Gate 2016 EE Set 2 - Question 47

For a DC-DC boost converter, \frac{V_{0}}{V_{S}}=\frac{1}{1-\alpha }, where \alpha is the duty cycle

\frac{400}{360}=\frac{1}{1-\alpha }

\Rightarrow 400 - 400\alpha =360

\Rightarrow \alpha =\frac{400-360}{400}

\Rightarrow \alpha = 0.1

Power = V_{S}I_{S}

4000=(360)I_{S}\Rightarrow \frac{4000}{360}

I_{S}=11.1A

"all devices are ideal"\Rightarrow Neglect ripple in I_{S}

Consider the following formula, I_{rms}=I_{S}=\left ( \frac{T_{ON}}{T}^{\frac{1}{2}} \right )

I_{rms}=I_{S}\sqrt{\alpha } = 11.1\sqrt{0.1}=3.5A