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Thought 21 - Battery Pack Wall Wart

Recently got a couple pieces of non-mobile electrics, and they take batteries? why? Couldn't they put a wall wart to go with it. (For those who do not know: wall wort = ugly black two prong ac adapter, it looks like a wart on your wall). I mean really mass produces wall warts are pretty cheap and would only effect the price by a dollar or two. But consuming batteries for something stationary is just wasteful.

Ok rant over. Time for a solution.

Here's what we came up with today for a plausible and quick solution.
First find a power supply close to the battery power needed 2 batteries = ~3 vdc, 3-4 batteries = ~5 vdc.
Then cut the end off of it keeping track of the plus and minus.
Next build a fake battery:
For a quick and dirty solution a PVC tube the same radius as the battery.
Cut into appropriate number of 'batteries'.
Epoxy nails in place for the plus and minus terminals.
Solder the ends of the wall wart to the nails.
Finally fill in the whole thing with epoxy.

Really you'd only need two 'batteries' for any solution just put time so that + and minus match up at per insertion diagram.

Possible problem #1: There are 2^(n-1) possible solution for placement of the 'batteries'. (n = number of batteries required) Each terminal has a + and a -, fortunately there's no circuit until the proper + and - are found.

Possible problem #2: Center tapped power. Basically the electric device has multiple voltage ranges and is using the whole set for one voltage say 5 v and only two to product 3.3 v logic. In this case the simple fake batteries will not work. The proper solution would be to create a wall wart with multiple isolated 1.5v ends. Unfortunately the amount of circuitry will not really fit into a AA.
This would look something like: transform from 120 to say 4 2-3 VAC rectify, filter, and linear regulate back down to 1.5 vdc. Then provide 'batteries' that can tether plug into the new wall wart. Though this might be a bit on the large size.

Anyway need to make a simple one, then we'll work a circuit for solution #2.

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