# Summary of Calculations ## **Given data** - \( V_{in} = 48\,\text{V} \) - \( V_o 15\,\text{V} \) - \( f_s = 100\,\text{kHz} \) - \( I_{o,\text{max}} = 10\,\text{A} \) The converter is a **buck converter** with a **bifilar inductor** (two identical inductors in parallel, so \( L_{eq} = \frac{L_1}{2} \)). --- ## **(a) Voltage and current ratings of diode** - **Diode reverse voltage rating:** \[ V_{D,\text{rev}} = V_{in} = 48\,\text{V} \] - **Diode average current:** \[ D = \frac{V_o}{V_{in}} = \frac{15}{48} = .3125 \] \[ I_{D,\text{avg}} = (1 - D) I_o = (1 - .3125)(10) = 6.875\,\text{A} \] - **Diode peak current:** Equal to peak inductor current: \[ I_{D,\text{peak}} = I_{L,\text{peak}} = I_o + \frac{\Delta I_L}{2} \] --- ## **(b) Slope of inductor current during charging** Given \( L_1 = L_2 = 500\,\mu H \Rightarrow L_{eq} = 250\,\mu H \): \[ \frac{di_L}{dt} = \frac{V_{in} - V_o}{L_{eq}} = \frac{48 - 15}{250 \times 10^{-6}} = 132,000\,\text{A/s} \] \[ = .132\,\text{A/μs} \] **Charging slope:** \(\boxed{.132\,\text{A/μs}}\) --- ## **(c) Critical inductance to maintain CCM down to 10% load** - **Boundary condition for CCM:** \[ \Delta I_L = 2 I_{L,\text{min}} \] - At 10% load: \[ I_{L,\text{min}} = .1 I_o = 1\,\text{A} \] - **Inductor ripple current:** \[ \Delta I_L = \frac{(V_{in} - V_o) D}{L_{eq} f_s} \] - **Solving for \(L_{eq}\):** \[ 2 \times 1 = \frac{(48 - 15) \times .3125}{L_{eq} \times 100 \times 10^3} \] \[ L_{eq,\text{crit}} = 51.6\,\mu H \] - Since \(L_{eq} = \frac{L_1}{2}\): \[ L_1 = 2 \times 51.6\,\mu H = 103.2\,\mu H \] **Critical \(L_1 = L_2 \approx \boxed{103\,\mu H}}\)** --- ## **(d) Peak switch current at full load for \(L_1 = L_2 = 200\,\mu H\)** - \( L_{eq} = 100\,\mu H \) \[ \Delta I_L = \frac{(48 - 15) \times .3125}{100 \times 10^{-6} \times 100 \times 10^3} = 1.03125\,\text{A} \] - Peak switch current: \[ I_{pk} = I_o + \frac{\Delta I_L}{2} = 10 + .5156 = \boxed{10.52\,\text{A}} \] --- ## **(e) Peak-to-peak output voltage ripple** - \( L_{eq} = 125\,\mu H \) (from \(L_1 = L_2 = 250\,\mu H\)) - \( C = 10\,\mu F \) \[ \Delta I_L = \frac{(48 - 15) \times .3125}{125 \times 10^{-6} \times 100 \times 10^3} = .825\,\text{A} \] - Output voltage ripple: \[ \Delta V_o = \frac{\Delta I_L}{8 f_s C} = \frac{.825}{8 \times 100 \times 10^3 \times 10 \times 10^{-6}} = .103\,\text{V} \] **Output ripple:** \(\boxed{\approx .103\,\text{V (p-p)}}\) --- ## **Final Summary** | Part | Quantity | Expression | Result | |--------|------------------------|-------------------|------------| | (a) | Diode reverse voltage \(V_{D,\text{rev}}\) | \(V_{in}\) | 48 V | | | Diode average current \(I_{D,\text{avg}}\) | \((1 - D) I_o\) | 6.875 A | | (b) | Inductor current slope \(\frac{di_L}{dt}\) | \(\frac{V_{in} - V_o}{L_{eq}}\) | .132 A/μs | | (c) | Critical \(L_1 = L_2\) for CCM | \(2 \times 51.6\,\mu H\) | 103 μH | | (d) | Peak switch current \(I_{sw,\text{peak}}\) | \(I_o + \frac{\Delta I_L}{2}\) | 10.52 A | | (e) | Output voltage ripple \(\Delta V_o\) | \(\frac{\Delta I_L}{8 f_s C}\) | .103 V |