Yes, I should have qualified what I wrote, my only excuse being that it was late at night (or early morning!) when I wrote it!
I should have added, at the end of the sentence you've just quoted, something along the lines of what I wrote subsequently, e.g. "... and even if it had no reservoir capacitor at all, the peak 'DC' voltage feeding the LED (which appears to be an important determinant of perceived brightness) would still be approaching the peak of the AC."
Agreed. I'm not sure why they bothered in that circuit, particularly given that, if I've done my sums right, the voltage across it, in Rocky's posted circuit will only be about 70V** (hence ripple likely to go right down to zero).
** talking of which, the 63V working voltage for their 4.7μF capacitor in that circuit seems a bit 'marginal'/iffy. By my reckoning, if one assumes a forward voltage of about 3.5V for each of the (white), LEDs, that's 52.5V for 15 of them. It seems that the current in that circuit will be limited to a bit under 30mA (essentially determined by the 220nF input capacitor), so there will be about 17V across the 560Ω resistor. That seems to be a total of about 69.5V.
Kind Regards, John
