Shut off circuit when under a certain voltage?
Use something like an 8.2 volt zener diode to conduct current to the base of a BJT when the supply voltage is about 9 volts. Depending on your load current the BJT may be able to drive the load adequately without further transistors. Two resistors are also required; one in series with the zener to limit base current and one to properly turn off the BJT when volts drop below about 8.5 volts: -







simulate this circuit &ndash; Schematic created using CircuitLab

The combined "blocking voltage" of an 8.2 volt zener and the base emitter junction is about 8.9 volts so round about this figure the base is going to start taking a little current and switching on the load. Given that the BJT will have a gain of around 100, 1mA into the base will yield about 100mA into the load.

If this circuit doesn't quite turn on enough at 9 volts then use a lower value zener possibly padded out with a forward connected diode in series. 

A 7.5 volt zener will start to feed about 1mA into the base at a battery voltage of about 8.3 volts. At 9 volts I reckon the base will be taking about 8mA and pretty much saturating the transistor for loads that consume less than a couple of hundred milli amps.