Determining H-Bridge DC-link decoupling capacitors
This is a attempt to answer part of my own question:


  how do I calculate the ripple current trough the capacitors in order to determine the required ripple current rating?


To calculate the capacitor ripple current for a given duty cycle D, [0, 1], we first determine the voltage across the load by calculating the voltage on each side of the bridge and subtracting one from the other:

Uload = (Uin * D) - (Uin * (1 - D))


Then determine the current trough the load:

Iload = Uload / Rload


Since the filter inductors are in series with the load, the average inductor current equals the load current.

Power on the load:

Pload = Uload * Iload


This power is the constant power supplied by the power supply for the given duty cycle, so the current supplied by the power supply is:

Iin = Pload / Uin


The difference between the current supplied by the power supply and the load/inductor current has to be supplied by the capacitor(s), during the active part of the period:

Icap-active = Iin - Iload


During the freewheeling part of the period, the inductor returns some of the energy it had stored to the capacitor(s). Since the current trough a inductor cannot instantly change, the current returned by the inductor equals Iload. So the capacitor recharges by the sum of Iin and Iload during freewheeling:

Icap-freewheel = Iin + Iload


The resulting current ripple is a square wave, we need to determine the RMS value which is of importance for electrolytic capacitors:

Icap-rms = √((Icap-active² * D) + (Icap-freewheel² * (1 - D)))


Example for Uin = 50V, Rload = 8ohm, 80% duty cycle (D = 0.8):

Uload = (50V * 0.8) - (50V * 0.2) = 30V
Iload = 30V / 8 ohm = 3.75A
Pload = 30V * 3.75A = 112.5W
Iin = 112.5W / 50V = 2.25A
Icap-active = 2.25A - 3.75A = -1.5A 
Icap-freewheel = 2.25A + 3.75A = 6A
Icap-rms = √(1.8 + 7.2) = 3A


Have a computer do all the work for each duty cycle in 1% steps and you get something like this, from which you can easily determine the peaks: