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?


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


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