Download presentation

Presentation is loading. Please wait.

Published byKeith Finn Modified about 1 year ago

1
Dr. Ali M. Eltamaly, King Saud University 1 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.

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.

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

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

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

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

8
Dr. Ali M. Eltamaly, King Saud University

9
Single-phase half-wave diode rectifier with resistive load.

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

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

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

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

14
Dr. Ali M. Eltamaly, King Saud University

17
Half wave diode rectifier with free wheeling diode

19
Dr. Ali M. Eltamaly, King Saud University

21
1.1 6.49518 1.12 4.87278 1.14 3.23186 1.16 1.57885 1.18 -0.079808 1.2 -1.73761

22
Dr. Ali M. Eltamaly, King Saud University

23
Single-Phase Full-Wave Diode Rectifier Center-Tap Diode Rectifier

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.

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

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

27
Dr. Ali M. Eltamaly, King Saud University

30
Single-Phase Full Bridge Diode Rectifier With Resistive Load

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 =

32
Dr. Ali M. Eltamaly, King Saud University

33
Example 5 solve Example 4 if the load is 30 A pure DC Input Power factor=

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
the total reduction per period is:

40
Dr. Ali M. Eltamaly, King Saud University the rms value of the supply current

41
Dr. Ali M. Eltamaly, King Saud University

45
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= V m =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= V m =650.54V

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

57
Dr. Ali M. Eltamaly, King Saud University

59
Three-Phase Full Wave Rectifier With Resistive Load

65
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= V m =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.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google