FET Battery backup switching

Can anyone tell me if this circuit will work. And want are the pitfalls.
It is for an offline battery backup circuit.

Dear tmd
Hi
I can see some problems at your circuit ,
If the power is on your circuit D19 is on , and if the power isn't , D19 is off . i think you should change it's polarity .
Regards
Goldsmith

---------- Post added at 15:55 ---------- Previous post was at 15:53 ----------

BTW : perhaps , i couldn't understand your exact , meaning , if you tell me a bit more explanation , thus i can more help you .

goldsmith said:

Dear tmd
Hi
I can see some problems at your circuit ,
If the power is on your circuit D19 is on , and if the power isn't , D19 is off . i think you should change it's polarity .
Regards
Goldsmith

---------- Post added at 15:55 ---------- Previous post was at 15:53 ----------

BTW : perhaps , i couldn't understand your exact , meaning , if you tell me a bit more explanation , thus i can more help you .

Click to expand...

Yes, Sorry. When the 12V is on, the PSU runs 15V out which runs via D19 to charge the battery. A small dropping resistor will be nedded to make the output voltage between 13.8V to 14.4V.
When the 12V drops off, the charge circuit also drops and the FET turns on, supplying the power from the battery to the 12V circuit. A steering diode will be needed to stop this feeding back to the 15V supply.
I hope this helps.

goldsmith said:

Dear tmd
Hi
I can see some problems at your circuit ,
If the power is on your circuit D19 is on , and if the power isn't , D19 is off . i think you should change it's polarity .
Regards
Goldsmith

---------- Post added at 15:55 ---------- Previous post was at 15:53 ----------

BTW : perhaps , i couldn't understand your exact , meaning , if you tell me a bit more explanation , thus i can more help you .

Click to expand...

When 12V in is on. The PSU runs 15V out through D19 and charges the battery. As this is pushes the gate higher than source/drain of the FET, it is off.
When 12V in is off. the PSU drops off, the pull down lowers the gate and turns on the FET. This then supplies the 12V out from the battery.
This is the intended application. Will it work??:???:

Hi again
Your circuit is still improper . why you are confusing yourself ? why source voltage is 12 volts ? battery can turn on the diode ! it means dissipations . why not two simple diodes ? why that mosfet ?

The reason for this circuit is simple. But the solution is not.
This is to supply power to a 12V circuit. Additional voltage is not an option due to excessive heat in the supplied circuit if the 12V rail goes above 13.2V.
The battery needs 14.4V to properly charge.
The battery must supply a high current (500mA to 4A) when there is no mains available for up to 2 hours (using an 8Ahr battery).
Off charge, the battery terminal voltage is 13.2V (low enough for the rest of the circuit). On charge, the terminal voltage is too excessive (14.4V).
The charging current does not need to be more than 500mA. There must not be a dip in the 12V supply below 11V when changing from mains 12V to the battery voltage.
The cost must be very low. A relay switching will produce an excessive dip in 12V. I have proven this circuit using a bipolar transistor but it's on resistance for battery supply is too excessive and burns out unless an large physical package is used (TO3) or generates too much heat.
Hence this circuit using a FET with a low on resistance.