Question:
In the DC–DC converter shown in the figure, the current through the inductor is continuous. The switching frequency is \(500\,\text{Hz}\).
The output voltage \(V_o\) across the load is assumed to be constant and ripple-free.
The peak value of the inductor current (in amperes) is __________ (rounded off to the nearest integer).
In the DC–DC converter shown in the figure, the current through the inductor is continuous. The switching frequency is \(500\,\text{Hz}\).
The output voltage \(V_o\) across the load is assumed to be constant and ripple-free.
The peak value of the inductor current (in amperes) is __________ (rounded off to the nearest integer).
Show Hint
For boost converters, calculate the ripple current using \( \Delta I_L = \frac{V_i D}{f_s L} \) and add it to the average current to find the peak value.
Updated On: Feb 3, 2026
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Solution and Explanation
Step 1: Determine the duty ratio
For a boost converter, the output voltage is related to the input voltage by:
\[ V_o = \frac{V_s}{1-\alpha} \]
\[ 40 = \frac{20}{1-\alpha} \]
\[ \alpha = 0.5 \]
Step 2: Calculate load current
\[ I_o = \frac{V_o}{R} = \frac{40}{10} = 4\ \text{A} \]
Step 3: Average inductor current
For a boost converter operating in CCM:
\[ I_L = \frac{I_o}{1-\alpha} \]
\[ I_L = \frac{4}{1-0.5} = 8\ \text{A} \]
Step 4: Inductor current ripple
The peak-to-peak inductor current ripple is:
\[ \Delta I_L = \frac{\alpha V_s}{fL} \]
\[ \Delta I_L = \frac{0.5 \times 20}{500 \times 2 \times 10^{-3}} = 10\ \text{A} \]
Step 5: Peak inductor current
\[ I_{\text{peak}} = I_L + \frac{\Delta I_L}{2} \]
\[ I_{\text{peak}} = 8 + \frac{10}{2} = 13\ \text{A} \]
Final Answer:
Peak inductor current = 13 A
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