Answered: Draw circuit for Boost Converter. Also…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

Draw circuit for Boost Converter. Also derive its Output equation. What will be the value of Duty Cycle
(D) if the input voltage is 50 V DC and desired output voltage is 20 V DC.

Expert Solution

Step 1

Boost converter: It is a DC-DC converter (chopper) that step-ups the input voltage level while the current is stepped down and gives the desired higher voltage at the output, thus it is also called a step-up converter. It is also called a fly-back converter.

So, the boost converter is a step-up converter having:

$o u t p u t v o l t a g e > i n p u t v o l t a g e$

Circuit diagram of boost converter:

It is a type of switched-mode power supply (SMPS) that contains a minimum of two semiconductors i.e., a diode (D) and a transistor (switch S), and one energy storing element either inductor (L), or capacitor (C), or combination of both.

Step 2

The minimum average voltage of this step-up converter is equal to the input voltage V_d but can be changed by changing the duty cycle.

The basic operation of this converter is the storage of energy (in an inductor) at a low supply voltage and its delivery at a higher voltage. For example, the transfer of energy takes place only during the off period of the switch from source to load. This is the reason it is known as a fly-back converter. A transistor (or switch) usually transfers the energy from an inductor (energy storing element) to the output circuit. A large capacitor is required at the load end for smoothening the output voltage (it acts as a filter) as the energy is supplied in the form of pulses. The basic and essential parts of the boost converter are shown in the circuit above, with "S" as the main-switching element.

Now, assuming that the circuit is operating for a long time and the inductor current varies from the minimum value (I₁) to the maximum value (I₂). The switch is closed at t=0, the inductor current is at its minimum. The inductor current rises up during the period $0 \leq t \leq t_{o n}$

Applying mesh analysis in the input loop then the differential equation will be:

$V_{d} - L \frac{d i_{s}}{d t} = 0$

Considering small current variation,

$V_{d} = L \frac{I_{2} - I_{1}}{t_{o n}} = L \frac{∆ I}{t_{o n}} . . . . . (1)$

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