To solve this problem, we need to determine the value of the filter inductance \( L \) for a Buck converter that will ensure continuous conduction mode (CCM) at one-third of the maximum output power. Given: - Input voltage, \( V_{in} = 20 \, \text{V} \) - Output voltage, \( V_o = 12 \, \text{V} \) - Maximum output power, \( P_o = 72 \, \text{W} \) - Switching frequency, \( f_s = 400 \, \text{kHz} \) First, calculate the maximum output current \( I_o \): \[ I_o = \frac{P_o}{V_o} = \frac{72 \, \text{W}}{12 \, \text{V}} = 6 \, \text{A} \] At one-third of the maximum output power, the output current \( I_{o, \text{min}} \) is: \[ I_{o, \text{min}} = \frac{I_o}{3} = \frac{6 \, \text{A}}{3} = 2 \, \text{A} \] The duty cycle \( D \) for the Buck converter is given by: \[ D = \frac{V_o}{V_{in}} = \frac{12 \, \text{V}}{20 \, \text{V}} = 0.6 \] The inductor current ripple \( \Delta I_L \) is given by: \[ \Delta I_L = \frac{V_{in} - V_o}{L} \cdot D \cdot \frac{1}{f_s} \] To ensure CCM, the peak-to-peak inductor current ripple \( \Delta I_L \) should be less than twice the minimum output current \( I_{o, \text{min}} \): \[ \Delta I_L \leq 2 \cdot I_{o, \text{min}} \] \[ \Delta I_L \leq 2 \cdot 2 \, \text{A} = 4 \, \text{A} \] Rearranging the inductor current ripple equation to solve for \( L \): \[ L = \frac{(V_{in} - V_o) \cdot D}{\Delta I_L \cdot f_s} \] Substitute the values: \[ L = \frac{(20 \, \text{V} - 12 \, \text{V}) \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}} \] \[ L = \frac{8 \, \text{V} \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}} \] \[ L = \frac{4.8 \, \text{V}}{1.6 \times 10^6 \, \text{A} \cdot \text{Hz}} \] \[ L = 3 \times 10^{-6} \, \text{H} \] \[ L = 3 \, \mu\text{H} \] Therefore, the value of the filter inductance \( L \) should be \( 3 \, \mu\text{H} \) to ensure the converter remains in CCM at one-third of the maximum output power.

To solve this problem, we need to determine the value of the filter inductance $L$ for a Buck converter that will ensure continuous conduction mode (CCM) at one-third of the maximum output power. Given: - Input voltage, $V_{in} = 20 \, \text{V}$ - Output voltage, $V_o = 12 \, \text{V}$ - Maximum output power, $P_o = 72 \, \text{W}$ - Switching frequency, $f_s = 400 \, \text{kHz}$ First, calculate the maximum output current $I_o$: $I_o = \frac{P_o}{V_o} = \frac{72 \, \text{W}}{12 \, \text{V}} = 6 \, \text{A}$ At one-third of the maximum output power, the output current $I_{o, \text{min}}$ is: $I_{o, \text{min}} = \frac{I_o}{3} = \frac{6 \, \text{A}}{3} = 2 \, \text{A}$ The duty cycle $D$ for the Buck converter is given by: $D = \frac{V_o}{V_{in}} = \frac{12 \, \text{V}}{20 \, \text{V}} = 0.6$ The inductor current ripple $\Delta I_L$ is given by: $\Delta I_L = \frac{V_{in} - V_o}{L} \cdot D \cdot \frac{1}{f_s}$ To ensure CCM, the peak-to-peak inductor current ripple $\Delta I_L$ should be less than twice the minimum output current $I_{o, \text{min}}$: $\Delta I_L \leq 2 \cdot I_{o, \text{min}}$ $\Delta I_L \leq 2 \cdot 2 \, \text{A} = 4 \, \text{A}$ Rearranging the inductor current ripple equation to solve for $L$: $L = \frac{(V_{in} - V_o) \cdot D}{\Delta I_L \cdot f_s}$ Substitute the values: $L = \frac{(20 \, \text{V} - 12 \, \text{V}) \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}}$ $L = \frac{8 \, \text{V} \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}}$ $L = \frac{4.8 \, \text{V}}{1.6 \times 10^6 \, \text{A} \cdot \text{Hz}}$ $L = 3 \times 10^{-6} \, \text{H}$ $L = 3 \, \mu\text{H}$ Therefore, the value of the filter inductance $L$ should be $3 \, \mu\text{H}$ to ensure the converter remains in CCM at one-third of the maximum output power.