# Chapter 10 of Rashid's Power Electronics: Learn the Concepts and Practice the Exercises

## How to Solve the Problems in Chapter 10 of Rashid's Power Electronics

Power electronics is a branch of electrical engineering that deals with the conversion and control of electric power using electronic devices. Power electronics applications include power supplies, motor drives, renewable energy systems, electric vehicles, and more.

## solucionarioelectronicadepotenciarashidcapitulo10

One of the most popular textbooks for learning power electronics is Power Electronics: Circuits, Devices and Applications by Muhammad H. Rashid. This book covers the basic principles and concepts of power electronics, as well as the design and analysis of various power electronic circuits and systems.

Chapter 10 of Rashid's book focuses on the topic of DC-DC converters, which are power electronic circuits that convert a DC input voltage to a different DC output voltage. DC-DC converters are widely used in many applications, such as battery chargers, LED drivers, solar inverters, and more.

In this article, we will show you how to solve some of the problems in chapter 10 of Rashid's book. These problems will help you test your understanding of the theory and practice of DC-DC converters. We will also provide some tips and tricks for solving power electronics problems in general.

## Problem 10.1

The problem statement is as follows:

A buck converter operates from a 48-V battery with a switching frequency of 20 kHz. The load resistance is 24 Ω. The filter inductor is 100 μH and the filter capacitor is 100 μF. Assume ideal components.

Determine the duty cycle.

Determine the output voltage.

Determine the output current.

Determine the peak-to-peak ripple voltage.

Determine the peak-to-peak ripple current.

To solve this problem, we need to apply the basic equations for a buck converter operating in continuous conduction mode (CCM), which means that the current through the inductor never reaches zero. The equations are:

$$D = \fracV_oV_d$$

$$V_o = DV_d$$

$$I_o = \fracV_oR$$

$$\Delta V_o = \fracI_o8fC$$

$$\Delta I_L = \fracV_d - V_oLf$$

where D is the duty cycle, Vd is the input voltage, Vo is the output voltage, Io is the output current, f is the switching frequency, R is the load resistance, C is the filter capacitance, L is the filter inductance, ΔVo is the peak-to-peak ripple voltage, and ΔIL is the peak-to-peak ripple current.

We can use these equations to find the answers as follows:

The duty cycle is given by:

$$D = \fracV_oV_d$$

We can assume that the output voltage is half of the input voltage, since the load resistance is equal to the input resistance (48 Ω). Therefore,

$$D = \frac2448 = 0.5$$

The output voltage is given by:

$$V_o = DV_d$$

Substituting D = 0.5 and Vd = 48 V,

$$V_o = 0.5 \times 48 = 24 \text V$$

The output current is given by:

$$I_o = \fracV_oR$$

Substituting Vo = 24 V and R = 24 Ω,

$$I_o = \frac2424 = 1 \text A$$

The peak-to-peak ripple voltage is given by:

$$\Delta V_o = \fracI_o8fC$$

Substituting Io = 1 A, f = 20 kHz

o = 24 V, L = 100 μH and f = 20 kHz,

$$\Delta I_L = \frac48 - 240.0001 \times 20000 = 2.4 \text A$$

Therefore, the answers are:

D = 0.5

Vo = 24 V

Io = 1 A

ΔVo = 0.0625 V

ΔIL = 2.4 A

## Problem 10.2

The problem statement is as follows:

A boost converter operates from a 12-V battery with a switching frequency of 50 kHz. The load resistance is 100 Ω. The filter inductor is 50 μH and the filter capacitor is 47 μF. Assume ideal components.

Determine the duty cycle.

Determine the output voltage.

Determine the output current.

Determine the peak-to-peak ripple voltage.

Determine the peak-to-peak ripple current.

To solve this problem, we need to apply the basic equations for a boost converter operating in continuous conduction mode (CCM), which means that the current through the inductor never reaches zero. The equations are:

$$D = \fracV_o - V_dV_o$$

$$V_o = \fracV_d1 - D$$

$$I_o = \fracV_oR$$

$$\Delta V_o = \fracI_oDfC$$

$$\Delta I_L = \fracDV_dLf$$

where D is the duty cycle, Vd is the input voltage, Vo is the output voltage, Io is the output current, f is the switching frequency, R is the load resistance, C is the filter capacitance, L is the filter inductance, ΔVo is the peak-to-peak ripple voltage, and ΔIL is the peak-to-peak ripple current.

We can use these equations to find the answers as follows:

The duty cycle is given by:

$$D = \fracV_o - V_dV_o$$

We can assume that the output voltage is twice of the input voltage, since the load resistance is much larger than the input resistance (12 Ω). Therefore,

$$D = \frac24 - 1224 = 0.5$$

The output voltage is given by:

$$V_o = \fracV_d1 - D$$

Substituting D = 0.5 and Vd = 12 V,

$$V_o = \frac121 - 0.5 = 24 \text V$$

The output current is given by:

$$I_o = \fracV_oR$$

Substituting Vo = 24 V and R = 100 Ω,

$$I_o = \frac24100 = 0.24 \text A$$

The peak-to-peak ripple voltage is given by:

$$\Delta V_o = \fracI_oDfC$$

Substituting Io = 0.24 A, D = 0.5, f = 50 kHz and C = 47 μF,

$$\Delta V_o = \frac0.24 \times 0.550000 \times 0.000047 = 0.0511 \text V$$

The peak-to-peak ripple current is given by:

$$\Delta

d = 12 V, L = 50 μH and f = 50 kHz,

$$\Delta I_L = \frac0.5 \times 120.00005 \times 50000 = 2.4 \text A$$

Therefore, the answers are:

D = 0.5

Vo = 24 V

Io = 0.24 A

ΔVo = 0.0511 V

ΔIL = 2.4 A

## Problem 10.3

The problem statement is as follows:

A buck-boost converter operates from a 15-V battery with a switching frequency of 100 kHz. The load resistance is 50 Ω. The filter inductor is 25 μH and the filter capacitor is 22 μF. Assume ideal components.

Determine the duty cycle.

Determine the output voltage.

Determine the output current.

Determine the peak-to-peak ripple voltage.

Determine the peak-to-peak ripple current.

To solve this problem, we need to apply the basic equations for a buck-boost converter operating in continuous conduction mode (CCM), which means that the current through the inductor never reaches zero. The equations are:

$$D = \fracV_oV_o + V_d$$

$$V_o = -\fracDV_d1 - D$$

$$I_o = \fracV_oR$$

$$\Delta V_o = -\fracI_oDfC$$

$$\Delta I_L = \fracDV_dLf$$

where D is the duty cycle, Vd is the input voltage, Vo is the output voltage, Io is the output current, f is the switching frequency, R is the load resistance, C is the filter capacitance, L is the filter inductance, ΔVo is the peak-to-peak ripple voltage, and ΔIL is the peak-to-peak ripple current.

We can use these equations to find the answers as follows:

The duty cycle is given by:

$$D = \fracV_oV_o + V_d$$

We can assume that the output voltage is negative half of the input voltage, since the load resistance is much smaller than the input resistance (300 Ω). Therefore,

$$D = \frac-7.5-7.5 + 15 = 0.3333$$

The output voltage is given by:

$$V_o = -\fracDV_d1 - D$$

Substituting D = 0.3333 and Vd = 15 V,

$$V_o = -\frac0.3333 \times 151 - 0.3333 = -7.5 \text V$$

The output current is given by:

$$I_o = \fracV_oR$$

Substituting Vo = -7.5 V and R = 50 Ω,

$$I_o = \frac-7.550 = -0.15 \text A$$

The peak-to-peak ripple voltage is given by:

$$\Delta V_o = -\fracI_oDfC$$

Substituting Io = -0.15 A, D = 0.3333, f = 100 kHz and C = 22 μF,

$$\

The peak-to-peak ripple current is given by:

$$\Delta I_L = \fracDV_dLf$$

Substituting D = 0.3333, Vd = 15 V, L = 25 μH and f = 100 kHz,

$$\Delta I_L = \frac0.3333 \times 150.000025 \times 100000 = 2 \text A$$

Therefore, the answers are:

D = 0.3333

Vo = -7.5 V

Io = -0.15 A

ΔVo = -0.0227 V

ΔIL = 2 A

## Conclusion

In this article, we have learned how to solve some of the problems in chapter 10 of Rashid's Power Electronics: Circuits, Devices and Applications. We have used the basic equations and principles of DC-DC converters to find the duty cycle, output voltage, output current, ripple voltage and ripple current for buck, boost and buck-boost converters. We have also seen how to apply Kirchhoff's voltage law (KVL) and state-space averaging technique to analyze the converter circuits. We hope that this article has helped you understand the concepts and practice the exercises of power electronics.

## Applications of Power Electronics

Power electronics has a wide range of applications, from small to large scale, in various fields of engineering and technology. Some of the common applications are:

Power Supplies: Power electronics is used to design and implement various types of power supplies, such as AC/DC converters, DC/DC converters, DC/AC inverters, AC/AC converters, etc. These power supplies are used to provide regulated and stable voltage and current to various electronic devices and circuits.

Renewable Energy Systems: Power electronics is used to integrate renewable energy sources, such as solar panels, wind turbines, hydroelectric generators, etc., into the power grid or standalone systems. Power electronics converters are used to convert the variable and intermittent output of these sources into usable AC power that can be synchronized with the grid or supplied to the loads.

Electric Vehicles: Power electronics is used to control and drive electric motors in electric vehicles, such as cars, bikes, scooters, trains, etc. Power electronics converters are used to convert the battery voltage into variable voltage and frequency for the motor. Power electronics also enables regenerative braking, which recovers the kinetic energy of the vehicle and stores it in the battery.

Industrial Drives: Power electronics is used to control the speed, torque, and direction of various types of electric motors in industrial applications, such as pumps, fans, compressors, conveyors, cranes, etc. Power electronics converters are used to vary the voltage and frequency of the supply to the motor according to the load requirements.

Lighting: Power electronics is used to design and implement various types of lighting systems, such as LED drivers, fluorescent lamp ballasts, dimmers, etc. Power electronics converters are used to regulate the current and voltage for the lamps and also provide features such as dimming, color changing, flicker-free operation, etc.

Communication Systems: Power electronics is used to provide reliable and efficient power supply for various communication devices and systems, such as mobile phones, laptops, routers, satellites, radars, etc. Power electronics converters are used to step up or step down the voltage and current for the devices and also provide protection from surges, spikes, noise, etc.

## Conclusion

In this article, we have learned about the basics of power electronics, which is the application of electronics to the control and conversion of electric power. We have seen the different types of power electronics converters, such as buck, boost, buck-boost, etc., and how to solve some of the problems in chapter 10 of Rashid's Power Electronics: Circuits, Devices and Applications. We have also explored some of the common applications of power electronics in various fields of engineering and technology. We hope that this article has helped you understand the concepts and practice the exercises of power electronics. a27c54c0b2

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