Lawrence: your description is one way of looking at it, but I prefer to avoid the term 'voltage doubler' here since to me, that term is better 'reserved' for what I call a 'conventional' voltage doubler (for example, that video clip that you recently sent in).
To me, I think of Joe's rectification scheme as a full-wave bridge plus a half-wave rectifier: in other words a 'one and a half-wave rectifier', where the word 'one' implies 'full-wave': i.e. one full cycle of the input waveform. There may well be a better choice of words to explain that scheme of course: semantics was never one of my strong points!
Going back a few posts, Joe did mention that the unregulated voltage did fall away fairly steeply as the load was increased (that was why I asked him to send in the full cct. diagram of this PSU) - and now we can see that it is the presence of that half-wave rectifier that explains that behavior. Nevertheless, it looks like a useful dodge to employ when you have a transformer whose secondary voltage is just not quite large enough for the final output voltage that is required - provided, of course, that the reduction in available current at higher O/P voltages is acceptable, compared to the higher current that would be available at a lower voltage if the bridge alone was used.
Al.
To me, I think of Joe's rectification scheme as a full-wave bridge plus a half-wave rectifier: in other words a 'one and a half-wave rectifier', where the word 'one' implies 'full-wave': i.e. one full cycle of the input waveform. There may well be a better choice of words to explain that scheme of course: semantics was never one of my strong points!
Going back a few posts, Joe did mention that the unregulated voltage did fall away fairly steeply as the load was increased (that was why I asked him to send in the full cct. diagram of this PSU) - and now we can see that it is the presence of that half-wave rectifier that explains that behavior. Nevertheless, it looks like a useful dodge to employ when you have a transformer whose secondary voltage is just not quite large enough for the final output voltage that is required - provided, of course, that the reduction in available current at higher O/P voltages is acceptable, compared to the higher current that would be available at a lower voltage if the bridge alone was used.
Al.
