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 [email protected] http://www.seul.org/cgi-bin/mailman/listinfo/geda-user
