Building my understanding of the issue from (First PSU – need help with capacitor size) (especially the comments/ripple wiki/several capacitor sizing webpages) the calculation for rectifying a full wave bridge rectifier at 50A 16V should be:
$$\frac{50A}{2 * 60Hz * 2V (Ripple)} = .208333$$
Converting from F to uF, I get $$.208333*10^6=208,333uF$$
So I should get a filter capacitor close to 0.21000uF at greater than 30V (Peak/rounded/safety margin)? This is my first time doing such calculations, are there any glaring errors? I know that the person several people struggled with some of the complexity, and I'm not sure if my interpretation of some of the variables (particularly dV) is correct.
Best Answer
Err, no, I think your last step there is off.
$$ \frac{50\text{A}}{2\cdot 60\text{Hz}\cdot 2\text{V}} = 0.2083\text{F} = 208333\mu\text{F} $$
That's a huge capacitor.
Or, more likely, huge bank of capacitors.(Edit: as it turns out, there are plenty of capacitors > 200mF and > 35V. However, they are expensive, and about the size of a Mason jar.)Are you sure you want to be designing a 50A bridge rectifier? There are likely better ways to design a 50A AC-DC converter.
Also, how crucial is the 16V output? I'm guessing the 16V accounts for voltage drop on a regulator, so does 15V regulated out work? Because there are power supplies that can provide 15V @ 50+ A (see this XP Power PSU for example).