Question

A buck converter operating in discontinuous mode is shown in figure below. If the switching frequency is 10 kHz and duty ratio = 0.4, then the output voltage is

Answer 1

$I_L=\frac{\frac{1}{2}{I}_{max}(D+D_1)T}{T}=\frac{1}{2}{I}_{max}(D+D_1)$

$\Delta I=I_{max}=\frac{V_0{D}_{1}T}{L}$

$I_L=\frac{V_0}{R}=\frac{1}{2}\left(\frac{V_0{D}_{1}T}{L}\right)(D+D_1)$

$D_1^2+D{D}_1-\frac{2L}{RT}=0$

$D_1^2+0.4D_1-0.2=0$

$D_1=0.29,-0.69$

$D_1=0.29$

$V_0=V_s\left(\frac{D}{D+D_1}\right)=24\times \frac{0.4}{0.4+0.29}=24\times \frac{0.4}{0.69}=13.91V$