Electronic – Ripple current at output capacitors

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.

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.