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Discontinuous Mode Analysis of Boost Regulator

A boost conve...

Question

$A boost converter has an input voltage of V_{S} = 5 V . The average output voltage V_{0} = 15 V and the average load current I_{0} = 0.5 A . The switching frequency is 25 k H z . If L = 150 μ H a n d C = 220 μ F, then the peak inductor current will be _____ A$
1.95

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Solution

The correct option is A 1.95
$Peak inductor current, I_{2} = I_{S} + \frac{Δ I}{2} Δ I = \frac{k V_{S}}{f L} = \frac{V_{S} (V_{0} - V_{S})}{f L V_{0}} V_{0} = \frac{V_{S}}{(1 - k)} k = 1 - \frac{5}{15} = \frac{2}{3} Δ I = \frac{2}{3} \times \frac{5}{25 \times 10^{3} \times 150 \times 10^{- 6}} = 0.89 A I_{S} = \frac{0.5}{(1 - \frac{2}{3})} = 1.5 A I_{2} = I_{S} + \frac{Δ I}{2} = 1.5 + \frac{0.89}{2} = 1.945 A$

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Similar questions

V_{S} = 5 V

V_{0} = 15 V

I_{0} = 0.5 A

L = 150 μ H a n d C = 220 μ F

V_{s} = 20 V, V_{0} = 50 V, R = 50 Ω, L = 20 μ H, D = 0.5.

V_{s} = 5 V

V_{0} = 15 V

I_{0} = 0.5 A

L = 150 μ H

C = 220 μ F

Δ V_{c}

V_{s}

Ω

μ

μ

μ

μ

V_{S}

I_{P}

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