i think it will be ok IF your supply is good and puts out seomthign at least ~8.3V all the time
cons that might not matter that much: when a DC supply is not present teh battery output is lowered by the diode. Diodes still leak current when reversed, so say dc supply is higher than the battery output, a small amount of current will flow into the battery, and slowly charge it
. The amount is quite small so i dont think it would ever explode or somethign like that.
cons that will really matter: if you plug ina supply and it sags the diode will probably burn out assuming the battery can source the current. Consider a battery that has 9V adn a supply that sags to 6 or 8V for whatever reason. You have 9v on the anode of the diode and something much less on the cathode. For forward biased diode the current is EXPONENTIALLY proportional to the voltage accross the junction. i believe the formula is something like
I_diode ~ I_saturation * e ^(V/Thermal Voltage) where I_sat and Thermal Voltages are some constants depending on fabrication technique and ambiant operating condition etc. and V is the voltage accros the diode. The thermal voltage is usually on the order of 25 mV. I_sat is usually in the order of pico or femto amps.
so if you just plug in some numbers, say I_sat is 10^-15 and Thermal V = 25mV. For the case of 8 volts you end up with a 1v across the diode and 235 amps flowing though the diode. Of course the battery wont source that, but it serves to show that you will likely burn the diode.
To me this is the only real problem with the setup, and if your supply is solid, it might not happen.
