Ripple current

In the context of an unregulated power supply (transformer, bridge
rectifier, filter caps) what exactly is the definition of ripple current?
Is it just the RMS current in and out of the Cap?

I am asking because I am trying to select filter caps for just such a supply.
I've built one or two small supplies before, but now I'm trying to design
a 12-Amp, 24-Volt one, and I don't want the caps to blow up or anything.

Mac

 

The one you mention is an important one, because it heats the
capacitors and also is the reason the capacitor voltage varies.

This implies that the capacitors discharge with a 12 amp current
between rectifier pulses (when they have to carry the entire load),
and charge up with more than this because the charge time is usually
shorter than the discharge time (just the peaks of the cycle). So you
should choose capacitors that have a total RMS ripple current rating
well above 12 amps at the ambient temperature inside the supply. The
ripple voltage you can tolerate at full current sets a lower limit on
total capacitance.

Here are some examples and formulas for ripple voltage:
http://www.geofex.com/Article_Folders/Power-supplies/powersup.htm

 

Unfortunately this page doesn't give any formulas regarding ripple voltage
and -current.
In the case of a bridge rectifier you can roughly calculate:

periodic peak ripple current = Ua0/sqrt(2Ri*Rload)
Ua0= output voltage @ 0current;
Ri= internal resistance = (Uload-Unominal)/Inominal; for a 135VA this is
around 0.07*Unom/Inom

peak to peak ripple voltage = (1- 4th root(Ri/2Rload))* Ia/(2*C*f)
the minimum output voltage is the calculated output voltage - 2/3 ripple
voltage.

ciao Ban
Bordighera, Italy

 

I read in sci.electronics.design that John Popelish <>

Probably about 24 A for this sort of supply. The diodes conduct for
about 60 degrees of each half-cycle, and discharge for the remaining 120
degrees. It's worth checking the current waveform and measuring the
conduction angle.

 

OK, so ripple current is really just RMS current, then? I can estimate
this pretty easily using a spreadsheet. After I build it I'll check to see
if I was on target with my estimate.

I'll use 12 Amps when the bridge diodes are reverse biased, and I = c
dv/dt when the bridge diodes are forward-biased. This should safely
over-estimate the current.

Thanks!

Mac

 

Thanks, Ban!

--Mac