IIRC, voltage is a matter of potential, and so long as your wall wart is of the big heavy variety, ie, uses a transformer instead of being a switched power supply, you could just use an 18v or 2x9v's the same way as you would 2 9v batteries. But test before you hook it to anything important :D
<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;">
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>
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