Let's solve the problem step by step. --- ## **Step 1: Write Down the Given Data** - **Switching frequency, \( f_s \) = 20 kHz = 20,000 Hz** - **Input voltage, \( V_i \) = 175 V to 250 V (use minimum for worst case)** - **Output voltage, \( V_o \) = 375 V (regulated)** - **Output power, \( P_o \) = 3.5 kW = 350 W** - **Output voltage ripple \( \Delta V_o < 5 \) V** --- ## **Step 2: Output Current Calculation** \[ I_o = \frac{P_o}{V_o} = \frac{350}{375} = 9.33~\text{A} \] --- ## **Step 3: Buck-Boost Converter?** Based on the circuit, this is a **boost converter**. --- ## **Step 4: Duty Cycle Calculation at Minimum Input** \[ D = 1 - \frac{V_i}{V_o} \] At minimum input (\( V_i = 175~V \)): \[ D_{max} = 1 - \frac{175}{375} = 1 - .467 = .533 \] --- ## **Step 5: Inductance Calculation for DCM** For DCM operation, the critical inductance is when the inductor current just returns to zero at the end of each cycle. The boundary (critical) inductance for DCM is given by: \[ L_{critical} = \frac{(1-D)^2 \cdot R}{2f_s} \] Where \( R = \frac{V_o}{I_o} = \frac{375}{9.33} = 40.2~\Omega \). Plug in values at \( D_{max} \) for the worst case: \[ L_{critical} = \frac{(1-.533)^2 \cdot 40.2}{2 \cdot 20,000} \] \[ = \frac{(.467)^2 \cdot 40.2}{40,000} \] \[ = \frac{.218 \cdot 40.2}{40,000} \] \[ = \frac{8.76}{40,000} \] \[ = .000219~\text{H} = 219~\mu H \] To ensure DCM, use slightly less than this value (e.g., \( L < 219~\mu H \)). --- ## **Step 6: Capacitance Calculation for Ripple** The output voltage ripple in a boost converter is approximately: \[ \Delta V_o = \frac{I_o D}{f_s C} \] Solve for \( C \): \[ C = \frac{I_o D}{f_s \Delta V_o} \] Given: - \( I_o = 9.33~A \) - \( D = .533 \) - \( f_s = 20,000~Hz \) - \( \Delta V_o = 5~V \) \[ C = \frac{9.33 \times .533}{20,000 \times 5} \] \[ = \frac{4.976}{100,000} \] \[ = .00004976~F = 49.76~\mu F \] --- ## **Step 7: Final Answers** ### **Inductance Required** \[ \boxed{L < 219~\mu H} \] ### **Capacitance Needed** \[ \boxed{C > 49.8~\mu F} \] --- ### **Summary Table** | Parameter | Value | |----------------------|---------------------| | Inductance \( L \) | \( < 219~\mu H \) | | Capacitance \( C \) | \( > 49.8~\mu F \) | --- **Let me know if you need further clarification on any step!**

# DC-DC Converter Analysis ## Problem Statement Determine the inductance required to maintain discontinuous conduction mode (DCM) operation in a DC-DC converter with the following parameters: - Switching frequency: \( f_s = 20 \, \text{kHz} \) - Input voltage range: \( V_i = 175 \, \text{V} \) to \( 250 \, \text{V} \) - Output voltage: \( V_o = 375 \, \text{V} \) - Output power: \( P_o = 3.5 \, \text{kW} \) - Maximum allowable output voltage ripple: \( \Delta V_o < 5 \, \text{V} \) ## Step 1: Data Summary ### Given Values - \( f_s = 20,000 \, \text{Hz} \) - \( V_i = 175 \, \text{V} \) (minimum for worst-case analysis) - \( V_o = 375 \, \text{V} \) - \( P_o = 3,500 \, \text{W} \) - \( \Delta V_o < 5 \, \text{V} \) ## Step 2: Calculate Output Current The output current \( I_o \) can be calculated as follows: \[ I_o = \frac{P_o}{V_o} = \frac{3500}{375} \approx 9.33 \, \text{A} \] ## Step 3: Determine Duty Cycle For a boost converter, the duty cycle \( D \) is calculated using the formula: \[ D = 1 - \frac{V_i}{V_o} \] At minimum input \( V_i = 175 \, \text{V} \): \[ D_{max} = 1 - \frac{175}{375} \approx 0.467 \] ## Step 4: Calculate Critical Inductance for DCM The critical inductance \( L_{critical} \) for DCM can be calculated with: \[ L_{critical} = \frac{(1-D)^2 \cdot R}{2f_s} \] Where \( R \) is calculated as: \[ R = \frac{V_o}{I_o} = \frac{375}{9.33} \approx 40.2 \, \Omega \] Now substituting values into the critical inductance formula: \[ L_{critical} = \frac{(1 - 0.467)^2 \cdot 40.2}{2 \cdot 20,000} = \frac{(0.533)^2 \cdot 40.2}{40,000} \] Calculating: \[ = \frac{0.284 \cdot 40.2}{40,000} \approx \frac{11.433}{40,000} \approx 0.000286 \, \text{H} = 286 \, \mu H \] Thus, for DCM operation: \[ L < 286 \, \mu H \] ## Step 5: Calculate Capacitance for Voltage Ripple To maintain the voltage ripple below \( 5 \, \text{V} \): \[ \Delta V_o = \frac{I_o D}{f_s C} \] Rearranging for \( C \): \[ C = \frac{I_o D}{f_s \Delta V_o} \] Substituting in the values: \[ C = \frac{9.33 \cdot 0.467}{20,000 \cdot 5} \] Calculating: \[ = \frac{4.36}{100,000} \approx 0.0000436 \, F = 43.6 \, \mu F \] ## Step 6: Final Results ### Inductance Required \[ \boxed{L < 286 \, \mu H} \] ### Capacitance Needed \[ \boxed{C > 43.6 \, \mu F} \] ## Summary Table | Parameter | Value | |----------------------|---------------------| | Inductance \( L \) | \( < 286 \, \mu H \) | | Capacitance \( C \) | \( > 43.6 \, \mu F \) | Feel free to ask for further clarifications or additional calculations!