22

222

FFV

V

V

VV

V

VRF

dc

rms

dc

dcrms

dc

ac

The PIV= 3 Vm=650.54V

Three-Phase Full Wave Rectifier With DC Load Current

ttttv �t

ItI o

s

13sin13

111sin

11

17sin

7

15sin

5

1sin

32)(

%3125

1

23

1

19

1

17

1

13

1

11

1

7

1

5

1))((

22222222

tITHD s

oS II3

2 oS II

3*2

1

%01.311/3*2

3/21))((

2

2

1

S

Ss I

ItITHD

Power Factor =S

S

S

S

I

I

I

I 11 )0cos(*

LL

oS

V

ILu

21cos 1

oo

rd fLILI

V 62

6

oLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan

63

2 2 uII oS

2sin

621

u

u

II oS

63

sin*3

2cos

63

2

2sin

62

2cos

2

1

uu

uu

uI

u

u

Iu

I

Ipf

o

o

S

S

Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC load current Find;Commutation time and commutation angle.DC output voltage.Power factor.Total harmonic distortion of line current.

oradu 61.14.2549.0

LL

oS

V

ILu

21cos 1

dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan

VVdcactual 43830300*008.*50*633000*35.1

9644.0

6

2549.0

3*2549.0

2549.0sin3

63

sin*3

u

u

upf

AuI

I ds 929.239

6

2549.0

3*

300*2

63

2 22

Au

u

II oS 28.233

2

2549.0sin*

2*2549.0*

300*3432*

2sin

2

341

9644.02

2549.0cos*

929.239

28.233

2cos*1

u

I

Ipf

s

S

%05.24128.233

929.2391

22

1

S

Si I

ITHD