Power Electronics

$$ \begin{aligned} &\textbf{Given:} \\ &t_{rr} = 3.9\ \mu s, \quad \frac{di}{dt} = 50\ \text{A/}\mu s, \quad SF = 0.3 \\[2mm] &\textbf{Formulas used:} \\ &t_{rr} = t_a + t_b \\ &SF = \frac{t_b}{t_a} \\ &t_{rr} = t_a(1 + SF) \\ &I_{rr} = t_a \times \frac{di}{dt} \\ &Q_{RR} = \frac{1}{2} I_{rr}\, t_{rr} \\[3mm] &\textbf{Step 1: Find } t_a \\ &t_a = \frac{t_{rr}}{1+SF} = \frac{3.9}{1+0.3} = \frac{3.9}{1.3} = 3\ \mu s \\[2mm] &\textbf{Step 2: Find } t_b \\ &t_b = SF \times t_a = 0.3 \times 3 = 0.9\ \mu s \\ &\text{Check: } t_a + t_b = 3 + 0.9 = 3.9\ \mu s \checkmark \\[2mm] &\textbf{Step 3: Peak Reverse Recovery Current, } I_{rr} \\ &I_{rr} = t_a \times \frac{di}{dt} = 3\ \mu s \times 50\ \text{A/}\mu s \\ &\boxed{I_{rr} = 150\ \text{A}} \\[2mm] &\textbf{Step 4: Storage Charge, } Q_{RR} \\ &Q_{RR} = \frac{1}{2} I_{rr}\, t_{rr} = \frac{1}{2} \times 150 \times 3.9\times10^{-6} \\ &Q_{RR} = \frac{1}{2} \times 585\times10^{-6} \\ &\boxed{Q_{RR} = 292.5\ \mu C} \\[3mm] &\textbf{Results Summary:} \\ &t_a = 3\ \mu s, \quad t_b = 0.9\ \mu s, \quad I_{rr} = 150\ \text{A}, \quad Q_{RR} = 292.5\ \mu C \end{aligned} $$