In a step up converter, the duty ratio is adjusted to regulate the output voltage $V_{0}$ at $48 V$ . The input voltage varies in a wide range from $12 V$ to $36 V$ . The maximum power output is $120 W$ . For stability reasons, it is required that the converter should always operate in a discontinuous current conduction mode. The switching frequency is $50 k H z$ . Assuming ideal components and $C$ as very large, the maximum value of inductor that can be used is ______ $μ H$

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Solution

To find the maximum value of L that keeps the current conduction discontinuous, we will take the inductor current at the edge of continuous conduction.
Current $I_{0} = \frac{P_{0}}{V_{0}} = \frac{120}{48} = 2.5 A$
$2 I_{L} = \frac{K V_{s}}{f L}$ [ the edge of discontinuous conduction]
$I_{L (a v g)} = I_{S (a v g)}$
from no loss circuit, $V_{0} I_{0} = V_{S} I_{S}$
$48 \times 2.5 = 12 \times I_{S}$
$I_{S (a v g)} = 10 A$
[ $V_{S} = 12 V$ because large value of inductor is required for $K = 0.75$ ]
$L = \frac{K V_{S}}{2 I_{s} f} = \frac{0.75 \times 12}{2 \times 10 \times 10^{5}} = 4.5 μ H$

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