wow, I must have made a mistake and thinking about something else since the answer to this question was the easiest google search I've ever made
If we assume that the ripple voltage is small in comparison to the rectifier input voltage, and using full wave rectifier, and don't mind looking at the charge/discharge curve is linear then:
Iload/f*C = Vripple
where Iload is Ampere, f is Hertz, C is Farads and Vripple in Volts peak-peak.
Note that it is only an approximation.
I find it difficult to provide more than 4-5000µF, or if space is made a 10,000µF can be fitted if the high cost for such a cap(the rectified voltage is a little below 39V, so 80V tolarent cap is needed?) then:
3A/100Hz*0.004700F = 6.38Vpk-pk
or
3A/100Hz*0.010000F = 3Vpk-pk
I wonder if I can get away with ~6Vpk-pk ripple at the input to my buck converter...
It will feed a linear regulator to be used as a lab supply, can a switching step-down regulator tolerate that high ripple. What do you think?
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By the way, as soon as my computer stops crashing I will check out that simulation.
I don't know where my head as been since I noticed that post only know....