Let's solve the problem step by step. --- # Given Data: - \( V_{in} = 48\,V \) - \( V_o = 15\,V \) - Switching frequency \( f_s = 100\,kHz \) - Maximum load \( I_{load} = 10\,A \) - Inductors are identical: \( L_1 = L_2 = 500\,\mu H \) - \( V_{in} \) and \( V_o \) values are provided - \( C = 10\,\mu F \) --- # Step 1: Determine the voltage and current ratings of the diode ## Diode voltage rating: - **Forward voltage drop** is usually around .7V for silicon diodes, but this is not explicitly specified. - The diode must withstand the maximum voltage across it, which during switching can approach \( V_{in} \) or \( V_o \). ## Diode current rating: - Maximum load current: \( I_{load} = 10\,A \) --- # Step 2: Calculate the inductor current during steady state In a buck converter, the inductor current \( I_L \) ripple is given by: \[ \Delta I_L = \frac{V_{L}}{f_s L} \] where \( V_{L} \) is the voltage across the inductor during the ON or OFF states. --- # Step 3: Calculate the inductor current ripple \( \Delta I_L \) ### For \( L_1 \) and \( L_2 \), the voltage across the inductor during ON and OFF states: - During ON state (when switch is closed): \[ V_{L_{ON}} = V_{in} - V_{o} = 48V - 15V = 33V \] - During OFF state: \[ V_{L_{OFF}} = -V_{o} = -15V \] ### Inductor current ripple during ON time: \[ \Delta I_{L_{ON}} = \frac{V_{L_{ON}}}{f_s L} = \frac{33\,V}{100 \times 10^3 \times 500 \times 10^{-6}\,H} = \frac{33}{50} = .66\,A \] Similarly, during OFF time: \[ \Delta I_{L_{OFF}} = \frac{|V_{L_{OFF}}|}{f_s L} = \frac{15\,V}{50} = .3\,A \] --- # Step 4: Determine the inductor current during steady state Average inductor current \( I_{L} \) is approximately equal to the load current: \[ I_{L} \approx I_{load} = 10\,A \] The ripple components are small compared to the load current, so: \[ I_{L_{peak}} = I_{L} + \frac{\Delta I_{L}}{2} \] \[ I_{L_{peak}} \approx 10 + \frac{.66}{2} = 10 + .33 = 10.33\,A \] Similarly, the minimum inductor current: \[ I_{L_{min}} \approx 10 - .33 = 9.67\,A \] --- # Step 5: Determine the voltage and current ratings for the diode - **Peak reverse voltage (\( V_{R} \))**: The diode must block the maximum voltage across it during OFF state: \[ V_{R} \approx V_{in} = 48\,V \] - **Peak forward current**: The diode must handle the peak inductor current: \[ I_{D, peak} \approx 10.33\,A \] --- # Summary: | Parameter | Approximate Value | |------------|-------------------| | Diode voltage rating | ≥ 48 V (preferably higher for safety margin) | | Diode current rating | ≥ 11 A (considering peak current and safety margin) | --- Let me know if you'd like me to proceed further with the remaining parts of the problem!