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Calculating the ripple current of the output filter of a full bridge smps? Answered

I have been designing an offline 100khz 720 watt full bridge converter for 24 volts dc out at 30 amps. I've calculated that at a minimum duty cycle of .3 and a voltage ripple of .03v I need a minimum of 7k uf. I read the output filter section of "The Power Supply Cookbook" and it gives  this equation to get the peak to peak current ripple: I(pk-pk)=(2*Iout)/(min duty cycle). with a duty cycle of .3 this gives me 200A ripple current. With four caps each would need to be about 2000uf and a ripple current of 50A, which would be expensive. I've only every built flyback and boost converters so I wonder if the equation for peak to peak current ripple is only meant for those topologies, or perhaps I've calculated wrong. Is there a different equation I need to use or have I done something wrong? Any tips for selecting capacitors and what chemistry to use (tantulum/aluminum)? Thanks in advance I'm new to forward mode converters!


No, its the same, it comes directly from the stated parameters.

Go for low ESR caps.....

Okay so the the lower ESR caps are going to have the higher RMS. Could I simply use a single 1uf ceramic cap in parallel with some low RMS electrolytics, or is it based on capacitance of each cap so that the ceramic cap would have little effect on the overall RMS?

looking at your numbers, 7000uf sounds awfully high at that frequency . As they say, show your working! !