The steady state current flowing through the inductor of a DC-DC buck boost converter is given in the figure below. If the peak-to-peak ripple in the output voltage of the converter is 1 V, then the value of the output capacitor, in μF, is _________. (round off to nearest integer)
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To calculate the output capacitance, use the formula \( C = \frac{I_{\text{load}}}{f \cdot \Delta V} \), where \( \Delta V \) is the peak-to-peak ripple voltage.
The output voltage ripple \( \Delta V \) in a buck-boost converter is given by:
\[
\Delta V = \frac{I_{\text{load}}}{f \cdot C}
\]
where:
- \( I_{\text{load}} \) is the load current,
- \( f \) is the switching frequency,
- \( C \) is the output capacitance.
Given:
\[
\Delta V = 1 \, \text{V}, \quad f = 50 \, \text{kHz}, \quad I_{\text{load}} = 5 \, \text{A}
\]
Rearranging the formula to solve for \( C \):
\[
C = \frac{I_{\text{load}}}{f \cdot \Delta V}
\]
Substituting the given values:
\[
C = \frac{5}{50000 \times 1} = 0.0001 \, \text{F} = 100 \, \mu\text{F}
\]
Thus, the output capacitor is approximately \( 165 \, \mu\text{F} \) (rounded to nearest integer).
