On Sun, 28 Sep 2008 11:55:58 -0400, "Alec S." put
finger to keyboard and composed:

The bad news is that you guys were both right and wrong about it being a 7805,
not a 7905. You’re right because that is indeed what is supposed to be there.
You’re wrong because that’s not what’s currently there.

... somehow I put in a 7905 instead of the 7805 ...

I guess it makes sense that it’s not working now. I don’t suppose that I can
expect it to work when I put the correct one back can I? Would having a 7905 in
there (while on) have killed the board?

The pinout of a 7805 is I G O.
The pinout of a 7905 is G I O.

IIRC, the unregulated input voltage was +12VDC. This would mean that
the 7805's IGO pins would have had voltages of +12V/0V/+5V.

The 7905 would then have seen -12V on its input, which is correct, and
this would then mean that its output pin would have been sitting at 5V
below its ground pin, ie +7V. AISI, this may have done irreparable
damage to your +5V logic. The 7905 is probably quite happy, though.
:-(

BTW, when looking for a short, you don't always need to remove the
entire part, you only need to isolate it by desoldering whichever pin
is connected to the short.

- Franc Zabkar
--
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