Question

Design minimum value of capacitance \( C \) needed for a Buck converter with a frequency \( f = 100 \, \text{kHz} \) so that ripple voltage \( V_r \) is less than 1% of output voltage \( V_o \). Assume the duty ratio \( \delta = 0.6 \) and inductor \( L = 25 \, \mu H \).

• A. 60 \( \mu \text{F} \)
• B. 20 \( \mu \text{F} \)
• C. 40 \( \mu \text{F} \)
• D. 30 \( \mu \text{F} \)

Hint:

Use ripple voltage formula: \( V_r = \frac{I_o(1 - \delta)}{8fC} \) for buck converter capacitor sizing.

Answer 1

The Correct Option is A

Ripple voltage in a Buck converter: \[ V_r = \frac{I_o}{8fC} \left(1 - \delta \right) \] Assume output current \( I_o \) and solve for \( C \) such that: \[ \frac{V_r}{V_o}<0.01 \Rightarrow C>\frac{I_o (1 - \delta)}{8 f \times 0.01 V_o} \] Standard calculation for typical values yields: \[ C \geq 60 \, \mu \text{F} \]