From: BobEngelhardt@comcast.net   
      
   On 8/29/2021 8:48 PM, Phil Allison wrote:   
   >   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   
   >   
   >   
     Thank you - that is helpful in 2 ways: pointing out what I missed &   
   was wrong about, and even more helpful is not having to correct my analysis.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)