1. Dr. Ali M. Eltamaly King Saud University
EE432 Power Electronics
Dr. Ali M. Eltamaly
King Saud University
Dr. Ali M. Eltamaly, King Saud University

2. Dr. Ali M. Eltamaly, King Saud University
Chapter 1
Introduction
1.1. Definition of Power Electronics
Power electronics refers to control and conversion of electrical power by power semiconductor devices wherein these devices operate as switches
Electronic power converter
Rectifier converting an AC voltage to a DC voltage,
Inverter converting a DC voltage to an AC voltage,
Chopper or a switch-mode power supply that converts a DC voltage to another DC voltage, and
Cycloconverter and cycloinverter converting an AC voltage to another AC voltage.
Dr. Ali M. Eltamaly, King Saud University

3. Dr. Ali M. Eltamaly, King Saud University
1.2 Rectification
uncontrolled and controlled rectifiers
DC-To-AC Conversion
Emergency lighting systems,
AC variable speed drives,
Uninterrupted power supplies, and,
Frequency converters.
DC-to-DC Conversion
Step-down switch-mode power supply,
Step-up chopper,
Fly-back converter, and ,
Resonant converter.
typical applications
DC drive,
Battery charger, and,
DC power supply.
Dr. Ali M. Eltamaly, King Saud University

4. Dr. Ali M. Eltamaly, King Saud University
1.5 AC-TO-AC Conversion
cycloconverter or a Matrix converter converts
Adjustable Speed Drives (ASD)
Dr. Ali M. Eltamaly, King Saud University

5. Diode Circuits or Uncontrolled Rectifier
Rectification: The process of converting the alternating voltages and currents to direct currents
Dr. Ali M. Eltamaly, King Saud University

6. 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.
Dr. Ali M. Eltamaly, King Saud University

7. rectification effeciency Dr. Ali M. Eltamaly, King Saud University
Performance Parameters
rectification effeciency
form factor
ripple factor
Dr. Ali M. Eltamaly, King Saud University

8. Dr. Ali M. Eltamaly, King Saud University

9. Single-phase half-wave diode rectifier with resistive load.
Dr. Ali M. Eltamaly, King Saud University

10. Dr. Ali M. Eltamaly, King Saud University
the load and diode currents
Dr. Ali M. Eltamaly, King Saud University

11. Dr. Ali M. Eltamaly, King Saud University
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.
(d) It is clear from Fig2.2 that the PIV is
Dr. Ali M. Eltamaly, King Saud University

12. Dr. Ali M. Eltamaly, King Saud University
Half Wave Diode Rectifier With R-L Load
Fig.2.3 Half Wave Diode Rectifier With R-L Load
Dr. Ali M. Eltamaly, King Saud University

13. Dr. Ali M. Eltamaly, King Saud University
Divide the above equation by L we get:
Dr. Ali M. Eltamaly, King Saud University

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17. Dr. Ali M. Eltamaly, King Saud University
Half wave diode rectifier with free wheeling diode
Dr. Ali M. Eltamaly, King Saud University

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20. Dr. Ali M. Eltamaly, King Saud University

21. Dr. Ali M. Eltamaly, King Saud University 
1.1 
1.12 
1.14 
1.16 
1.18 
1.2 
Dr. Ali M. Eltamaly, King Saud University

22. Dr. Ali M. Eltamaly, King Saud University

23. Dr. Ali M. Eltamaly, King Saud University
Single-Phase Full-Wave Diode Rectifier
Center-Tap Diode Rectifier
Dr. Ali M. Eltamaly, King Saud University

24. Dr. Ali M. Eltamaly, King Saud University
PIV of each diode =
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.
Dr. Ali M. Eltamaly, King Saud University

25. Dr. Ali M. Eltamaly, King Saud University
The PIV is
Dr. Ali M. Eltamaly, King Saud University

26. Dr. Ali M. Eltamaly, King Saud University
Center-Tap Diode Rectifier With R-L Load
Dr. Ali M. Eltamaly, King Saud University

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30. Dr. Ali M. Eltamaly, King Saud University
Single-Phase Full Bridge Diode Rectifier With Resistive Load
Dr. Ali M. Eltamaly, King Saud University

31. Dr. Ali M. Eltamaly, King Saud University
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.
The PIV=300V
Input power factor =
Dr. Ali M. Eltamaly, King Saud University

32. Dr. Ali M. Eltamaly, King Saud University

33. Dr. Ali M. Eltamaly, King Saud University
Example 5 solve Example 4 if the load is 30 A pure DC
Input Power factor=
Dr. Ali M. Eltamaly, King Saud University

34. Full Bridge Single-phase Diode Rectifier with DC Load Current

36. Example 5 solve Example 4 if the load is 30 A pure DC
Input Power factor=

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

38. Dr. Ali M. Eltamaly, King Saud University

39. Dr. Ali M. Eltamaly, King Saud University
the total reduction per period is:
Dr. Ali M. Eltamaly, King Saud University

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the rms value of the supply current
Dr. Ali M. Eltamaly, King Saud University

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45. Dr. Ali M. Eltamaly, King Saud University
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.
Dr. Ali M. Eltamaly, King Saud University

46. 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.

48. Three-Phase Half Wave Rectifier

50. ThePIV of the diodes is
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.

51. The PIV=
Vm=650.54V

52. Three-Phase Half Wave Rectifier With DC Load Current and zero source induct
New axis

55. Example 8 Solve example 7 if the load current is 100 A pure DC
The PIV=
Vm=650.54V

56. Dr. Ali M. Eltamaly, King Saud University
Three-Phase Half Wave Rectifier With Source Inductance
Dr. Ali M. Eltamaly, King Saud University

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59. Three-Phase Full Wave Rectifier With Resistive Load

65. Example 10 The rectifier shown in Fig. 2
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 .

66. The PIV=
Vm=650.54V

67. Three-Phase Full Wave Rectifier With DC Load Current

72. 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.