I just would like to ask about the small-ripple approximation and output ripple of buck/boost converters:

I understand that in a buck converter, there is an LC filter at the output. If correctly designed, the cutoff frequency of the LC filter will be much less than the switching frequency, hence the output voltage should have minimal ripple and we can approximate this to zero ripple at the output. This approximation is valid as long as the switching period is much smaller than the natural time constants of the circuit - hence we only see the linear portion of the inductor current and capacitor voltage response and not a full second-order sinusodial solution. enter image description here

This all makes sense. However, the notes further go on to say that: enter image description here

Now, here is my confusion. There is still a current ripple! I assume this current ripple will go to the capacitor and cause a voltage ripple at the output. This ripple is at the switching frequency! This is against our small-ripple approximation made above

Does this mean that our LC filter hasn't worked? I'm having a hard time distinguishing the sources of ripples

  • 1
    \$\begingroup\$ There is still a current ripple! Correct, also realize that the output capacitor has a certain series resistance (ESR). That resistance with the current ripple causes a voltage ripple at the output. That's perfectly normal and also needed for the feedback loop in a DCDC converter. Some converters become unstable when the ESR of the output capacitor is too high or too low. So yes the ripple is small but never zero. \$\endgroup\$ Commented May 23, 2021 at 15:46
  • \$\begingroup\$ Doesn't that contradict the fact that we assumed with the small ripple approximation that the output voltage = DC level and ignored the ripple. It's kind of confusing - we use the small-ripple approximation to simplify our calculations and say that the output voltage has no ripple but the result from those simplified calculations tell us that there is an output voltage ripple. \$\endgroup\$ Commented May 23, 2021 at 17:16
  • \$\begingroup\$ "This ripple is at the switching frequency! This is against our small-ripple approximation made above". Just because the current ripple and the voltage ripple are at the same frequency doesn't mean that the small voltage ripple assumption is wrong. That depends on the amplitude of the voltage ripple and not directly on the frequency. The assumption \$V_{ripple}<<V\$ still holds. The contribution of \$V_{ripple}\$ to the current ripple is still negligible compared to the contribution of \$V\$, the average output voltage. \$\endgroup\$
    – AJN
    Commented May 23, 2021 at 17:44
  • \$\begingroup\$ The current ripple at the switching frequency has two components: one caused by the average output voltage and switching frequency and the second caused by the ripple ourput voltage. The second one is assumed to be negligible due to the low magnitude of Vripple compared to Vaverage. \$\endgroup\$
    – AJN
    Commented May 23, 2021 at 17:46
  • \$\begingroup\$ "this current ripple will go to the capacitor and cause a voltage ripple at the output. This ripple is at the switching frequency!". Yes; but, its contribution to the output voltage ripple magnitude is given by \$\frac{1}{C}\int{I_L}dt\$. Considering only the fundamental frequency \$I_L \sin(\omega t)\$, the voltage ripple is \$\frac{I_L}{C}\frac{1}{\omega} \cos(\dots)\$. For a switching frequency of, say, \$\omega = 10^4 rad/s\$ the attenuation of the voltage signal compared to current signal for a sufficiently large capacitor makes the ripple voltage contribution negligible. \$\endgroup\$
    – AJN
    Commented May 23, 2021 at 17:54

1 Answer 1


You are mixing two notions: one is the output voltage ripple which, for analysis considerations, is considered zero, due to the assumed perfect filter, and the current ripple, which must be present in the inductor (and capacitor) due to the inherent switching behaviour of the converter.

To exemplify, here is a simple pulsed voltage source with an equivalent filter and load:


The LC filter is ideal, no parasitics, which makes the output (black trace) have a very small ripple compared to the input (blue trace). The current through the inductor (red trace), however, must have a ripple, or it would be impossible to function. The current through the load (green trace) is also shown, for comparison. Since I(R1) is virtually flat, you can assume that I(C1) is just like I(L1), except with zero average. This snapshot in time is in the steady-state.

As Bimpelrekkie mentions in the comments, if the LC elements would have been non-ideal, then the waveforms would be less than ideal, too, but not that much different:


The series resistances are some 25 mΩ, which is a modest value, and there are no series inductances, parallel capacitances, etc. The difference that an idealized filter makes is in the analysis. For comparison, here are the two transfer functions for the 1st picture, and for the 2nd:

$$\begin{align} H_{\mathrm{ideal}}(s)&=\dfrac{\dfrac{1}{LC}}{s^2+\dfrac{1}{RC}s+\dfrac{1}{LC}} \tag{1} \\ H_{\mathrm{simple-real}}(s)&=\dfrac{\dfrac{1}{L}\dfrac{Rr_C}{R+r_C}s+\dfrac{1}{LC}\dfrac{R}{R+r_C}}{s^2+\dfrac{1}{LC}\dfrac{(R+r_C)r_LC+Rr_CC+L}{R+r_C}s+\dfrac{1}{LC}\dfrac{R+r_L}{R+r_C}} \tag{2} \end{align}$$

Which one do you think is more favourable for analysis? I'll let you build the transfer function for the rest of the parasitics.

Maybe if you look at it from a different point of view: it's a switching application, therefore V(i) is a pulse. Let's consider the ideal case. The current through L1 is a triangle, and basic analysis says that:

$$i_L(t)=i_C(t)+i(R(t)\; \Rightarrow\; i_C(t)=i_L(t)-i(R(t)\tag{3}$$

If the output voltage is constant (flat), then \$i_R(t)\$ is constant, which leaves \$i_C(t)\$ be the same triangle as \$i_L(t)\$, only without any DC component.

Your confusion is at this point: you think that because the capacitor current has ripple, the voltage has ripple, but you forget that the current through the capacitor generates a voltage across it based on the integral of the current through it (also mentioned by AJN in the comments). And that integral means a pole at DC, thus the DC is infinite compared to the higher frequencies (still ideal case). Here is what the voltage across the capacitor would be if there were no other components:


Look, in particular, at the value on the Y-axis. In the upper plot there's the derivative of the current through L1 (or the voltage across it), compared to the output voltage, showing that in the ON time L1 supplies the power, while on the decreasing slope C1 does, by mainaining the voltage. Again, since this is an ideal case, there is no ripple.

If not, try a reductio ad absurdum: basic analysis tells you that V(o)=V(i)-V(i,o). Since the input is a pulse, the current through the inductor is triangular (see the formulas in the graphs you posted). The voltage across the inductor is the derivative of the current, and the derivative of a triangle is a pulse. The amplitude will be a function of the ratio between the output and the input voltage, or the ON time. This you can already see in the last picture, top plot (where I should have plotted V(i,o), not the other way around). And since V(i) is a pulse, and V(i,o) is a pulse, the difference is the DC, or V(o). And since the difference between the derivative of the voltage across L1 and the amplitude of V(i) is the duty cycle, a, you get that V(o)=a*V(i). No ripple.

  • \$\begingroup\$ I see. But the LC filter is telling us the output voltage should be constant with no ripple. However, the inductor current ripple at the switching frequency flows to the capacitor, thereby causing an output voltage ripple - contradicting the fact that the LC filter guaranteed us a constant output voltage. Assuming 0 ESR and no parasitics. \$\endgroup\$ Commented May 23, 2021 at 17:24
  • \$\begingroup\$ @jaurunjljgrtutkwcy You didn't read everything: "Since I(R1) is virtually flat, you can assume that I(C1) is just like I(L1), except with zero average". If you do some basic Kirchoff, what do you get? I(L1)=I(R1)+I(C1). The very plots show you that I(R1) is almost flat. \$\endgroup\$ Commented May 23, 2021 at 17:59
  • \$\begingroup\$ I've updated my answer, maybe now it helps. \$\endgroup\$ Commented May 23, 2021 at 18:24
  • \$\begingroup\$ The LC filter is not "telling us" anything -- we are assuming that the output voltage ripple is zero to make the math easier when determining the inductor current ripple. If you're getting wrapped around the axle on this, tell yourself that the output voltage ripple is negligible -- meaning, you can ignore it for the purpose of determining the inductor ripple. Basically, as soon as the difference in inductor ripple because of output ripple gets less than the difference in inductor ripple because the components don't match with your ideal -- you can ignore it. \$\endgroup\$
    – TimWescott
    Commented May 23, 2021 at 18:50
  • 1
    \$\begingroup\$ Ohhh! I see it now! Your second explanation got it! So, an inductor current ripple does not mean that there will be an output voltage ripple. If we add an ESR, then we will see the effect of the current ripple flowing in the capacitor ESR at the output voltage node. In that case though, we no longer have an ideal LC filter! \$\endgroup\$ Commented May 23, 2021 at 20:11

Your Answer

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

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