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 voltage (\( V_o \)) across the load is assumed to be constant and ripple-free. The peak inductor current in amperes is (rounded off to the nearest integer).
\includegraphics[width=0.5\linewidth]{64.png}
In the DC-DC converter shown in the figure, the current through the inductor is continuous. The switching frequency is \( 500 \, \text{Hz} \). The voltage (\( V_o \)) across the load is assumed to be constant and ripple-free. The peak inductor current in amperes is (rounded off to the nearest integer).
\includegraphics[width=0.5\linewidth]{64.png}
\includegraphics[width=0.5\linewidth]{64.png}
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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: Jan 23, 2025
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Solution and Explanation
Step 1: Define the relation for peak-to-peak inductor current.
For a boost converter, the inductor current ripple is given by:
\begin{align*}
V_0 &= \frac{V_S}{1-\alpha}
40 &= \frac{20}{1-\alpha}
\alpha &= 0.5
I_0 &= \frac{V_0}{R} = \frac{40}{10} = 4 \text{ A}
\end{align*} Boost: \begin{align*} I_L &= \frac{I_0}{1-\alpha} = \frac{4}{1-0.5} = 8 \text{ A}
\Delta I_L &= \frac{\alpha V_S}{fC} = \frac{0.5 \times 20}{500 \times 2 \cdot 10^{-3}} = 10
I_{mx} &= (i_L)_{\text{peak}}
&= I_L + \frac{\Delta I_L}{2} = 8 + \frac{10}{2} = 13 \text{ A} \end{align*}
40 &= \frac{20}{1-\alpha}
\alpha &= 0.5
I_0 &= \frac{V_0}{R} = \frac{40}{10} = 4 \text{ A}
\end{align*} Boost: \begin{align*} I_L &= \frac{I_0}{1-\alpha} = \frac{4}{1-0.5} = 8 \text{ A}
\Delta I_L &= \frac{\alpha V_S}{fC} = \frac{0.5 \times 20}{500 \times 2 \cdot 10^{-3}} = 10
I_{mx} &= (i_L)_{\text{peak}}
&= I_L + \frac{\Delta I_L}{2} = 8 + \frac{10}{2} = 13 \text{ A} \end{align*}
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