Power Supply Filter Capacitor Size

[Chris MacDonald]

I recently acquired something interesting and useful from "Art of
Electronics", not an uncommon occurrence of course.  The topic starts on
page 329.  Please correct me if I have gone astray anywhere, I'm
certainly not an expert here.

One can compute the size of filter capacitor needed for a power supply
with the aid of the formula:

C ~= ( I * T ) / R

where:

C is capacitance of filter cap in farads
I is the maximum current in amps the power supply needs to provide
T is the time period in seconds between peaks of incoming AC voltage
(typically .008 for full-wave rectified 60hz line AC (.017 for half-wave
rectified)
R is the maximum acceptable ripple voltage allowable (in volts)

Ripple voltage is the amount of "voltage sag" allowable between peaks of
the incoming rectified DC.  Typically a voltage regulator follows the
filter capacitor and will have a minimum input voltage for proper
regulation.  The maximum ripple would be the peak rectified voltage
minus this minimum input voltage.

Here's an example.  A circuit will draw 120mA at 15 volts, so with some
padding, a regulated power supply is desired which will supply 200mA at
15 volts.  A 15 volt AC transformer is used through a half-wave
rectifier, this provides a peak voltage of 25 volts from a 120 volt
source.  Allowing for a 12% sag in line voltage (too optimistic for
some?), peak voltage might drop to 22 volts.  The voltage regulator used
requires that its input voltage remain 2 volts higher than the regulated
output, so 17V.  Thus the maximum allowable ripple is 22 minus 17 or 5
volts.  

C = ( I * T ) / R
C = ( 0.2 * 0.017 ) / 5
C = 0.00068 farads or 680uF

Allowing for a 20% tolerance of the capacitor 850uF would be a minimum
choice, 1000uF might be better.

Two questions I have, does the regulator have to work harder if it's
input voltage is highly modulated?  Is it important/desirable to have
low input ripple voltages to the regulator for any reason?

One other neat thing I was able to do with this was to approximately
determine the current draw of a circuit by measuring the filter
capacitor ripple voltage of its power supply using I = ( R * C ) / T.

In any case it's probably much easier, better, and safer to just buy a
high quality power supply and save yourself a lot of aggravation both in
building one and in debugging supply-induced problems.  Further, I am
not an expert, just someone crazy and/or dumb enough to think that
building a power supply might be a neat idea.  Building power supplies
or messing with line voltage can be dangerous, even lethal.  I hereby
disclaim all liability for direct, indirect, incidental, or
consequential damages or injury that result from any use of the above
information.  

On that cheery note, I hope someone finds this interesting and I hope I
haven't made any grievous errors!

-Chris M.