Power Electronics Lecture (7) Prof. Mohammed Zeki Khedher Department of Electrical Engineering...

Power ElectronicsLecture (7)

Prof. Mohammed Zeki Khedher

Department of Electrical Engineering

University of Jordan

1

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

Comparision between Single Phase Rectifiers

Half wave Full wave Fullwave• center-tap bridge• Peak repetitive reverse voltage VRRM 3.14Vdc 3.14 Vdc 1.57

Vdc• Rms input voltage per transformer leg Vs 2.22 Vdc 1.11 Vdc 1.11

Vdc• Diode average current IF.AV. 1.00 Idc 0.50 Idc 0.50

Idc• Diode rms current IF.RMS. 1.57 Idc 0.785 Idc 0.785 Idc• Form factor of diode current IF.RMS. 1.57 1.57 1.57• Rectification ratio 0.405 0.81 0.81• Form factor 1.57 1.11 1.11• Ripple factor 1.21 0.482 0.482• Transformer rating primary VA 2.69 Pdc 1.23 Pdc 1.23 Pdc• Transformer rating secondary VA 3.49 Pdc 1.75 Pdc 1.23 Pdc• Output ripple frequency fr 1 fi 2 fi 2 fi

Multi-phase Rectifier

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 A 3-phase star rectifier 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 Full Wave Rectifier With Resistive Load

Example 10 The 3-phase bridge rectifier 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

Single phase rectifier 3 phase rectifier

Performance:

3-phase bridge rectifier with RL load

Condition for continuous load current