Power Supply Filter Capacitor Size

[harrybissell at prodigy.net]

The (linear) regulator has to work as hard as the voltage drop across it. Less ripple will mean a capacitor voltage that is on the average higher, so the thermal load will be more. So the "greater" ripple will have less thermal efffect on the chip (this might not stay true at incredibly high junction temperatures, much higher than any DIY'er wants to see !!!)

OTOH The regulator has a "ripple rejection" spec that is usually related to db and frequency. Ripple is usually a sawtooh signal so it has higher frequency components. Even if it was a sine wave (at 60 or 120 hz), more ripple comming in spells more ripple in the output.

So I conclude (IMHO) the less ripple the better in almost every case.

You could trade ripple for cap size if you really didn't care about the ripple. 

More Cap is better (but watch out for the inrush current in the rectifier diodes...

:^) Harry


 ---- On Dec  7 Chris MacDonald <macdonald at evenfall.com> wrote: 
> 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.