Problemas De Electronica De Potencia Andres Barrado Pdf Universidad De Valencia -

$I_{o} = P_o/V_o = 30/15 = 2A$. $I_{L,avg} = I_o = 2A$. $\Delta i_L = 0.3 \times 2A = 0.6A$. Fórmula: $\Delta i_L = \frac{V_o (1-D)}{f_s L} \Rightarrow L = \frac{V_o (1-D)}{f_s \Delta i_L} = \frac{15 \times 0.5}{150\times10^3 \times 0.6} = 83.3\mu H$.

= $V_o/V_{in} = 15/30 = 0.5$ (CCM ideal). $I_{o} = P_o/V_o = 30/15 = 2A$

However, remember Professor Barrado’s own advice (often included in the PDF’s preface): "No se aprende electrónica de potencia mirando soluciones. Se aprende intentando, fallando, y comparando con las soluciones correctas." (You don’t learn power electronics by looking at solutions. You learn by trying, failing, and comparing with the correct solutions.) and comparing with the correct solutions.)