Electronic – LC Filter for a Full Wave Bridge Rectifier

capacitorfull-bridgehigh-currentinductorrectifier

According to what I've read on a few websites and PDFs (http://www.irjes.com/Papers/vol2-issue6/Version-1/E02064249.pdf is an example) using an inductor-capacitor filter to assist AC to DC rectification in ripple management is much better than using either component by themselves. (In the example PDF capacitor was sized down by 75% & inductor was sized down by 94%)

How can I size the inductor and capacitor such that I can get a certain voltage ripple after being rectified by a full wave bridge rectifier? (Even in the paper they recommended a trial and error method, essentially guessing 25% for the capacitor and manually finding the inductance for the inductor)

I haven't taken differential equations yet, but I'm past Calculus III if you prefer to answer in a non algebraic way. (Although algebra is preferred)

I'm running 1kW (split phase, plans on moving to 3-phase) through this bridge (full wave) for high power electrochemistry: $$972W = 18V * 54A$$

Best Answer

Have a look at:

http://www.circuitstoday.com/filter-circuits

Ripple Factor = Vac rms/Vdc = (√2/3)(Xc/XL) = (√2/3)(1/[2wc])(1/[2wL]) = 1/(6√2w2LC)