From: peterxpercival@hotmail.com   
      
   tabbypurr@gmail.com wrote:   
   > On Saturday, 20 July 2019 18:35:38 UTC+1, Peter Percival  wrote:   
   >> Phil Allison wrote:   
   >>> Ralph Mowery wrote:   
   >>>   
   >>>   
   >>>>   
   >>>> All of that is well and good.   
   >>>>   
   >>>> However, is that going to be peak, average, rms or just what that the   
   >>>> origional question wanted.  Is there going to be a capacitor to smooth   
   >>>> out the voltage, or is the full wave rectified DC going to be used ?   
   >>>>   
   >>>> Lots of unasked questions in the origional question.   
   >>>>   
   >>>   
   >>>   
   >>> ** The original Q is a pile of utterly ambiguous drivel.   
   >>>   
   >>> If Q posters would simply reveal the PURPOSE of their question, ambiguity   
   would disappear and useful answers become possible.   
   >>>   
   >>> But no, they want to play at being smart and be in control the answers by   
   NOT revealing any such damn thing.   
   >>>   
   >>>   
   >>>   
   >>> ....  Phil   
   >>   
   >> Oh dear.   
   >>   
   >> I am looking at Fig 3.6 on p 42 of Marston's /110 operational amplifier   
   >> projects for the home constructor/.  The circuit depicted is that of a   
   >> power supply delivering 3-30V at 0-1A.  It is to be supplied with "+40   
   >> to 45V (unregulated)".  The text has nothing to say about where that   
   >> comes from.  I have decided to use a bridge rectifier attached to the   
   >> secondary of a transformer with 240(ish)V primary; and I wish to know   
   >> what secondary I need.   
   >>   
   >> The "2amps" in my OP was a guess of mine that if the PS delivers 1amp,   
   >> then 2amps in would be more than enough.   
   >   
   > A bit of an old post, but one that never got a fair answer.   
   > To get 40v out, you need 40x 0.707 ac input = 28.3v rms plus the diode drop   
   of the bridge which under load will be 2x 1-2v. If we guess 1.5v diode drops   
   under load then you'd need 31.3v in. That much has mostly been covered   
   already, though a lot of EEs    
   mistakenly think diodes drop 0.65v all the time.   
   >   
   > However there are 2 other factors not so far addressed.   
   > First, the reservoir cap will charge to its peak v but then decay until   
   recharged. You need to supply a bit more V to counter the decay for 100th of a   
   second.   
   > Second, there is mains voltage variation to consider. If you want the psu to   
   be reliable & meet mains voltage specs it needs to get enough V to the   
   regulator when V_mains is at the lowest it is permitted to go, which is a fair   
   bit under nominal voltage.   
   >   
   > Finally you should only need a 1A supply, parasitic drain is trivial in any   
   normal 1A V reg.   
   >   
   > And really finally, check the reg can dissipate the power at full load when   
   mains V is max, I_out is max & V_out is min. And ensure you have sound   
   short/overload protection.   
   >   
   >   
   > NT   
   >   
   Thank you.  Much obliged!   
      
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