Re: Capacitor Sizing in Power Supplies

Hi,

You have it correctly. Doing violence to a differential equation one
comes up with C=(I dt)/dv.

I is your required current in amps
dt is the time between charge restoring pulses, (.0083 sec at 60Hz)
dv is your ripple voltage.
C is capacitance in Farads.

Where does the "80,000" come from? First, the .00833 seconds (1/120)
is rounded off to .008, then multiplied by 1 milliom to have "C" read
in microfarads.

Next, a good value for ripple voltage is to have it be 10% of the
supply voltage, so 8000 is multiplied by 10, resulting in 80,000 as
the multiplier.

So a 2VDC supply at 20A having a .2 volt ripple would indeed require
800,000 uF.

Mariss

--- In CAD_CAM_EDM_DRO@egroups.com, jmw@c... wrote:
> I seem to recall a formula in some Gecko docs re capacitor sizing
on
> roll your own power supplies (the xformer, bridge, cap variety).
And
> I think this formula was of the form (80,000*amps)/volts = uF of
cap.
> (The 80,000 is related somehow to the number of millisecs per half
> cycle of 60 cycle AC and the time it takes the cap to charge?)
>
> Suppose you wanted to build a 2VDC @ 20A supply. Without getting
into
> why one might want to do that, this works out to 80,000*20/2 = 800k
> uF. That's a lot of capacitor, albeit at pretty low voltage.
>
> Is this a non-linear situation or would 800k be the proper sized
cap?
>
> (Mariss, if I'm misquoting you--sorry.)
>
> Thanks in advance.
>
> --Jack