Electronic – capacitor ripple current in buck converter


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?

Best Answer

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.