How do I get an 18v power supply?.. Two 9v's in series?? [picture]

Started by elenore19, February 19, 2011, 02:59:20 PM

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elenore19

I'm building a pedal that I realize calls for 18v input voltage.
Can I just have two 9v jacks hooked up in series? The pedal would require two power plugs...but yeah.
Would this work? I'd prefer not to have to order new parts.

Thanks for the help!
-Elliot

EDIT:

Something like this?

waltk

I guess any two 9V sources connected in series would give you 18V, but you would have to be sure they didn't have a common ground (Otherwise, it would just short out your power suppies).  Your drawing wouldn't work as it stands.  I presume you intended that the red wire on the left jack, and the green wire on the right jack would not be there.

It's pretty common to hook up two batteries in series this way to get 18V.  The other way is to have a single 9V supply, and make a voltage doubler using a MAX1044.  Figure 9 in this PDF: http://pdfserv.maxim-ic.com/en/ds/ICL7660-MAX1044.pdf is one way to do it.

ppatchmods

It would be better to just add a max1044 voltage doubler. You can find it in the layout gallery.
...beat me to it!
When your life is over, will any of this STUFF really matter?

petemoore

  If ''board'' = V+...
  That shows two somthings in series.
  two batteries in series equals 2x voltage.
  If there's a center tap...
  connect a black from battery clip 1, and a red from battery clip 2, twist them together = this is the Vbias = 1/2v of split supply = centertap = Vb.
  After that is connected, there is 1 black {goes to V-}, and one red {goes to V+.
  Since it is a partial schematic and it shows the center tap of the two somethings having a third node connection...
  Otherwise for single ended 18vdc supply, twist 1 red from B-clip, 1 black from the other B-clip, twist/solder/insulate, what's left is the + red and black -, with fresh batteries, between them is a floating 18vdc potential.
  As far as 1 battery Vs. 2 batteries, the simple list of priority comparisons isn't simple, but leaving the batteries in when not in use pretty well 2x's the battery losses, how much power the doubler chip consumes seems pretty tame in comparison.
Convention creates following, following creates convention.