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Is there a rule(preferably a mathematical one) for determining the value of the input filter cap for DC power supplies.I get a lot of different opinions and more than one "rule of thumb" on this.
The equation I derived for this is C= (I*t)/((delta)V). C= Capacitance I=continuous Current, t = time the capacitor is the sole source for I ~8.3ms for a bridge rectified AC source, and (delta)V the Voltage ripple you can tolerate on the Cap (delta being the mathmatical symbol for change- couldn't find it on char map).
So much depends on how sensitive the circuit being powered is to noise.
Audio, video, rf and some digital circuits get twitchy about power supply ripple or noise. I have heard several "rule of thumb" estimates ranging from 1000mf per amp (minimal) to 4700 mf per amp (good on about anything).
One thing to try is build the circuit and try it, if you get p.s. noise then add more capacitance (up to about 4700 per amp).
If you have any worries about the power supply introducing noise to the circuit then you should build or buy an active regulator.
You can build one exactly to your needs with a simple op-amp circuit or buy one cheaply such as the LM78xx or LM79xx series regulators. You can drive pass transistors with these if you need more current.
The notion of using the 78xx series for your supply is nice but it will only help with low frequency noise. If you are only concerned about the AC ripple at line frequency then it may cut it, but you can be sure that high frequency noise will sail right through the regulator. use some ceramic capacitors in parallel with your filter caps and place them as close to the supply point as possible (right on pins). It will help.
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