At 10:24 AM 3/18/2004 -0800, you wrote:
>Spehro Pefhany wrote:
>
>>...Ripple p-p =  I/(C * 120)  (for full-wave and 60Hz)
>>Eg. 1A, you need 1V of ripple, C = I*1V/120 = *1/120)F = 8,300uF.
>
>I think you got this wrong, should be:
>     C =  I/(V*120)
>so smaller ripple requires bigger cap, but the answer is the same
>because V = 1V!

You're right.


>>Why do you need 1A? That's an awful lot. At 100mA you might be able to
>>use a cheap and small 1000uF part.
>
>C = I/(V*120) = 0.1/120 = 830 uF

Yes, of course, it's 1/10, but I qualified it ("may") because the
assumptions were just picked out of the air, and 1000uF is more available
than the next closest standard value (820uF). Another gotcha is the ripple-
current rating, which may force a larger capacitor than would otherwise be
required.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

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