yea, just use an 18v wall wart. what would have been 0 would be -9, and what would have been +18 would be 9. Like Laurie said, it's a matter of relative potential.<br><br><div><span class="gmail_quote">On 2/27/06, <b class="gmail_sendername">
amokan</b> <<a href="mailto:amokan@gmail.com">amokan@gmail.com</a>> wrote:</span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">So, are there wallwarts that will do +/- ? Am I better off using an AC wallwart and then building one of the many power supply circuits available online?
<br><br>I really just want something cheap and that only involves 1 plug. I also want to avoid anything much more than a regulator in my circuit to keep it small.
<br><br>I've never really put much thought into this until now. All of my circuits are done on a breadboard with a bench supply and by the time I get them in a frac, they just plug into my main supply that "magically" works :)
<br><br>Thanks again everyone. Sorry to drag this thread on.<br><br><br><br><div><span class="gmail_quote">On 2/27/06, <b class="gmail_sendername">Laurie Biddulph</b> <<a href="mailto:elby_designs@ozemail.com.au" target="_blank" onclick="return top.js.OpenExtLink(window,event,this)">
elby_designs@ozemail.com.au
</a>> wrote:</span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
<div>If the outputs are truly isolated from the mains power (specifically that
is not connected to the Earth pin - which they should be) then you can connect
the positive of one wallwart to the 0V of the other. This (double) 0V connection
is connected to the 0V connection of your module and you then have +ve (from the
free positive lead) and -V (from the free 0V lead).</div>
<div> </div>
<div>What you won't get is tracking between the outputs so one might go up while
the other droops etc. If you are using these to power regulators in your circuit
then this will be less of a concern. You will also need to power outlets and if
your wallwarts are large then it may not be possible to get 2 on to them
(double) outlet.</div>
<div> </div>
<div>The size of a wallwart is a possible indication of wether it is a
transformer or switching design but not always. A low power transformer
unit will be quite small. Usually there is next to no weight in the switching
designs. Above should hol true for these as well as the outputs should be
totally floating.</div>
<div><br>Best Regards</div>
<div> </div>
<div>(Mr) Laurie Biddulph<br>Mobile: 0404 846 943</div>
<div> </div>
<div>Elby Designs<br>ABN: 70 022 727 605<br><a href="http://www.elby-designs.com" target="_blank" onclick="return top.js.OpenExtLink(window,event,this)">http://www.elby-designs.com</a></div>
<div> </div>
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<blockquote style="border-left: 2px solid rgb(0, 0, 0); padding-right: 0px; padding-left: 5px; margin-left: 5px; margin-right: 0px;">
<div style="font-family: arial; font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adjust: none; font-stretch: normal;">----- Original Message ----- </div>
<div style="background: rgb(228, 228, 228) none repeat scroll 0% 50%; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-family: arial; font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adjust: none; font-stretch: normal;">
<b>From:</b>
<a title="amokan@gmail.com" href="mailto:amokan@gmail.com" target="_blank" onclick="return top.js.OpenExtLink(window,event,this)">amokan</a> </div>
<div style="font-family: arial; font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"><b>To:</b> <a title="synth-diy@dropmix.xs4all.nl" href="mailto:synth-diy@dropmix.xs4all.nl" target="_blank" onclick="return top.js.OpenExtLink(window,event,this)">
synth-diy</a> </div>
<div style="font-family: arial; font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"><b>Sent:</b> Monday, February 27, 2006 5:20
PM</div>
<div style="font-family: arial; font-style: normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adjust: none; font-stretch: normal;"><b>Subject:</b> [sdiy] wall wart confusion
</div>
<div><br></div>I didn't pay attention back in high school electronics when it
came to power supplies so please forgive me if this is a newbie
question...<br><br>I've built bipolar supply kits for modular use but never
really understood _how_ they work. Negative supplies are a mystery to me. I
know there is a difference between ground and the negative rail, but I don't
know what the difference is. A 9v battery is in-fact bipolar, yes?
<br><br>Long story short, I'm working on modules that will run in a stomp box
format and I'm trying to figure out the cheapest & easiest way to power
them. I know I could use the dual 9v battery solution to power them (assuming
the circuits could run at 9v) but I'd really like to just use a standard 9v or
12v dc wallwart that I have laying around the house. How do I get a negative
voltage off of a standard 12v wall wart? Is it already there normally, or just
a positive and a ground? <br><br>Sorry for rambling. If there is a web
resource explaining this, please let me
know.<br><br>Thanks!<br></blockquote>
</span></div></span></div></blockquote></div><br>
</blockquote></div><br><br clear="all"><br>-- <br><a href="http://www.nomesh.net">http://www.nomesh.net</a><br><a href="http://www.radioresponse.org">http://www.radioresponse.org</a><br><a href="http://www.freebsd.org">http://www.freebsd.org
</a>