Simply reproducing the filter twice, one for each polarity of rectifier will not work. If you can float the load or the source then splitting the circuit into two and using a bridge rectifier in each will work OK. The attached shows what I mean. Cheers, Andy. www.signality.co.uk On 16 June 2011 23:05, myken <myken@xxxxxx> wrote: > Thanks DJ, > > I had the same thought that Vx was floating somewhere unwanted, that's why I > added the resistor (which didn't work). > Gazing at this problem for a couple of days make me miss the obvious, just > split the filter. Brilliant. > I'll give it a try. > > Robert. > > On 16/06/11 20:48, DJ Delorie wrote: >> >> When you put two capacitors in series, there's no way to know what the >> voltage between them will be. You have three with a common central >> connection Vx. V1 acts to charge the node, the loads act to discharge >> it, so an unequal load means unequal discharging and thus nonzero >> average node voltage. >> >> Since D1 and D2 may have different average currents through them, Vx >> will adjust until the average current through R2 is the same as the >> net current throuth the two diodes. >> >> Can you split the filter into two filters, one for each load? or at >> least move C1 to the Vx side of the filter, and split it into two >> capacitors? >> >> >> _______________________________________________ >> geda-user mailing list >> geda-user@xxxxxxxxxxxxxx >> http://www.seul.org/cgi-bin/mailman/listinfo/geda-user >> > > > > _______________________________________________ > geda-user mailing list > geda-user@xxxxxxxxxxxxxx > http://www.seul.org/cgi-bin/mailman/listinfo/geda-user >
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