converting PSU design from bipolar to mono-polar

warning: newb question ahead:


I'm breadboarding this for 12v relay power.
Removing the (-) side components just to get the (+) positive voltage for the relays. This part I can do, no problem...

...but then it occurred to me I'd have more available current with a paralleled secondary. My transformer is tiny at only 165mA so more current would be a big help - but there's no way to wire the bridge rectifier with a paralled secondary. I guess that's a necessary element to this design - to get 9v boosted for 12v regulation.

Should I just built this as is - leaving off the negative side components? And hope I can drive enough relays? (relays want 30mA) Or get a higher voltage secondary tx? Any thoughts or guidance will be appreciated. Thanks, Kato

Leave off the negative side components. Disconnect the transformer secondary centertap. Conect the - (minus) terminal of the bridge rectifier to the 0V line. This'll give you about 23 volts to the input of the 12V regulator.
(18-1.2)*1.414 = 23.75

Hey thanks for the response nyDave.

What I love about DIY, is that I can just hook it up and find out.
In the intervening time since I posted, I got out the alligator clips...

18vct transformer:

• 10.5vAC on each secondary, no load.

• secondary wired in parallel --> connected to bridge rectifier --> 9.something vDC out of rectifier. (answered my question.)

• bridge rectifier wired per schematic: 18.6vDC after rectifier.

Is a load required (or cap reservoir) for the magical voltage increase provided by rectification?

(I'm going to connect it the way you suggested...)

Hey Dave, just wanted to say, thanks for your post.

It worked. Success. 115vAC in, 12vDC out. (actually 11.96v.)