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).
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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
Top Questions on Current
A single-phase half-controlled bridge converter supplies an inductive load with ripple-free load current. The triggering angle of the converter is \( 60^\circ \). The ratio of the rms value of the fundamental component of the input current to the rms value of the total input current of the bridge is (rounded off to 3 decimal places).
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\(\omega\) (rad/s) Magnitude (dB) Phase (degrees) 0.5 -7 -40 1.0 -10 -80 2.0 -18 -130 10.0 -40 -200 System diagram:
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