Correct. The simple method I mentioned will not charge
batteries - it will merely run off of the permanent supply
and kick-in the battery back-up if ever the permanent
supply is removed/lost. Without complete circuit schematics,
I'm making an assumption that you have some, probably
unregulated, permanent dc power available from rectified 120V
AC. With this voltage being greater then the regulated 9V battery
back-up supply by the diode built-in voltage (approx. 0.7V), you
may connect both of the supplies through the diode. You would
connect the back-up supply to the anode and the permanent ~10V
supply and rest of the circuit to the cathode. As long as permanent
supply is ~9.7V, the diode is reversed-biased (non-conducting),
hence all power to the circuit is supplied by permanent supply. If
permanent supply voltage drops, the diode "turns on" and the circuit
will now be supplied by the back-up supply....
You can also get fancy and insert a pushbotton switch that'll break
the permanent supply and bypass the detecting mechanism- testing the
battery back-up.

--
sincerely,

Russell Powell

Artisan Components
SR. FAE - U.S. Central

469-438-6589





"JeffM" wrote in message
om...
The battery is a non-rechargable type and is for backup only.

Put in a diode so that when the AC is cut,
the battery passes juice to the detector
but the AC supply never tries to charge the battery.