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I'm working with Buck converters and Boost converters separately.

Can someone tell me the reason why the output ripple voltage (and current) exists in the Output Capacitors and how does the input ripple voltage (and current) occurs in the input capacitors?

I think since the current in the inductor is ramping up and down, this current flows in and out of the output capacitor which is why we get output ripple current in the capacitor. And if its an Electrolytic capacitor, we have its ESR, which gives the Output ripple voltage? Is my understanding correct? If so, could you provide a formula for it which might help in intuitive understanding?

But I am not able to understand how input ripple voltage and current occurs at the input capacitors? Please help with that.

Any proper references or links to the formulas for their calculation of ripple current and voltage and the input and output of the capacitors?

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  • \$\begingroup\$ Well you wouldn't call it input ripple if it was on the output... \$\endgroup\$
    – Hearth
    Commented Apr 9, 2021 at 4:47
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    \$\begingroup\$ If current is flowing in and out of a capacitor, it will have ripple voltage even if it is an ideal capacitor with zero ESR. The reason is that the capacitor rule is I = C dv/dt. If you rearrange that to solve for dv/dt, it says dv/dt = I/C. So if I is not zero, there will be changing voltage. \$\endgroup\$
    – user57037
    Commented Apr 9, 2021 at 5:34
  • \$\begingroup\$ Oh, then ESR does not involve with the concept of ripple voltage? \$\endgroup\$
    – user220456
    Commented Apr 9, 2021 at 6:47
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    \$\begingroup\$ Well, the ESR will affect the ripple voltage. But even a theoretical capacitor with zero ESR will have some ripple voltage. ESR will increase ripple voltage but may reduce ripple current in the capacitor. \$\endgroup\$
    – user57037
    Commented Apr 9, 2021 at 8:39
  • \$\begingroup\$ Both ESR and capacitance will affect ripple. Basically it means that a capacitor with no ESR defines the minimum capacitance for given ripple, but due to ESR, capacitance needs to be larger so keep the total ripple from capacitance and ESR below your ripple limit. \$\endgroup\$
    – Justme
    Commented Apr 9, 2021 at 8:53

2 Answers 2

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Both capacitors have ripple because inductor current rises or falls.

You have the output ripple figured out, either the inductor current is large enough so current can charge the cap, or the inductor current is low so cap must discharge into the load.

Same goes for the input capacitor. Either switch is on so current is drawn from it to raise inductor current, or switch is off so no current is drawn from it which is the phase when inductor current falls.

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  • \$\begingroup\$ Thank you for the answer. Can you provide with the formulas for calculating the output and input ripple (Voltage and Current) ? I couldn't see a proper consistent formula of ripple for the dc-dc converters. Like, is the calculation same for the buck and boost converter? \$\endgroup\$
    – user220456
    Commented Apr 9, 2021 at 4:55
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    \$\begingroup\$ Almost all converter chip datasheets have the formulas, I have not memorized them and I would suppose the formulas are different for different tyoe of converters. \$\endgroup\$
    – Justme
    Commented Apr 9, 2021 at 5:11
  • \$\begingroup\$ Few of the datasheets have the formulas. And that too for output capacitors only. Many of them don't have the formula for the input capacitors. \$\endgroup\$
    – user220456
    Commented Apr 9, 2021 at 5:14
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The current in the input capacitor depends on the switching cell structure and the operating conditions. The first thing to realize with a perfect decoupling or with a front-end filter is that all the high-frequency ac pulses are delivered by the capacitor while the dc current flows in the input source. It is an approximation of course but this is what is implied when calculating ripple currents. The current absorbed by a buck converter, for instance, is that of the power switch. That current is made of a dc and an ac component. The quadratic sum of these currents leads to the input rms current: \$I_{SW,rms}=\sqrt{I_{in,dc}^2+I_{in,ac}^2}\$.

In this formula, the dc current is simply the input current obtained knowing the delivered power and the input voltage: \$I_{in,dc}=\frac{P_{out}}{V_{in}}=\frac{V_{out}I_{out}}{V_{in}}=MI_{out}\$. The rms current in the capacitor is thus derived as \$I_{C,rms}=\sqrt{I_{SW,rms}^2-I_{in,dc}^2}\$.

To determine the rms current in the power switch, you can have a look at the formula I derived in the second edition of my book on power supplies. If you simulate or measure the current circulating in the power switch of a buck converter operated in continuous conduction mode (CCM), you should see this waveform:

enter image description here

and if you follow the calculation steps, the rms current in the switch is obtained with the following expression:

\$I_{SW,rms}=\frac{\sqrt{D(12I_{out}^2+\Delta I_L^2)}}{2\sqrt{3}}\$.

Let's check this formula with a quick calculation around a 10-A buck converter:

enter image description here

In this example, the rms current in the input capacitor is quite high and amounts to nearly 5 A. The SIMPLIS simulation confirms this value:

enter image description here

If your converter has a different structure, you will proceed in a similar way to first obtain the input current signature and extract the ac value of it.

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