High frequency switching in the diodes was my thinking also. When the=20 diodes in the rectifier turns on (capacitor voltage is lower than two=20 diode drops, the circuit starts to draw current from the transformer=20 which results in slightly higher voltage drop for the diodes and=20 lower output voltage from the transformer. This will produce short=20 time, high frequency, relatively high current on/off switching in the=20 diodes until the transformer voltage has rised high enough to keep=20 the diodes on. This can produce lots of EMI. Another way to reduce=20 this is to put small capacitors in parallel with each rectifier=20 diodes. Since this switching is much higher than the 100Hz ripple (but at=20 100Hz intervals) it should be much easier to filter out than the=20 100Hz ripple in itself. Note that only using a capacitor as the filter element for the power=20 supply will reduce the maximum output DC current. The inductor in series with the supply filter capacitor will reduce=20 ripple current in the capacitor, lower the peak output voltage and=20 increase the maximum output DC current after the filter. See /Ruben > stages. I > would also look at possibly using an inductor plus capacitors on the > input > and increasing the filtering capacitance. If you increase the > capacitance > without adding inductance, then the current drawn from the input > supply > will become ultra-short, very high current pulses only right around > the > peaks of the AC waveform, which is bad for thermal reasons and > possibly for > hum production, too, since it may introduce high-frequency > components to > the ripple which are not well rejected by the regulator. The > inductors not > only perform additional filtering but they also smooth out the diode > bridge > conduction. >=20 --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .