b) the value of the output filter capacitor, C, so that the output voltage ripple to be lower than 0.5 V. Use the inductance you calculated in part (a).

To determine the value of the output filter capacitor, $C$, to achieve a specified output voltage ripple lower than $0.5 \, V$, we can follow these steps:

1. Break Down the Problem

1. Identify the relevant formulas for output voltage ripple in an LC filter circuit.
2. Use the inductance value calculated in part (a).
3. Find the relationship between ripple voltage, capacitance, and load current.

2. Relevant Concepts

The output voltage ripple ($\Delta V$) for a buck converter or similar topology can be approximated by the formula:

$\Delta V = \frac{I_{out}}{f \cdot C}$

Where:

• $I_{out}$ = output load current
• $f$ = switching frequency
• $C$ = output capacitance

3. Analysis and Detail

To ensure the output voltage ripple, $\Delta V$, remains lower than $0.5 \, V$, we rearrange the formula to solve for the capacitance $C$:

$C = \frac{I_{out}}{f \cdot \Delta V}$

4. Verify and Summarize

• Substitute the known values of $I_{out}$, $f$, and $\Delta V$ into the rearranged formula.
• Calculate the minimum capacitance required.

Let’s assume you have the following values:

• $I_{out} = 2 \, A$ (example current)
• $f = 100 \, kHz$ (example frequency)
• $\Delta V < 0.5 \, V$

Substituting these values into the equation:

$C = \frac{2 \, A}{100,000 \, Hz \cdot 0.5 \, V}$

Final Answer

Calculating:

$C = \frac{2}{100,000 \cdot 0.5} = \frac{2}{50,000} = 0.00004 \, F = 40 \, \mu F$

The required value of the output filter capacitor, $C$, should be at least 40 µF to keep the output voltage ripple under 0.5 V.