From: pallison49@gmail.com   
      
    bobenge...@gmail.com wrote:   
   =======================   
   >   
   > 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).   
   >   
      
   ** Yep two:   
      
   1.   The drop out voltage of a standard 12v reg IC is not 12.8 -  more like   
   13.8.   
   2.   The AC supply varies and may be be 10% low on occasion.   
      
      
   So, you need a 15VAC tranny.   
      
   OR you can make a "voltage doubler" supply with two diodes and two electros.   
      
   That will give about 33V DC  -   minus the same deductions.   
      
      
      
      
   .....  Phil   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)