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9V Battery and DC Supply Circuit Wanted

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mpuckett

New Member
Hi,
I have just built a solid state amplifier from runoffgroove.com called Ruby and it works and sounds great! I am most definitely a newbie though and while I can follow directions and solder ok I really have no idea how to do any kind of circuit design.

The design calls for a 9V battery. I would like to build something that would allow both a battery and a 9VDC power supply and have the battery automatically disconnected if the 9VDC supply is plugged in. So it would run off the battery normally until the 9VDC power supply is plugged in which would switch the battery off and run off the power supply instead.

Does anyone have a simple circuit diagram that can accomplish this or could point me to a site that might? I have tried Googling this already which is what led me here and so far have not found what I am looking for.
 

MikeMl

Well-Known Member
Most Helpful Member
A lot of the coaxial power jacks that go with the plugs you find on AC plug-in power supplies (I call 'em wall-arts) have a normally closed switch contact which is opened by the act of plugging in. You could wire the battery though the switch contact so that the battery powers the circuit when nothing is plugged in.

Otherwise, two silicon diodes with their cathodes wired to the circuit (load), and each anode connected to battery and wall-wart, respectively would work, too.
 
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mpuckett

New Member
Thanks Mike,
I didn't know they made switched coaxial connectors like this, that could work very well.

Regarding the diode solution, would the battery still be supplying any output if the wall wart is plugged in or does the diode completely block it?
 

MikeMl

Well-Known Member
Most Helpful Member
As long as the Wall-Wart output voltage is slightly higher than the battery voltage, then no current comes out of the battery. The battery diode prevents the Wall-Wart from "charging" (backfeeding) the battery. The Wall-Wart diode prevents the battery from discharging backwards into the Wall-Wart. If the 0.6V forward drop across the diode(s) is a problem, use Schottky diodes; their forward drop is only ~0.3V
 

audioguru

Well-Known Member
Most Helpful Member
The Ruby amplifier uses a little LM386 IC. If the 9V battery is brand new then the output power at clipping is a whopping 0.45W.
If the volume is turned up too high then the output is 0.9W square-waves.
Peanuts power.
 

mpuckett

New Member
MikeMI,
Thanks. I knew the purpose of the diodes was to prevent the backfeeding which was why I asked about them stopping the battery drain in the forward direction. I could use a 12V wall wart and be sure that it was always higher than the battery though.

Regarding the 0.6V drop across the diode being a problem, does that mean that the battery would always be supplying 0.6V across it? That is probably a problem then that will lead me to look at the other option then.

Audioguru,
Peanuts power is exactly what I am looking for in this application.
 

Hero999

Banned
I'd use a 6V transformer with a rectifier to power the amp and connect the 9V battery in parallel with a diode in series.

The 0.6V drop of the diode shouldn't be a problem.
 
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