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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.

I want to know much large should be the capacitor.
We say large but how many farads is not mentioned.

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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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