Current Flowing Indicator.

[Mark Hennessy]

Now Mark is definitely going to be after me.

Ho ho!

I've spent the afternoon out at a kids party, which fortunately was held at a social club with an open bar. So, I'm not really about to think to deeply about all this

Having said that, if a power supply isn't happy, then they are normally pretty quick to let you know. So, if the extra cap isn't warming up, and is definitely upping the supply voltage, then all is probably well. At a glance, I think I can imagine how it might be working, but that can wait for the morning...

All the best,

Mark (staggers off to the beer fridge, knowing tomorrow is going to hurt!)

[Joe]

OK, I've had a play.

I rigged a transformer up with four 1N4007s in a bridge with the following results:-

AC - 12.6V
DC - 10.1V
DC with smoother - 15.2V
DC doubled as in my PSU - 21.1V
DC doubled as per other diagrams - 25V
DC doubled as per other diagrams plus smoother across +ve and -ve - 26.5V

All capacitors 220μF and all measurements made with DMM using a 180Ω resistor as a load.

- Joe

[pwdrive]

Hi Joe, those voltages seem about right for each scenario.
The capacitor connections shown in your last diagram don't seem right to me, anyone have any thoughts?
If you don't intend fitting a switch to switch from V to 2 x V then you could dispense with two of the rectifiers.

Lawence.

[Skywave]

The capacitor connections shown in your last diagram don't seem right to me, anyone have any thoughts?
Lawrence.

Yes: I did some elementary cct. analysis. The 'extra' capacitor gets charged on alternate half-cycles from the transformer. During the 'in-between' alternate half-cycles, that capacitor is discharged through the load. Hence, the output voltage is higher than if that cap. was not there. I have met this arrangement before, although infrequently.

Al.

[pwdrive]

Hi Al, I still don't see how the left hand capacitor shown in Joe's last handrawn diagram can obtain it's charge, the two capacitors appear to be connected as series opposing as opposed to series aiding.

Lawrence.

[pwdrive]

By series opposing I was trying to say that the electrolytic capacitors are connected back to back, my appologies.

Lawrence.

[Skywave]

Hi Lawrence,
The attached drawing gives my analysis . . . .
(It's easier to read that if it is opened in a New Tab)

   

Al. / Nov. 12, 2012 //

[pwdrive]

Thanks for the drawings and explanation Al, it has made things clearer for me now.
So if I have got this straight the difference between Joe's full wave doubler circuit and the more conventional full wave double circuit is that in Joe's the voltage doubling is effectively due to a series combination of the transformer secondary voltage and the charge on the capacitor where as in the more conventional circuit it is due to the series combination of the charges on the two capacitors.
In Joe's experiments the former method outputs a lower voltage where as the latter outputs a higher voltage.

Lawrence.

[Skywave]

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.

[pwdrive]

I hope I havn't confused Joe in all this.

Lawrence.