On Fri, 31 May 2002, John Dammeyer wrote:

>If the formula is for calculating the droop between peaks then shouldn't
>the peak voltage of the transformer be used?
>
>  i.e. 9VAC is 9*1.414 = 12.76V peak.  If we are drawing 1A during the
>8.3ms then we get 8.3mCoulumbs/12.7cV = .652mF or 652 uF.  I would think
>that using 1000uF per 1Amp does work out nicely.  Then even with the
>diode drops it works out with the tolerance of the capacitor.

The ripple calculated like this will be higher than actually measured
because when you let the voltage droop so much the conduction angle is
significant and the formula does not really apply as is. The actual
conduction of the diodes will supply effective output power for up to 50%
of the time of a period, with the cap supplying the rest. You would need
to integrate the waveform in segments (the conduction angle segment is a
part of a sine and the discharge in between is a triangle or a part of an
exponential curve - for resistive loads - and even this is not correct
because the cap also sustains the load during the period of conduction
after the peak) to get the proper value.  There are tables of caps to be
used per ripple and per current desired. I do not have one but there are
older books that had them. Maybe an older edition of ARRL ?

Additionally, if the load is not constant current, but depends on the
voltage (resistor etc) then the same ripple calculated above will be even
lower than with a constant current load. The difference between a
resistive and a constant current load becomes large when the ripple is
large. The cap/ripple tables give values for resistive loads only.

Peter

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