> On 10/03/2008, piclist wrote: >> Michael Cunningham wrote: > > .... >> > "When you convert a AC to DC by a full wave rectifier >> > bridge you have to multiply the output voltage by 1.41 >> > to obtain the VA" A bit awkwardly put. If I have a 10VAC (RMS) input and full-wave rectify that with ideal diodes assembled into an ideal bridge rectifier, and draw 1A from the output, the input current will be 1A. 10W in, 10W out. However, if I put a BFC (big fat capacitor) on the output, the output voltage will rise to 14.14V (minus a bit because of ripple), and if I draw 1A from it, I'll be getting 14.1W out. The input power (obviously) is also 14.1W. Typically you want to use a transformer rated 1.6 to 1.8 x the DC output current of the power supply because of the I^2R heating of the transformer coils. In this case, if we draw 1A (14.1W), we'd want to use a transformer rated at 1.6-1.8A at 10V, or 16-18VA Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" s...@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist