After a few quick calculations regarding the power supply filter capacitance, I think you should spec a different transformer. With the transformer you have specified: Vfl = 16.968Vp, I used this for my calculations even though you are not using full rated current draw for the transformer. Vout = 12.0V. According to National Semiconductor the voltage accros the regulator could be 2.5V worse case. I used Vf = 0.7V for the diode. Subtract all the drops from the transformer voltage, this leaves you with the maxinum ripple voltage accros the capacitor. dV = 1.768V. Assuming a full load current the filter capacitor for the 12 volt supply should be >47000 mfd. The 5V supply will require an even larger capacitor.
I may be under estimating the transformer ratings, and over estimating the dropout voltage, but is that what a good designer should do? On October 29, 2004 06:16 pm, Robert Riemer wrote: > Some technical tips: > > 1) Use fewer 500 mfd capacitors. There may be too much capacitance. If not, > use fewer larger value capacitors. Probabaly the same price and size > approx.. The value of capacitance depends on the transformer voltage, amount > of allowable output ripple, and current drawn from the supply. If the > capacitance is too large, the regulator could dissapate too much power. If > it is too small the output ripple may be large, our the regulator can drop > out of regulation. > > 2) The transformer is potentially unevenly loaded. There is a split 12v > supply but not 5v. > > 3) The terminal labled +Ureg (unregulated) should be as variable, not > unregulated. > >
