# 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, $$I_{rms}=I_{S}=\left ( \frac{T_{ON}}{T}^{\frac{1}{2}} \right )$$

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