# capacitor voltage ripple in buck converter

I am struggling with output voltage ripple calculation of the buck converter below. Firstly, I don't understand the following statement from the lecture.
Can anyone explain it?

If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current flows through the capacitor.

The images are from the lecture here (pages 39-40).

If the capacitor voltage ripple is small, then essentially all of the ac component of inductor current flows through the capacitor.

Paint a scenario and examine the currents: -

• Ripple is 50 mV p-p, nominally triangular and 100 kHz
• Output load is 10 ohm
• Output capacitor is 100 uF

The AC ripple current through the load is simply 5 mA p-p - this is the baseline for comparison. The RMS is the peak value (2.5 mA) x 0.577 = 1.443 mA

For the capacitor, we have to calculate the slope of the voltage. It rises 50 mV in 5 us so that's a rate of 10 kV/s. Going back to basics, Q=CV and differentiating we get: -

$\dfrac{dq}{dt} = C\dfrac{dv}{dt}$ which of course equals current.

Therefore current is 10,000 x 100uF = +/-1 amp and square in shape. RMS is 1 A.

1 amp is a lot bigger than 1.443 mA and "C" is always chosen to minimize ripple so, as C gets bigger, then the ripple gets smaller (hence a lower AC current through the load resistor). Ultimately the AC ripple current through the resistor tends towards zero and the ripple current in the capacitor remains at a constant.

Capacitor ripple current remains constant because it is defined by the inductor and input voltage to the regulator and, to make this analysis clearer it makes sense to consider the input voltage to be constant.

• Great, I understand most of it. Just one question. Why output voltage ripple is nominally triangular? I see books draw it as a sinusoidal wave. Dec 12, 2016 at 16:28
• It depends on the type of capacitor. Electrolytics are usually dominated by the ESR and it will look triangular certainly whereas with a ceramic it will tend to look more sinusoidal: e2e.ti.com/cfs-file.ashx/__key/… Dec 12, 2016 at 18:03
• Yeah, I see. Another point is that you said capacitor ripple current remains constant as C gets bigger because it is defined by the inductor and input voltage to the regulator. How to prove it mathematically? Let's call iL, iC and iR (R as a subscript not the multiplication i*R) are ripple currents through inductor, capator and load respectively. Then iL = iC + iR. From here, how to prove iC is constant regardless of C? Dec 13, 2016 at 4:22
• Ripple current comes from the inductor. Inductor current rises up to a maximum each cycle. That is defined by duty cycle, input and output voltage levels, and inductor value. As the input voltage is assumed to be constant and, as the output voltage is relatively flat (the whole point of the question), inductor ripple current is a well defined thing. Given that the whole point of this question is about proving the ripple current in the capacitor is seriously bigger than that which flows in the resistor it's no giant step to say inductor ripple current equals capacitor ripple current. Dec 13, 2016 at 8:38

As can be seen from the inductor current waveform graph $I_L=I_{average}+I_{ac}$ where $I_{average}$ is the dc component (the average dashed line) and $I_{ac}$ ($\Delta{I_L}$)is the component that makes the inductor current deviates from the average line.

What that statement means is that when the output voltage ripple is small, $I_o=I_{average}$ and $I_{ac} = I_C$. This is because $I_o=V_o/R$, and since $V_o$ is constant, $I_o$ is also constant, meaning only the current through the capacitor $I_C$ is changing.

• If you are assuming that Vo is constant, then the voltage across the output capacitor is also constant. This means that there is no current flowing through the capacitor. Dec 12, 2016 at 17:13
• 12.1 12 11.9v for load may be consider to be constant Nov 9, 2018 at 0:47