How many microfarads do I need for smoothing?

Question

I am rectifying the mains voltage which is 120V @ 5A with a full wave rectifier. How many volts and microfarads will my capacitor need to smooth out my rectified DC?

Are there any online calculators for this?

Thank you in advance!

Answer 1

Capacitor voltage has to be able to handle the ripple. Are you transforming down first? If you're going directly from mains voltage, the caps should be rated at least that high, maybe double for safety (What is the voltage out?). A common configuration is called a pi filter which has two caps across the output leads, and an inductor in series with one output lead (Usually the positive) between the caps. Looks like the symbol for pi sort of. Bigger values are better for the caps, typically in thousands of microFarads. Try this online calculator, the link is partway down the page:

http://www.smeter.net/filters/lpf_hpf.php

Answer 2

Rectifying the mains (120V) will produce a peak DC voltage of 170 Volts, but since the mains RMS vlaue can peak, itself, to 125 or 130 volts, you need rectifiers with PRV values at a MINIMUM of 185 volts, but even that should be over-rated by a factor of 1.5, so diodes with a PRV of 300 Volts (or higher) is necessary.

The diodes should also be rated for twice the current (10 Amps).

As for sizing the capacitors, you really need a ripple specification at the load to properly calculate the capacitance.

As a 'ball-park' estimate of 100 mV ripple across a 5 amp (34 Ohm) load (that's about 850 watts, btw) -- for a capacitor to start at 170 volts and lose 0.1 volts in 8.33 millisec (1/120 Hz) is 99.941%. ln(0.99941) = -0.0006

8.33 ms / 0.0006 = 14.1 seconds
this is the 'time constant' for 120 Hz and 100 mV of ripple.

With an "R" of 34 Ohms, and a time constant of 14.1, that makes "C" = 0.4 Farads, which is a *very* large capacitor, especially at 200 or 300 WVDC. That's a large bank of capacitors.

Of course, if the ripple specification were, say 1 volt, or several volts, then the value of C goes down.

.

Answer 3

Well, that depends on if the wave is rectified, how much ripple voltage you want and what your load resistance is.

Vp = peak voltage
Vr = ripple voltage
f = frequency (60 Hz for US)
R = load resistance

If the wave is half-rectified: C = Vp/(Vr*f*R)
If you fully rectify the wave: C is half of the above value.

A second smaller capacitor would reduce the ripple further, which you may or may not need.

I would suggest using a bridge rectifier and a couple of capacitors.