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

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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.

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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.

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

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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

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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.

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

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Dr. Ali M. Eltamaly, King Saud University

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

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

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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

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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

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

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Dr. Ali M. Eltamaly, King Saud University

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Half wave diode rectifier with free wheeling diode

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Dr. Ali M. Eltamaly, King Saud University

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1.1

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Dr. Ali M. Eltamaly, King Saud University

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

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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.

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

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

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Dr. Ali M. Eltamaly, King Saud University

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Single-Phase Full Bridge Diode Rectifier With Resistive Load

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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 =

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Dr. Ali M. Eltamaly, King Saud University

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

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Full Bridge Single-phase Diode Rectifier with DC Load Current

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

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

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Dr. Ali M. Eltamaly, King Saud University

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the total reduction per period is:

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

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Dr. Ali M. Eltamaly, King Saud University

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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.

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Three-Phase Half Wave Rectifier

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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.

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The PIV= V m =650.54V

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

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

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

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Three-Phase Full Wave Rectifier With Resistive Load

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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.

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The PIV= V m =650.54V

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Three-Phase Full Wave Rectifier With DC Load Current

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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.

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