# How does a buck converter's output ripple waveform become sinusoidal?

Below image is from Fundamentals of Power Electronics. The ripple component of the inductor current will be flowing through the capacitor as shown in the ic(t) graph. This one I understood.

May I know how the ripple waveform becomes sinusoidal? What is the mathematical equation behind it?

• It's not really sinusoidal; in the ideal case it would be piecewise parabolic. Which is visually indistinguishable from sinusoidal, but mathematically distinct. Mar 7 at 4:01

The buck converter can be thought as a low-impedance square-wave generator feeding a $$\LC\$$ filter. This filter will average the square signal and theoretically route the dc component to the load while all the ac flows into the capacitor: You can see how the capacitor is affected by a parasitic resistance $$\r_C\$$ (its equivalent series resistance or ESR) which makes the total ac voltage across the load $$\R_L\$$ equal to $$\v_{ac}(t)=v_{r_C}(t)+v_C(t)\$$. If the ohmic term $$\r_C\$$ is of very low value, then the total voltage is that of the capacitor and is almost sinusoidal. Then, if the parasitic resistance no longer becomes negligible, the total ac waveform turns into piecewise parabolic as mentioned by Hearth in his comment. The below picture is excerpted from my book on the subject: The relationship between voltage across a capacitor and the current through it is:

$$I(t) = C \frac{dV(t)}{dt}$$

or

$$V(t) = \frac{1}{C}\int I(t) dt + V(0)$$

where $$\V(0)\$$ is the voltage across it at time $$\t=0s\$$. The important point here is to see that current through the capacitor is proportional to the rate of change of voltage across it.

Essentially, the capacitor it is integrating current through it, the ripple current which is also passing through the inductor. Since the current waveform is roughly triangular, you can expect the integration operation to "round it off", forming something that looks like a sinusoid, but isn't.

Intuitively, at the peaks of the triangular current waveform, rate of change of voltage across the capacitor is at a maximum, and this is visible as a region of steepest slope in the curve of voltage.

Conversely, where the current waveform crosses zero, that voltage rate of change is at a minimum, 0, where the plot is flat and horizontal. Those points are marked by the vertical dotted lines in your graph.

In between those extremes, current is steadily rising or falling, and slope of voltage is smoothly transitioning between steep and flat.

The result is the sinusoid-like smooth curve of voltage. The only waveform you can integrate to obtain a sinusoid, is another sinusoid, which that current waveform most definitely is not. Therefore the voltage across the capacitor is not sinusoidal.

The best way to see this is with a simple simulation. Here I have a source of triangular waveform current, being pushed and pulled through a capacitor: simulate this circuit – Schematic created using CircuitLab

This is the voltage across the capacitor (orange) with a true sinusoid (blue) behind it, to show how they are not quite the same, and below that is the current through the capacitor: