That's a very interesting experiment, David and most informative! I'd never given ripple current much thought- up until now
I suppose during the negative half cycle, the entire HT current has to be supplied from the resevoir and smoothing caps. In this case, then, the ripple is roughly 4 times the HT current so during these peaks the power dissipated by the dropper will be 1A ripple + .25A HT current. 1.25 squared * 100 ohms (dropper resistance) =156 Watts!!
Looking at the scope trace, it's only doing this for about 1/5 of a cycle, the rest of the time the HT current is being supplied by the capacitor, so no power will be lost in the dropper. Whether we can divide the 156W peak by 5 as an approximation of the average power, I don't know. Even so, that would still give 31W. Mind you, that is what my 100 ohm dropper seems to be trying to dissipate!
Out of pure interest, if we were to use a full wave bridge, the ripple would be significantly reduced. Would this in turn reduce the power dissipated by the dropper?
Many thanks for doing the experiment
All the best
Nick