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GATE 2015 EE – Question 42

Power Electronics · DC to DC conversion: Buck, Boost and Buck-Boost Converters · 2 marks · Multiple choice

A self commutating switch SW, operated at duty cycle $\delta$ is used to control the load voltage as shown in the figure.

[Figure: A boost converter with source $V_{dc}$ in series with an inductor L, a switch SW across the line, a diode D, a capacitor C and a load $R_L$. $V_L$ is the inductor voltage and $V_C$ is the capacitor voltage.]

Under steady state operating conditions, the average voltage across the inductor and the capacitor respectively, are

Diagram for GATE 2015 EE question 42
  1. $V_L = 0$ and $V_C = \frac{1}{1 - \delta}V_{dc}$
  2. $V_L = \frac{\delta}{2}V_{dc}$ and $V_C = \frac{1}{1 - \delta}V_{dc}$
  3. $V_L = 0$ and $V_C = \frac{\delta}{1 - \delta}V_{dc}$
  4. $V_L = \frac{\delta}{2}V_{dc}$ and $V_C = \frac{\delta}{1 - \delta}V_{dc}$

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Show answer and explanation

Correct answer: (A) $V_L = 0$ and $V_C = \frac{1}{1 - \delta}V_{dc}$

Explanation

In steady state the average voltage across an inductor is zero. The circuit is a boost converter, whose output voltage is $\frac{V_{dc}}{1 - \delta}$.