Auto shut down circuit

I have to design a AUTO SHUTDOWN circuit , It will be used in my inverter , when battery voltage reaches 10.2v Low battery alarm starts beeping and the load is instantly cutoff using relay , the low battery sensing circuit still draws some power from battery , so if kept for long time the battery will be fully discharged , so i want to make a circuit which shutsdown the complete system after 5 mins from low battery event. please help me how to start ?

You can make use of comparator IC like LM339 for doing so. You can use 555 timer for 5 mins. If you need more help let us know

But lm393 will also consume power after low battery event .i think you are telling me about low battery cutoff which i've already made.i want to incorporate shutdown feature now.

If you build this circuit,you realise that you must have a manually operated switch to restart the inverter after a low battery shut down?
Two obvious methods:- one is to feed the battery via normally open relay contacts to the circuit. connect the relay coil downstream of the contacts, put a driving transistor between the coil and earth. Or put a NPN transistor in series with the 12V feed collector to battery, emitter to circuit. Connect the base via a current limiting resistor (1K?) to the collector of the driving transistor, emitter to earth. Now in either case if there is no current going into the driving transistor, then the battery supply is cut off. You need to put a push button switch across the relay contacts or collector/base of the series transistor, to get a supply voltage to get some current into the base of the driving transistor.
Frank

i made the circuit.It's working now . i have used a p mos and a relay to achieve the same.