In almost all DC-DC converter books, they introduce the small-ripple approximation to make the derivation of the the Vout-Vin relationship easy.

Usually, they assume that the output of the converter is a constant DC with no higher order components.

Then, they derive the inductor current ripple and capacitor voltage ripple with the assumption that the output is constant DC.

That seems like a contradiction - they use the small-ripple approximation on the output voltage (to say it is fixed DC) but then they use this constant output voltage in the derivations to calculate inductor current ripple and capacitor voltage ripple? It is almost like they are only half applying the small-ripple approximation, they only apply to the output voltage, not to the cap and inductor waveforms!

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    \$\begingroup\$ You can find an explanation here. Using a flat dc voltage for \$V_{out}\$ rather than a pulsating level to analyze a converter drastically simplifies the analysis. You have to keep in mind that small-signal analysis is carried at steady-state and the goal of an ac-dc converter is to deliver a continuous dc voltage with the smallest output ripple, usually 1% or so of the regulated voltage. So it is fair to consider it continuous for simplifying analyses. \$\endgroup\$ Jun 24 at 5:47
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    \$\begingroup\$ You can also have a look at an answer I gave on SE here regarding the small-ripple approximation. \$\endgroup\$ Jun 24 at 6:55
  • \$\begingroup\$ That's why it's called an approximation! \$\endgroup\$
    – user253751
    Jun 24 at 9:19
  • \$\begingroup\$ @VerbalKint Oh. So it is valid to apply the small ripple approximation to the output voltage since the DC is more dominant and it won't make a huge difference to the accuracy of the calculation of inductor current ripple and capacitor voltage ripple, is that correct? \$\endgroup\$ Jun 24 at 17:27
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    \$\begingroup\$ Oui, this is correct. You can imagine how complex it would be to factor in the change in the output voltage when computing the inductor slope cycle by cycle. A quick simulation will show you that the slopes don't change at steady-state despite a little ripple. \$\endgroup\$ Jun 24 at 19:39

1 Answer 1


If there was no AC ripple, ( ie. no AC feedback) It would be unstable with under/overshoot ringing to step load response since it operates in a switched-mode rather than linear. This is used to compute the ripple voltage with DCR, ESR and dV= Ic*(dt/C+ESR) in the output caps and ripple current in L.

Typically current sense feedback improves stability to drive current boost and voltage feedback for average DC. Otherwise, voltage ripple is partially differentiated to simulate load current changes from dV/dt=Ic/C and also enhance loop phase margin aka Lead/Lag network. To regulate average DC over several cycles DC linear analysis is used.

  • \$\begingroup\$ FACT: If any feedback system saturates with no load regulated ripple, the loop gain is zero and it has no feedback. So for AC stability some ripple is essential and none would be unstable. directed towards some uneducated person (-1) \$\endgroup\$ Jun 24 at 11:20

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