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I'm designing a step-down buck converter and I'm a little bit worried about ripple current which flows through the output capacitors.

In my research on the internet I found that ripple current is given by Vrms_output/ESR. If Ι have 3.5A maximum output current and I want transient responce this mean, 0A to 3.5A = 3.5A ripple current. But this value (transient step), Ι think isn't my normal ripple current.

In my calculation corresponding to equation above if I want a ripple 10mV and select ceramic capacitor which gives me low ESR (< 5mΩ), this is equiavalent to 2A ripple current and I think that isn't normal!

If that is true, can a ceramic capacitor handle this magnitude of current?

Is this amount of current (ripple current, Irms) identical with Inductor ripple current of regulator?

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2 Answers 2

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I'm not sure I'm following your line of thought, but it sounds like you are attacking the problem backwards from the usual way.

Normally, you'll first set a limit on ripple current (\$\Delta{}I_L\$), and choose an inductor to meet that requirement.

Then you will choose your output capacitor to achieve the ripple voltage that you want.

For idealized components, you will have

$$\Delta{}V_{out} = \Delta{}I_L \sqrt{R_{ESR}^2 + \left(\frac{1}{8 f_{sw} C_{out}}\right)^2}$$

So the lower the ESR and the higher the capacitance, the lower the ripple voltage will be.

This ripple voltage is ripple due the inherent transients in each cycle of the switching circuit, when operating with a constant load current.

Ripple or ring due to load transients is more due to the voltage control loop than due to these considerations. So to minimize that you need to model your control loop and ensure you have sufficient phase margin to avoid an unwanted ring response to load transients.

Is this amount of current (ripple current, Irms) identical with Inductor ripple current of regulator?

Yes. If the load current is constant (or near enough) then the deviations of inductor current from the average value (which is the load current) have nowhere to go but in and out of the output capacitor.

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  • \$\begingroup\$ I have choose the inductor already. My designing contains the TI's LMR14030 (ti.com/lit/ds/symlink/lmr14030.pdf), and i deside to choose the ripple current at 0.4 Imax, such as datasheet examples. And it leads me to a 3.3uH. Finally i was found a minimum capacitance >80uF, and ESR < 7mΩ. But i cunfused because all these calculation i did with given ripple output voltage, and calculated ripple current caused by maximum current. And with ohm's law Vrms/ESR ripple current comes very high and this confused me much more. \$\endgroup\$
    – MrBit
    Commented Jan 26, 2016 at 18:46
  • \$\begingroup\$ Remember you can parallel capacitors to reduce the effective ESR. For example, eight 10-uF capacitors with 50 mOhm ESR would meet your needs. (But remember to derate your capacitors so maybe use eight capacitors of 22 uF each with 30 or 40 mOhm ESR) \$\endgroup\$
    – The Photon
    Commented Jan 26, 2016 at 18:51
  • \$\begingroup\$ Why to derate capacitors? You mean quantities? I prefer to split a capacitance in two-three quantities to meet the initial capacitance and required ESR \$\endgroup\$
    – MrBit
    Commented Jan 26, 2016 at 19:00
  • \$\begingroup\$ Because capacitor performance tends to change with temperature, applied voltage, etc. Ceramics are better than electrolytics, but still you might have to allow for the actual capacitance to be 50% lower than the nominal value at high temperatures or with voltage applied. Check your datasheet carefully if you want to be sure about the parts you choose. \$\endgroup\$
    – The Photon
    Commented Jan 26, 2016 at 19:19
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The ripple current applied to your output capacitors is your peak-to-peak inductor current. Nothing less, nothing more.

Recall:

\$ V_{L} = L \dfrac{di}{dt} \$

Solving for \$ di \$:

\$ di = \dfrac{V_L \cdot dt}{L} \$

Your ripple current is defined by the volt-seconds applied to the inductor, divided by its inductance.

The output capacitance doesn't affect the ripple current; its ESR controls the ripple voltage. The ripple current is unaffected by the ESR. (The inductor just doesn't care!)

Generally, ceramic capacitors can handle very high ripple currents compared with general-purpose electrolytic capacitors. There are certain types of electrolytic capacitors intended for high-frequency high-ripple applications; they will generally advertise their ESRs at 100kHz or more.

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  • \$\begingroup\$ Capacitance doesn't affect the ripple current? Why then in calculations of capacitance we use ripple current? \$\endgroup\$
    – MrBit
    Commented Jan 26, 2016 at 18:51
  • \$\begingroup\$ Ripple current affects what capacitor you need to achieve a given ripple voltage, but changing the capacitor does not change the ripple current. \$\endgroup\$
    – The Photon
    Commented Jan 26, 2016 at 18:52
  • \$\begingroup\$ Therefore, ripple current affect to capacitance or ESR which required to keep the voltage deviation into the allowable levels? \$\endgroup\$
    – MrBit
    Commented Jan 26, 2016 at 19:03
  • \$\begingroup\$ ESR affects the ripple voltage. The ripple current is constant - it is unaffected by the capacitance. \$\endgroup\$
    – Adam Lawrence
    Commented Jan 26, 2016 at 20:22
  • \$\begingroup\$ How ESR and capacitance related? \$\endgroup\$
    – MrBit
    Commented Jan 26, 2016 at 20:48

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