1. Diode Circuits or Uncontrolled Rectifier
Rectification: The process of converting the alternating voltages and currents to direct currents

2. 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.

3. rectification effeciency
Performance Parameters
rectification effeciency
form factor
ripple factor

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

6. the load and diode currents

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

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

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

11. 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.

12. The PIV is

13. Single-Phase Full Bridge Diode Rectifier With Resistive Load

14. Input power factor = The PIV=300V
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 =

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

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

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

19. 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.

21. Three-Phase Half Wave Rectifier

23. 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.

24. The PIV=
Vm=650.54V

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

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

29. Three-Phase Full Wave Rectifier With Resistive Load

35. 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 .

36. The PIV=
Vm=650.54V

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

42. 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.