Q33GATE 2002MCQ

In the chopper circuit shown in figure, the input dc voltage has a constant value $V_{s}$ . The output voltage $V_{o}$ is assumed ripple-free. The switch S is operated with a switching time period T and a duty ratio D. what is the value of D at the boundary of continuous and discontinuous conduction of the inductor current $i_L$ ?

🚀 Solve in Practice Mode

📖 Explanation

As the given chopper is buck chopper.

\begin{aligned} V_0&=\alpha V_s\\ \Delta I&=\frac{V_0T_{off}}{L}=\frac{\alpha V_s(1-\alpha )}{Lf}\\ \Delta I&=2I_L=2I_0\\ \Delta I&=2I_0=\frac{\alpha V_s(1-\alpha )}{Lf} \end{aligned}

$\frac{\Delta L}{2}=I_{L(avg)}=I_{0(avg)}$ this is applicable only at the boundary of continuous and discontinuous

\begin{aligned} L_c &=\frac{\alpha V_s(1-\alpha )}{2I_0f} \\ L_c&=\frac{\alpha V_sR(1-\alpha )}{R2I_0f} \\ (\because & \text{ multiplying and dividing by R}) \\ L_c&=\frac{V_0 R(1-\alpha )}{RI_02f} \\ L_c&= \frac{I_0R(1-\alpha )}{I_02f}\\ L_c&=\frac{R(1-\alpha )}{2f} \\ \therefore &= \text{ Boundary value of duty cycle is,}\\ \frac{2fL}{R}&=1-\alpha \\ \alpha &= 1-\frac{2fL}{R}=1-\frac{2L}{RT} \end{aligned}