From: BobEngelhardt@comcast.net   
      
   I'm going to make a 12v DC power supply with a linear reg.  I will   
   half-wave rectify 12v AC and have a smoothing cap. The load is only 10ma.   
      
   So: 12v RMS is 17v peak, minus the diode drop of 0.7 is 16.3v peak.   
   Using a 47uF cap, the ripple will be 3.5v p-p.  So the min voltage into   
   the regulator will be 16.3-3.5 = 12.8.  Ripple calculated from Vpp =   
   i/fC (.010/(60*47e-6).   
      
   Am I missing anything?   
      
   Thanks, Bob   
      
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