John Fields wrote in
:

On Thu, 8 Oct 2009 09:18:59 -0700, DaveC wrote:

peak = RMS * sqrt(2) = 110 * 1.414 ~ 155V

JF

This is for half-wave rect.?

No, it's for AC and for a full-wave rectified sine wave.

What the deal is is that if 120VDC is placed across, say, a 120 ohm
resistor then the resistor will dissipate 120 watts and generate heat.

In order to get the same amount of heat generated by an AC voltage
connected across the resistor, (called the root-mean-square, or 'RMS'
voltage) it'll have to go above the steady DC value, on its peaks,
because that's the only way it can make up for the valleys which fall
below the steady DC voltage.

For AC and full-wave rectified AC, the number of peaks and valleys are
the same, and to get either of those voltages to heat up the resistor
the same amount requires that the peaks rise to the steady DC voltage
multiplied by the square root of two, and that voltage is called the
'peak' voltage.

For half-wave rectification it's an entirely different story because
half of the half sine waves are missing.



So what IS the story, exactly? Right now you've focussed on the results of a
load (the resistor), so you've said things that might be taken to discredit
the rest of us who specified that either peak has an absolute value of
voltage across a capacitor that when unloaded, must be considered for its
safety. While a half-wave rectified form is a more complex wave whose RMS
value needs a different calculation, in this case it's the separation between
peaks that matters.

Many PSU capacitor failures seems to be a result of people underspecifying
the working voltage while assuming they'll be 'safely' loaded to prevent the
peak from dictating terms more than the RMS. It's not a safe assumption.