Output capacitance in analysis of Buck Converter

Question

In an analysis of buck converter, we make few assumptions. Those are:

• The circuit is operating in steady state.
• The inductor current is continuous.
• The capacitor is very large and the output voltage is held constant.
• The switching period is \$T\$; the switch is closed for time \$DT\$ and opens for time \$(1-D)T\$
• The components are ideal.

Here we say that the capacitor is very large to keep output constant.

How large, in Farads, should the capacitor be?

Answer 1

That answer will depend on the acceptable voltage ripple at the output, and the current draw at the output. A larger capacitor, will result in a lower voltage ripple for a set current. Similarly, a larger capacitor will allow for a larger current draw for a set acceptable ripple voltage.

Depends on you design. A general equation to use is C = I/(f*Vpp).

This site provides a nice explanation with some graphics http://www.skillbank.co.uk/psu/smoothing.htm