A buck converter has input voltage of 3.3 V and output voltage of 1.2 V. The converter feeds a communication system with a output of 6 A. The output voltage ripple must not exceed 2%. The value of inductor required such that the peak-to-peak variation in inductor current does not exceed 40 percent of average value will be _________ $μ H$ .
(Take switching frequency of the converter as 500 kHz)
0.637

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Solution

The correct option is A 0.637
For buck converter,
The duty ratio, $D = \frac{V_{0}}{V_{s}} = \frac{1.2}{3.3} = 0.364$
The switching frequency,
$f_{S} = 500 k H z$
Output current of converter,

$I_{L} = I_{0} = 6 A$
Allowable variation in output current,

$Δ i_{L} = \frac{40}{100} \times 6 = 2.4 A$
The critical value of inductance required,

$L = (\frac{V_{S} - V_{0}}{Δ {i_{L}}_{f}}) D$

$L = \frac{(3.3 - 1.2) (0.364)}{(2.4) \times (500000)} = 0.637 μ H$

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