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 /=η rectification effeciency

dcrms VVFF /=

11 22

222

−=−=−

== FFVV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

form factor

ripple factor

121

2

21

21

2−=

−=

S

S

S

SSi

II

IIITHD

121

2

21

21

2−=

−=

S

S

S

SSv

VV

VVVTHD

FaactorntDisplacemeFactorDistortionII

IVIV

IVPPF

S

S

SS

SS

SS

*

coscos1

111

=

=== φφ

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

∫ ==π

πωω

π 0

sin21 m

mdcVtdtVV

2sin

21

0

22 mmrms

VtdtVV == ∫π

ωωπ

RV

RVI mdc

dc π==

RV

RVI mrms

rms 2==

the load and diode currents RVII m

DS 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.

ππ

πωω

π

πmm

mdcVVtdtVV =−−== ∫ ))0cos(cos(

2)sin(

21

0 RV

RVI mdc

dc π==

2)sin(

21

0

2 mmrms

VtVV == ∫π

ωπ R

VI mrms 2

=

%53.40

2*

2

*

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dc ππη

57.12

2 ====π

πm

m

dc

rmsV

V

VVFF

211.1157.11 22 =−=−== FFVVRF

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

( ) ( )

+−=

−φ

ω

φφωω tansinsin)(t

m etZ

Vti

( ) ( ) 0sinsin)( tan =

+−=

−φ

β

φφββ eZ

Vi m

)cos1(*2

sin*2 0

βπ

ωωπ

β

−== ∫ mmdc

VtdtVV

)2sin(1(5.0*2

)sin(*21

0

2 ββπ

ωπ

β

−+== ∫VmdwttVV mrms

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

πωω

π

πm

mdcVtdtVV 2sin1

0

== ∫ RVI m

dc π2

=

( )2

sin1

0

2 mmrms

VtdtVV == ∫π

ωωπ R

VI mrms 2

=

PIV of each diode = mV2

RVII m

DS 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

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dc ππη

11.1222

2 ====π

πm

m

dc

rmsV

V

VVFF

483.0111.11 22 =−=−== FFVVRF

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.

VVtdtVV mmdc 956.1902sin1

0

=== ∫ πωω

π

πA

RVI m

dc 7324.122==

π

( ) VVtdtVV mmrms 132.212

2sin1

2/1

0

2 ==

= ∫

π

ωωπ

%06.81===rmsrms

dcdc

ac

dc

IVIV

PPη 11.1==

dc

rms

VVFF

482.011 22

222

=−=−=−

== FFVV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac The PIV=300V

Input power factor = 1cosRe==

SS

SS

IVIV

PowerApperantPoweral φ

Full Bridge Single-phase Diode Rectifier with DC Load Current

[ ]

[ ] .............,5,3,14

cos0cos2

cos2

sin*20

0

==−=

−== ∫

nforn

In

nI

tnn

ItdtnIb

oo

oon

ππ

π

ωπ

ωωπ

ππ

)..........9sin917sin

715sin

513sin

31(sin*4)( +++++= tttttIti o ωωωωω

π

%46151

131

111

91

71

51

31))((

2222222

=

+

+

+

+

+

+

=∴ tITHD s

π24

1o

SII =

%34.4814

21

241))((

2

2

2

1=−

=−

=−

=∴

π

πo

o

S

Ss I

IIItITHD

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

%90===rmsrms

dcdc

ac

dcIVIV

PPη 11.1==

dc

rmsVVFF

482.011 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

The PIV=Vm=300V

AII oS 01.27

230*4

24

1 ===ππ

Input Power factor= =PowerApperant

PoweralRe

LagI

IIV

IV

S

S

SS

SS 9.01*30

01.27cos*cos* 11 ===φφ

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

−= −

m

osV

ILu ω21cos 1 oSoS

rd ILfILV 42

4−=

−=

πω

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 42tan −=−=

π

−=

322 2 uII o

sπ

π 2sin*

28

1u

uII o

S π=

( )

−

=

32

sin2uu

upfππ

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 ==

VV actualdc 9703200*005.0*50*415556*2=−=

π

.285.015556

200*005.0*50**2*21cos21cos 11 radV

ILum

os =

−=

−= −− πω

( ) ( ) 917.0

3285.

2285.0

285.0sin*2

32

sin*22

cos*1 =

−

=

−

=

=

ππππ uu

uuIIpf

S

S

AuII oS 85.193

3285.0

2200*2

322 22

=

−=

−=

ππ

ππ

Auu

II oS 46.179

2285.0sin*

285.0*2200*8

2sin*

28

1 =

==

ππ

%84.40146.17985.1931

22

1=−

=−

=

S

Si I

ITHD

Three-Phase Half Wave Rectifier

mm

mdc VVtdtVV 827.0233sin

23 6/5

6/

=== ∫ πωω

π

π

π RV

RVI mm

dc*827.0

**233

==π

( ) mmmrms VVtdtVV 8407.08

3*321sin

23 6/5

6/

2 =+== ∫ πωω

π

π

π

RVI m

rms8407.0

=RV

RVII mm

Sr 4854.03

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 Ω 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 VVV 827.0233

==π R

VRVI mm

dc0827

233

==π

mrms VV 8407.0=R

VI mrms

8407.0=

%767.96===rmsrms

dcdc

ac

dcIVIV

PPη

%657.101==dc

rmsVVFF

%28.1811 22

222=−=−=

−== FF

VV

VVV

VVRF

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

321 3/

3/0

oo

ItdIa ∫−

==π

π

ωπ

[ ]

harmonicstrepleanallfor

nfornI

nfornI

tnnIdwttnIa

o

o

oon

0

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

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

sincos*1 3/3/

3/

3/

=

=−=

==

== −−∫

π

π

ωπ

ωπ

ππ

π

π

−−++−−++= ...8sin

817sin

715sin

514sin

412sin

21sin3

3)( ttttttIItI OO

s ωωωωωωπ

%24.1090924.119

*21

23

3/1))((2

2

2

1==−=−

=−

=

π

πO

o

S

Ss I

IIItITHD

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

VVVV mm

dc 613.310827.0233

===π

AIdc 100=

VVV mrms 759.3158407.0 == AIrms 100=

%37.98100*759.315100*613.310

====rmsrms

dcdc

ac

dcIVIV

PPη

%657.101==dc

rmsVVFF

%28.1811 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

The PIV= 3 Vm=650.54V

Three-Phase Full Wave Rectifier With Resistive Load

1 3 5

4 6 2

b

c

IL

VLIs

Ip

a

LLmLLm

mdc VVVVtdtVV 3505.1654.12333sin33 3/2

3/

===== ∫ ππωω

π

π

π

RV

RV

RV

RVI LLLLmm

dc3505.123654.133

====ππ

( ) LLmmmrms VVVtdtVV 3516.16554.14

3*923sin33 3/2

3/

2==+== ∫ π

ωωπ

π

π

RVI m

rms6554.1

=

RV

RVI mm

r 9667.03

6554.1==

RVI m

S 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 .

VVVV mm

dc 226.621654.133===

πA

RV

RVI mm

dc 0613.31654.133===

π

VVVV mmrms 752.6216554.14

3*923

==+=π

AR

VI mrms 0876.316554.1

==

%83.99===rmsrms

dcdc

ac

dcIVIV

PPη

%08.100==dc

rmsVVFF

%411 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

The PIV= 3 Vm=650.54V

Three-Phase Full Wave Rectifier With DC Load Current

[ ]

....,.........15,14,12,10,9,8,6,4,3,2,0

),.....3(132),3(

112),3(

72),3(

52,32

cos2sin*2

1311751

6/56/

6/5

6/

==

==−=−==

−== ∫

nforb

IbIbIbIbIb

tnn

ItdtnIb

n

ooooo

oon

πππππ

ωπ

ωωπ

ππ

π

π

++−−= ttttvtItI o

s ωωωωωπ

13sin13111sin

1117sin

715sin

51sin32)(

%31251

231

191

171

131

111

71

51))((

22222222

=

+

+

+

+

+

+

+

=tITHD s

oS II32

= oS IIπ

3*21 =

%01.311/3*23/21))(( 2

2

1=−=−

=

πS

Ss I

ItITHD

Power Factor =S

S

S

SII

II 11 )0cos(* =

−= −

LL

oSV

ILu ω21cos 1o

ord fLILIV 6

26

==π

ω

oLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan −=−=

−=

632 2 uII o

Sπ

π

=

2sin62

1u

uII o

S π

( )

−

=

−

=

=

63

sin*32

cos

632

2sin62

2cos

21

uu

uuuI

uu

Iu

IIpf

o

o

S

S

ππππ

π

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

oSV

ILu ω21cos 1

dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan −=−=

VVdcactual 43830300*008.*50*633000*35.1 =−=( ) ( ) 9644.0

62549.0

3*2549.0

2549.0sin3

63

sin*3=

−

=

−

=ππππ uu

upf

AuII ds 929.239

62549.0

3*300*2

632 22

=

−=

−=

ππ

ππ

Auu

II oS 28.233

22549.0sin*

2*2549.0*300*3432*

2sin

234

1 =

=

=

ππ

9644.02

2549.0cos*929.23928.233

2cos*1 =

=

=

uIIpf

s

S

%05.24128.233

929.239122

1=−

=−

=

S

Si I

ITHD