CHAPTER ONE : INTRODUCTION

1. What is the power electronics.
Applications of solid-state electronics in the electrical power field are steadily increasing; the term power electronics has been used since 1960s, after the introduction of the silicon controlled rectifier (SCR) by general electric. Power electronics has shows rapid growth in recent years with the development of power semiconductor devices that can switch large current efficiently at high voltages. Since these devices offer high reliability and small size. The scope of applications of power electronic systems as regulating power supply, variable speed of AC and DC machines, high voltage DC transmission lines, frequency changer and load matching in solar system.
The broad field of electrical engineering can be divided into three major areas, namely, electric power, electronic, and control. Power electronics deal the application of power semiconductor devices, such as thyristors and transistors, for the conversion and control of electrical energy at high power levels. This conversion is usually from AC to DC or vice versa, while the parameters controlled are voltage, current, or frequency. For example, a DC/DC converter has constant input and performs an output of variable voltage and current.
The process of power conversion or control is realized by controlling the adequate operation of power switches devices. The control signals necessary are generated by electronic or digital means, so the power flow is controlled by electronic means. That is why the power electronics terms is used for these system. Figures (1) and (2) below demonstrate the combination between power, electronic, control, and various branches.



Figures (1) and (2)Relation between power, electronics, and control (power electronics)

2. Why the power electronics.
Transfer of electric power from a source to a load can be controlled by varying the supply voltage (by using a variable transformer) or by inserting a regulator (such as a rheostat, variable reactor or switch). Semiconductor devices used as switches have the advantage of being relatively small, inexpensive, and efficient, and they can be used to control power automatically. An addition advantage of using a switch as a control element (compared to using adjustable resistance provided by rheostat or potentiometer) can be presented as follows:

a) A Rheostat as a control device.
Figure (3) shows a rheostat controlling a load. When R1 is set to zero resistance, full power is delivered to the load. When R1 is set for maximum resistance, the power delivered is close to zero. When R1 equal to RL , 50% of the power is consumed in the load and 50% in R1, so the efficiency is 50%. Moreover, the rheostat must be physically larger than the load to dissipate additional power.
In industrial applications where the power to be controlled is large, the efficiency of conversion is important. Poor efficiency means large losses, an economical consideration, and it also generates heat that must be removed from the system to prevent overheating.

Figure (3) A rheostat controlling a load.

EX.1 A DC source of 100 V is supplying 10 ohms resistive load. Find the power delivered to the load (PL), the power loss in the rheostat (PR), the total power supplied by the source (PT), and the efficiency (?) if the rheostat is set at : a)0 ohms, b)10 Ohms c)100 Ohms.

Sol.
a) Voltage across the load (VL) = 100 V
Power supplied to the load (PL) = 1002/10 = 1 KW
Power dissipated in the rheostat (PR) = 0 W
Power supplied by the source (PT) = PL + PR = 1 KW
Efficiency (? ) = PL/PT * 100 = 100 %

b) Voltage across the load (VL) = 10 * 100 / 20 = 50 V
Power supplied to the load (PL) = 502/10 = 250 W
Power dissipated in the rheostat (PR) = 502/10 = 250W
Power supplied by the source (PT) = PL + PR = 500 W
Efficiency (? ) = PL/PT * 100 = 50 %

c) Voltage across the load (VL) = 10 * 100 / 110 = 9.09 V
Power supplied to the load (PL) = 9.092/10 = 8.26 W
Power dissipated in the rheostat (PR) = 90.912/100 = 82.64 W
Power supplied by the source (PT) = PL + PR = 90.9 W
Efficiency (? ) = PL/PT * 100 = 9.08 %

It is clear from this example that the efficiency of power transfer from the source to the load is decrease as the rheostat resistance increase.

b) A switch as a control device.
In Figure (4), a switch is used to control the load. When the load is ON, maximum power is delivered to the load. The power loss in the switch is zero since it has no voltage across it. When the switch is OFF, 0 power is delivered to the load, again, the switch has no power loss since there is no current through it. The efficiency is 100 % because the switch does not waste power in either of its two positions.


Figure (4) A switch controlling the load.

The problem of this method is that unlike a rheostat, a switch cannot be set at intermediate positions to vary the power. However, we can create the same effect by periodically turning the switch ON and OFF. If we need more power, the switch is set ON for longer periods and OFF for shorter periods. When less power is needed, it is set OFF longer.

EX.2 A DC source of 100 V is supplying 10 ohms resistive load through a switch. Find the power delivered to the load (PL), the power loss in the switch (PS), the total power supplied by the source (PT), if the switch is : a) Closed, b) Open, C) Closed 50% of the time, & d) Closed 20% of the time.

Sol.
a) Voltage across the load (VL) = 100 V
Power supplied to the load (PL) = 1002/10 = 1 KW
Power loss in the switch (PS) = 0 W
Power supplied by the source (PT) = PL + PS = 1 KW

b) Voltage across the load (VL) = 0 V
Power supplied to the load (PL) = 0 W
Power loss in the switch (PS) = 0 W
Power supplied by the source (PT) = PL + PS = 0 W

c) With the switch closed 50% of the time (see figure 5)
Average Voltage across the load (VL) = 50 V
Power supplied to the load (PL) = 502/10 = 250 W
Power loss in the switch (PS) = 0 W
Power supplied by the source (PT) = PL + PS = 250 W

d) With the switch closed 20% of the time
Average Voltage across the load (VL) = 20 V
Power supplied to the load (PL) = 202/10 = 40 W
Power loss in the switch (PS) = 0 W
Power supplied by the source (PT) = PL + PS = 40 W


Figure (5) Load voltage using ON & OFF periods.

As this example shows, all the power supplied by the source is delivered to the load. The efficiency of power transfer is 100%. Don t forget that the switch is assumed to be ideal, but in practice the power loss in the electronic switches such as any type transistors or thyristors are very low.