I am having problems understanding ripple calculation for buck converter here (from page 39).

First, the author makes an assumption that the capacitance is large enough so nearly all [not all] of the inductor current ripple flows through the capacitor, and very little flows through the load. This assumption is valid.

However, in the calculation in the next pages, the author is calculating as if ALL inductor ripple current flows through the capacitor, no current flows through the load. Then he calculates for ripple voltage.

However, if we use the assumption that ALL inductor ripple current flows through the capacitor, no current flows through the load then the output ripple voltage should be ZERO. It doesn't make sense to me. There is a contradiction here.

What do you think about this?


1 Answer 1


If the current flows through the capacitor then it will charge the capacitor and the voltage will rise, that is the voltage ripple that is being calculated. That is why it is first calculated as the charge transferred in a cycle, since a capacitor has a direct relationship between charge and voltage.

  • \$\begingroup\$ But what the author assumed is that ALL ripple current flows through the capacitor --> No current flows through the load R --> The voltage ripple across the resistor is 0. However, the ripple across the capacitor is not zero. That is the contradiction I am talking about. \$\endgroup\$
    – emnha
    Mar 1, 2017 at 7:24
  • \$\begingroup\$ Some ripple current will flow through the load but for the sake of modelling in calculations you can ignore it if the capacitor is large enough (and has low enough ESR). There will be a steady current flowing through the load and when the current ripple comes along it will mostly flow into the capacitor since it has much lower impedance, but as the voltage across the capacitor rises it will increase the current through the load. This voltage increase is usually too small to create a significant current increase in a well designed circuit so you can usually exclude it from calculations. \$\endgroup\$
    – TWiz
    Mar 1, 2017 at 9:25
  • \$\begingroup\$ So in short there is a voltage and current ripple on the resistor, the voltage ripple is equal to that across the capacitor, but the current ripple is much lower because the load has a much greater impedance then the capacitor, unless the capacitor is way too small. \$\endgroup\$
    – TWiz
    Mar 1, 2017 at 9:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.